Student[LinearAlgebra] Examples

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Student[LinearAlgebra] Examples

Eigenvalues and Eigenvectors

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Example 1: Diagonalize a Matrix

Diagonalize $A = [\begin{array}{r} - 1 & - 12 \\ 1 & 6 \end{array}]$ by finding and applying an appropriate transition matrix $P$ .

Data entry

•	Control-drag the matrix. Context Panel: Assign to a Name≻ $A$

$[\begin{array}{r} - 1 & - 12 \\ 1 & 6 \end{array}]$ $\overset{assign to a name}{\to}$ $A$

Obtain the transition matrix $P$ , whose columns are the eigenvectors of $A$

•	Write the name $A$ . Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Eigenvalues, etc≻Eigenvectors

•	Context Panel: Select Element≻2

•	Context Panel: Assign to a Name≻ $P$

$A$ = $[\begin{array}{c} −1 & −12 \\ 1 & 6 \end{array}]$ $\overset{eigenvectors}{\to}$ $[\begin{array}{c} 3 \\ 2 \end{array}], [\begin{array}{c} −3 & −4 \\ 1 & 1 \end{array}]$ $\overset{select entry 2}{\to}$ $[\begin{array}{c} −3 & −4 \\ 1 & 1 \end{array}]$ $\overset{assign to a name}{\to}$ $P$

Diagonalize $A$ by applying $P$

•	Write the appropriate product of matrices. Use dot (period) for matrix multiplication. Context Panel: Evaluate and Display Inline

$P^{- 1} &period; A &period; P$ = $[\begin{array}{c} 3 & 0 \\ 0 & 2 \end{array}]$

Example 2: Singular Values of a Matrix

Obtain the singular values of $A = [\begin{array}{r} - 1 & - 12 \\ 1 & 6 \end{array}]$ , and verify the results from first principles

Data entry

•	Control-drag the matrix. Context Panel: Assign to a Name≻ $A$

$[\begin{array}{r} - 1 & - 12 \\ 1 & 6 \end{array}]$ $\overset{assign to a name}{\to}$ $A$

Obtain the singular values

•	Write the name $A$ . Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Eigenvalues, etc≻Singular Values

$A$ = $[\begin{array}{c} −1 & −12 \\ 1 & 6 \end{array}]$ $\overset{singular values}{\to}$ $[\begin{array}{c} \frac{\sqrt{194}}{2} + \frac{\sqrt{170}}{2} \\ \frac{\sqrt{194}}{2} - \frac{\sqrt{170}}{2} \end{array}]$

From first principles

•	Enter the product of the transpose of $A$ with $A$ . Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Eigenvalues, etc≻Eigenvalues

•	Context Panel: Assign to a Name≻V

$A^{%T} &period; A$ = $[\begin{array}{c} 2 & 18 \\ 18 & 180 \end{array}]$ $\overset{eigenvalues}{\to}$ $[\begin{array}{c} 91 + \sqrt{8245} \\ 91 - \sqrt{8245} \end{array}]$ $\overset{assign to a name}{\to}$ $V$

•	Expression palette: square-root operator Apply to each component of the vector V, whose components are the eigenvalues of $A^{T} A$

$\sqrt{V_{1}}$ = $\frac{\sqrt{194}}{2} + \frac{\sqrt{170}}{2}$

$\sqrt{V_{2}}$ = $\frac{\sqrt{194}}{2} - \frac{\sqrt{170}}{2}$

Example 3: Jordan Form

Obtain a transition matrix that puts $A = [\begin{array}{r} 5 & - 5 & - 4 \\ - 4 & 8 & 5 \\ 7 & - 11 & - 7 \end{array}]$ into Jordan form.

Maple can return the required transition matrix. The calculations below proceed from first principles.

•	Context Panel: Assign to a Name≻ $A$

$[\begin{array}{r} 5 & - 5 & - 4 \\ - 4 & 8 & 5 \\ 7 & - 11 & - 7 \end{array}]$ $\overset{assign to a name}{\to}$ $A$

•	Context Panel: Student Linear Algebra≻ Solvers and Forms≻Jordan Form

(Consequently, there is one chain of length 3 corresponding to the eigenvalue 2.)

$[\begin{array}{r} 5 & - 5 & - 4 \\ - 4 & 8 & 5 \\ 7 & - 11 & - 7 \end{array}]$ $\overset{Jordan form}{\to}$ $[\begin{array}{c} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{array}]$

Obtain the null spaces of $C = A - 2 I$ and $C^{2}$

•	Context Panel: Assign to a Name≻ $C$

(Note that Maple tolerates $A - 2$ as a short form of $A - 2 I$ , where $I$ is the identity matrix.)

$A - 2$ = $[\begin{array}{c} 3 & −5 & −4 \\ −4 & 6 & 5 \\ 7 & −11 & −9 \end{array}]$ $\overset{assign to a name}{\to}$ $C$

•	Context Panel: Evaluate and Display Inline Context Panel: Student Linear Algebra≻Vector Spaces≻Null Space

$C$ = $[\begin{array}{c} 3 & −5 & −4 \\ −4 & 6 & 5 \\ 7 & −11 & −9 \end{array}]$ $\overset{null space}{\to}$ $\{[\begin{array}{c} \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{array}]\}$

$C^{2}$ = $[\begin{array}{c} 1 & −1 & −1 \\ −1 & 1 & 1 \\ 2 & −2 & −2 \end{array}]$ $\overset{null space}{\to}$ $\{[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}], [\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]\}$

Select a vector in ${&reals;}^{3}$ that is not in the null space of $C^{2}$ and verify this choice

•	Context Panel: Assign to a Name≻b[3]

$⟨1, 0, 0⟩$ $\overset{assign to a name}{\to}$ $b_{3}$

•	Context Panel: Student Linear Algebra≻ Standard Operations≻Determinant

(Non-vanishing of the determinant shows $b_{3}$ is not a member of the null space of $C^{2}$ )

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}]$ $\overset{determinant}{\to}$ $−1$

Construct the remaining members of the one chain of linearly independent generalized eigenvectors

•	Context Panel: Evaluate and Display Inline Context Panel: Assign to a Name≻b[2]

$C &period; b_{3}$ = $[\begin{array}{c} 3 \\ −4 \\ 7 \end{array}]$ $\overset{assign to a name}{\to}$ $b_{2}$

•	Context Panel: Evaluate and Display Inline Context Panel: Assign to a Name≻b[1]

(Note that $b_{1}$ is an eigenvector.)

$C &period; b_{2}$ = $[\begin{array}{c} 1 \\ −1 \\ 2 \end{array}]$ $\overset{assign to a name}{\to}$ $b_{1}$

Construct the transition matrix whose columns are the vectors $b_{1}, b_{2}, b_{3}$

•	Context Panel: Evaluate and Display inline

•	Context Panel: Select Elements≻Combine into Matrix

•	Context Panel: Assign to a Name≻ $Q$

$[b_{1}, b_{2}, b_{3}]$ = $[[\begin{array}{c} 1 \\ −1 \\ 2 \end{array}], [\begin{array}{c} 3 \\ −4 \\ 7 \end{array}], [\begin{array}{c} 1 \\ 0 \\ 0 \end{array}]]$ $\overset{combine into Matrix}{\to}$ $[\begin{array}{c} 1 & 3 & 1 \\ −1 & −4 & 0 \\ 2 & 7 & 0 \end{array}]$ $\overset{assign to a name}{\to}$ $Q$

Verify that $Q$ puts $A$ into Jordan form

•	Context Panel: Evaluate and Display Inline

$Q^{- 1} &period; A &period; Q$ = $[\begin{array}{c} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{array}]$

Solution of Linear Systems

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Example 1: Solve a Completely Determined Linear System

Solve the completely determined system consisting of the equations

$x + y + z = 1, x - y - 2 z = 3, 5 x + 2 y - 7 z = 9$

Simply solve the equations

•	Control-drag the equations. Context Panel: Solve≻Solve

$x + y + z = 1, x - y - 2 z = 3, 5 x + 2 y - 7 z = 9$ $\overset{solve}{\to}$ $\{x = \frac{29}{15}, y = - \frac{4}{5}, z = - \frac{2}{15}\}$

Convert to a linear system

•	Control-drag the equations.

•	Context Panel: Student Linear Algebra≻ Constructions≻Generate Matrix≻Augmented (Complete dialog as per Figure 1.)

•	Context Panel: Student Linear Algebra≻ Solvers and Forms≻Linear Solve

Figure 1

$x + y + z = 1, x - y - 2 z = 3, 5 x + 2 y - 7 z = 9$ $\overset{to Matrix form}{\to}$ $[\begin{array}{c} 1 & 1 & 1 & 1 \\ 1 & −1 & −2 & 3 \\ 5 & 2 & −7 & 9 \end{array}]$ $\overset{linear solve}{\to}$ $[\begin{array}{c} \frac{29}{15} \\ - \frac{4}{5} \\ - \frac{2}{15} \end{array}]$

Example 2: Least-Squares Solution of an Overdetermined System

Obtain a least-squares solution to the overdetermined system consisting of the equations

$x + y + z = 1, x - y - 2 z = 3, 5 x + 2 y - 7 z = 9, 3 x - 7 y + 9 z = - 4$

•	Control-drag the equations and press the Enter key.

•	Context Panel: Student Linear Algebra≻Constructions≻Generate Matrix≻Matrix-Vector pair (Complete dialog as per Figure 1, in Example 1.)

•	Context Panel: Student Linear Algebra≻Solvers and Forms≻Least Squares

$x + y + z = 1, x - y - 2 z = 3, 5 x + 2 y - 7 z = 9, 3 x - 7 y + 9 z = - 4$

$x + y + z = 1, x - y - 2 z = 3, 5 x + 2 y - 7 z = 9, 3 x - 7 y + 9 z = −4$

$\overset{to Matrix form}{\to}$

$[\begin{array}{c} 1 & 1 & 1 \\ 1 & −1 & −2 \\ 5 & 2 & −7 \\ 3 & −7 & 9 \end{array}], [\begin{array}{c} 1 \\ 3 \\ 9 \\ −4 \end{array}]$

$\overset{least squares}{\to}$

$[\begin{array}{c} \frac{10207}{11574} \\ \frac{93}{1286} \\ - \frac{7777}{11574} \end{array}]$

Example 3: Minimum-Norm Least-Squares

Obtain the minimum-norm least-squares solution of the system $[\begin{array}{r} 7 & 0 & 5 & - 10 \\ 1 & - 5 & 1 & 9 \\ 10 & - 8 & 11 & - 1 \\ 4 & - 6 & 9 & 1 \end{array}] x = [\begin{array}{r} 1 \\ 2 \\ 3 \\ 4 \end{array}]$ .

Obtain the minimum-norm least-squares solution

•	Control-drag the system, editing it to a sequence of matrix and vector.

•	Context Panel: Student Linear Algebra≻Solvers and Forms≻Least Squares Check the "Optimized" box in the "Specify options for Least Squares" dialog

Work from first principles: obtain the general solution and minimize its norm:

Obtain the general solution

•	Control-drag the system, editing it to a sequence of matrix and vector.

•	Context Panel: Student Linear Algebra≻Solvers and Forms≻Least Squares Free-Variable Name≻ $s$

•	Context Panel: Evaluate at a Point≻ $s$

•	Context Panel: Assign to a Name≻X

$[\begin{array}{r} 7 & 0 & 5 & - 10 \\ 1 & - 5 & 1 & 9 \\ 10 & - 8 & 11 & - 1 \\ 4 & - 6 & 9 & 1 \end{array}], [\begin{array}{r} 1 \\ 2 \\ 3 \\ 4 \end{array}]$ $\overset{least squares}{\to}$ $[\begin{array}{c} s_{1} \\ 3 s_{1} + \frac{139}{518} \\ \frac{7 s_{1}}{5} + \frac{3574}{6475} \\ \frac{7 s_{1}}{5} + \frac{3133}{12950} \end{array}]$ $\overset{evaluate at point}{\to}$ $[\begin{array}{c} s \\ 3 s + \frac{139}{518} \\ \frac{7 s}{5} + \frac{3574}{6475} \\ \frac{7 s}{5} + \frac{3133}{12950} \end{array}]$ $\overset{assign to a name}{\to}$ $X$

Obtain the norm and minimize it

•	Write the name X and press the Enter key.

•	Context Panel: Student Linear Algebra≻Standard Operations≻Norm≻Euclidean

•	Context Panel: Differentiate≻With Respect To≻ $s$

•	Context Panel: Conversions≻Equate to 0 (This step is optional.)

•	Context Panel: Solve≻Solve

•	Context Panel: Assign to a Name≻ $S$

$X$

$[\begin{array}{c} s \\ 3 s + \frac{139}{518} \\ \frac{7 s}{5} + \frac{3574}{6475} \\ \frac{7 s}{5} + \frac{3133}{12950} \end{array}]$

$\overset{Euclidean-norm}{\to}$

$\frac{\sqrt{2334418800 s^{2} + 642796560 s + 72985218}}{12950}$

$\overset{differentiate w.r.t. s}{\to}$

$\frac{4668837600 s + 642796560}{25900 \sqrt{2334418800 s^{2} + 642796560 s + 72985218}}$

$\overset{equate to 0}{\to}$

$\frac{4668837600 s + 642796560}{25900 \sqrt{2334418800 s^{2} + 642796560 s + 72985218}} = 0$

$\overset{solve}{\to}$

$\{s = - \frac{10341}{75110}\}$

$\overset{assign to a name}{\to}$

$S$

•	Expression palette: Evaluation template Evaluate X at the solution in S

•	Context Panel: Evaluate and Display Inline

$\binom{X}{} | \binom{}{S}$ = $[\begin{array}{c} - \frac{10341}{75110} \\ - \frac{5434}{37555} \\ \frac{26981}{75110} \\ \frac{1847}{37555} \end{array}]$

Example 4: Stepwise Row Reduction and Back-Substitution

If the linear system $A x = y$ is expressed by the augmented matrix $[\begin{array}{r} 5 & - 6 & - 2 & 3 \\ - 4 & 3 & - 3 & - 1 \\ 1 & 1 & 0 & 2 \end{array}]$ , row-reduce to upper triangular form and solve for x.

•	Control-drag the matrix. Context Panel: Student Linear Algebra≻ Standard Operations≻Row-Reduced Form

•	Context Panel: Select Elements≻Restrict Columns (Complete dialog as per Figure 2. The return is then a vector and not a one-column matrix.)

Figure 2

$[\begin{array}{r} 5 & - 6 & - 2 & 3 \\ - 4 & 3 & - 3 & - 1 \\ 1 & 1 & 0 & 2 \end{array}]$ $\overset{row-reduced form}{\to}$ $[\begin{array}{c} 1 & 0 & 0 & \frac{59}{47} \\ 0 & 1 & 0 & \frac{35}{47} \\ 0 & 0 & 1 & - \frac{28}{47} \end{array}]$ $\overset{restrict columns}{\to}$ $[\begin{array}{c} \frac{59}{47} \\ \frac{35}{47} \\ - \frac{28}{47} \end{array}]$

Stepwise row reduction can be done via the Context Panel system, as per Figure 3.

Figure 3 Elementary row operations via the Context Panel system

The elementary row operations are also available in two tutors that can be accessed from the Context Panel (Student Linear Algebra > Tutors) . These are the Gaussian Elimination and Gauss-Jordan Elimination tutors..

Matrix Factorizations

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Example 1: LU Decomposition

Obtain the LU decomposition of the matrix $[\begin{array}{r} 1 & 6 & 5 \\ - 2 & 2 & 4 \\ 5 & 2 & - 6 \end{array}]$ .

•	Control-drag the given matrix. Context Panel: Student Linear Algebra≻Solvers and Forms≻LU Decomposition

$[\begin{array}{r} 1 & 6 & 5 \\ - 2 & 2 & 4 \\ 5 & 2 & - 6 \end{array}]$ $\overset{LU decomposition}{\to}$ $[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}], [\begin{array}{c} 1 & 0 & 0 \\ −2 & 1 & 0 \\ 5 & −2 & 1 \end{array}], [\begin{array}{c} 1 & 6 & 5 \\ 0 & 14 & 14 \\ 0 & 0 & −3 \end{array}]$

The returned matrices are $P, L, U$ , with $P$ being the matrix that tracks permutations of the rows; $L$ being the unit lower triangular factor; and $U$ being the upper triangular factor. By default, Maple returns the Doolittle, not the Crout, factorization.

Example 2: QR Decomposition

Obtain the QR decomposition of the matrix $[\begin{array}{r} 1 & 6 & 5 \\ - 2 & 2 & 4 \\ 5 & 2 & - 6 \end{array}]$ .

•	Control-drag the given matrix. Context Panel: Student Linear Algebra≻Solvers and Forms≻QR Decomposition

$[\begin{array}{r} 1 & 6 & 5 \\ - 2 & 2 & 4 \\ 5 & 2 & - 6 \end{array}]$ $\overset{QR decomposition}{\to}$ $[\begin{array}{c} \frac{\sqrt{30}}{30} & \frac{2 \sqrt{5}}{5} & \frac{\sqrt{6}}{6} \\ - \frac{\sqrt{30}}{15} & \frac{\sqrt{5}}{5} & - \frac{\sqrt{6}}{3} \\ \frac{\sqrt{30}}{6} & 0 & - \frac{\sqrt{6}}{6} \end{array}], [\begin{array}{c} \sqrt{30} & \frac{2 \sqrt{30}}{5} & - \frac{11 \sqrt{30}}{10} \\ 0 & \frac{14 \sqrt{5}}{5} & \frac{14 \sqrt{5}}{5} \\ 0 & 0 & \frac{\sqrt{6}}{2} \end{array}]$

Example 3: Singular-Value Decomposition

Obtain the singular-value decomposition of the matrix $[\begin{array}{r} 1 & 6 & 5 \\ - 2 & 2 & 4 \\ 5 & 2 & - 6 \end{array}]$ .

•	Control-drag the given matrix. Context Panel: Student Linear Algebra≻Solvers and Forms≻Singular Value Decomposition≻Singular Value Decomposition (U,S,Vt)

$[\begin{array}{r} 1 & 6 & 5 \\ - 2 & 2 & 4 \\ 5 & 2 & - 6 \end{array}]$ $\overset{singular value decomposition (U,S,Vt)}{\to}$ $[\begin{array}{c} 0.5988918686 & −0.7180631732 & −0.3545614298 \\ 0.4864104839 & −0.02555662302 & 0.8733565706 \\ −0.6361865833 & −0.6955085456 & 0.3339678021 \end{array}], [\begin{array}{c} 10.00874977 \\ 7.104651034 \\ 0.5906452392 \end{array}], [\begin{array}{c} −0.3551754316 & 0.3290919543 & 0.8749565122 \\ −0.5833492223 & −0.8094006799 & 0.06763300640 \\ −0.7304478746 & 0.4863836187 & −0.4794547714 \end{array}]$

The return consists of the factor $U$ , the vector of singular values, and the transpose of the factor $V$ . If $S$ is a diagonal matrix whose diagonal elements are the singular values, then $A = U S V^{T}$ .

Queries

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Example 1: Positive Definite Matrix

Is the symmetric matrix $[\begin{array}{r} 7 & 4 & 1 \\ 4 & 5 & 3 \\ 1 & 3 & 6 \end{array}]$ positive definite?

•	Control-drag the given matrix. Context Panel: Student Linear Algebra≻Queries≻ Is Definite?≻Positive Definite?

$[\begin{array}{r} 7 & 4 & 1 \\ 4 & 5 & 3 \\ 1 & 3 & 6 \end{array}]$ $\overset{is positive definite?}{\to}$ $true$

Typically, definiteness is assigned to bilinear forms $x^{T} A x$ derived from the symmetric matrix $A$ . If $A$ is not symmetric, the associated bilinear form can be represented by $x^{T} B x$ , where $B = (A + A^{T}) / 2$ , the "symmetric part of $A$ " is symmetric. Hence, Maple assigns definiteness to the symmetric part of a nonsymmetric matrix on the grounds that the matrix represents a bilinear form.

Example 2: Similar Matrices

Show that the matrices $A = [\begin{array}{r} - 5 & - 2 & - 2 \\ 3 & 4 & - 1 \\ 4 & - 2 & - 8 \end{array}]$ and $B = [\begin{array}{r} - 15 & - 1530 & - 2334 \\ 11 & 1433 & 2191 \\ - 5 & - 931 & - 1427 \end{array}]$ are similar by finding a matrix $C$ for which $C A = B C$ .

•	Write the sequence of matrices A and B Context Panel: Student Linear Algebra≻Queries≻Similar?

•	Context Panel: Select Element≻2

•	Context Panel: Assign to a Name≻C

$[\begin{array}{r} - 5 & - 2 & - 2 \\ 3 & 4 & - 1 \\ 4 & - 2 & - 8 \end{array}], [\begin{array}{r} - 15 & - 1530 & - 2334 \\ 11 & 1433 & 2191 \\ - 5 & - 931 & - 1427 \end{array}]$ $\overset{is similar?}{\to}$ $true, [\begin{array}{c} 1 & 0 & 0 \\ - \frac{486581}{522111} & \frac{533}{522111} & - \frac{1071}{348074} \\ \frac{316730}{522111} & \frac{98}{522111} & \frac{3001}{1044222} \end{array}]$ $\overset{select entry 2}{\to}$ $[\begin{array}{c} 1 & 0 & 0 \\ - \frac{486581}{522111} & \frac{533}{522111} & - \frac{1071}{348074} \\ \frac{316730}{522111} & \frac{98}{522111} & \frac{3001}{1044222} \end{array}]$ $\overset{assign to a name}{\to}$ $C$

Data entry

•	Control-drag each matrix. Context Panel: Assign to a Name≻ $A$ (or $B$ , as appropriate)

$[\begin{array}{r} - 5 & - 2 & - 2 \\ 3 & 4 & - 1 \\ 4 & - 2 & - 8 \end{array}]$ $\overset{assign to a name}{\to}$ $A$

$[\begin{array}{r} - 15 & - 1530 & - 2334 \\ 11 & 1433 & 2191 \\ - 5 & - 931 & - 1427 \end{array}]$ $\overset{assign to a name}{\to}$ $B$

Test for similarity and find $C$

•	Write a sequence of the names $A$ and $B$ , then press the Enter key.

•	Context Panel: Student Linear Algebra≻Queries≻Is Similar?

•	Context Panel: Select Element≻2

•	Context Panel: Assign to a Name≻ $C$

$A, B$

$[\begin{array}{c} −5 & −2 & −2 \\ 3 & 4 & −1 \\ 4 & −2 & −8 \end{array}], [\begin{array}{c} −15 & −1530 & −2334 \\ 11 & 1433 & 2191 \\ −5 & −931 & −1427 \end{array}]$

$\overset{is similar?}{\to}$

$true, [\begin{array}{c} 1 & 0 & 0 \\ - \frac{486581}{522111} & \frac{533}{522111} & - \frac{1071}{348074} \\ \frac{316730}{522111} & \frac{98}{522111} & \frac{3001}{1044222} \end{array}]$

$\overset{select entry 2}{\to}$

$[\begin{array}{c} 1 & 0 & 0 \\ - \frac{486581}{522111} & \frac{533}{522111} & - \frac{1071}{348074} \\ \frac{316730}{522111} & \frac{98}{522111} & \frac{3001}{1044222} \end{array}]$

$\overset{assign to a name}{\to}$

$C$

Verify similarity

•	Context Panel: Evaluate and Display Inline

$C &period; A$ = $[\begin{array}{c} −5 & −2 & −2 \\ \frac{2428078}{522111} & \frac{326169}{174037} & \frac{985481}{522111} \\ - \frac{1577354}{522111} & - \frac{212023}{174037} & - \frac{645562}{522111} \end{array}]$

$B &period; C$ = $[\begin{array}{c} −5 & −2 & −2 \\ \frac{2428078}{522111} & \frac{326169}{174037} & \frac{985481}{522111} \\ - \frac{1577354}{522111} & - \frac{212023}{174037} & - \frac{645562}{522111} \end{array}]$

Example 3: Orthogonal Matrix

Construct a (nontrivial) 3×3 orthogonal matrix.

•	Enter a list of three linearly independent vectors and press the Enter key.

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Gram-Schmidt≻normalized

•	Context Panel: Select Elements≻Combine into Matrix

•	Context Panel: Assign to a Name≻ $Q$

$[⟨- 4, 1, 6⟩, ⟨5, 3, 1⟩, ⟨7, - 8, 9⟩]$

$[[\begin{array}{c} −4 \\ 1 \\ 6 \end{array}], [\begin{array}{c} 5 \\ 3 \\ 1 \end{array}], [\begin{array}{c} 7 \\ −8 \\ 9 \end{array}]]$

$\overset{Gram-Schmidt (normalized)}{\to}$

$[[\begin{array}{c} - \frac{4 \sqrt{53}}{53} \\ \frac{\sqrt{53}}{53} \\ \frac{6 \sqrt{53}}{53} \end{array}], [\begin{array}{c} \frac{13 \sqrt{318}}{318} \\ \frac{5 \sqrt{318}}{159} \\ \frac{7 \sqrt{318}}{318} \end{array}], [\begin{array}{c} \frac{\sqrt{6}}{6} \\ - \frac{\sqrt{6}}{3} \\ \frac{\sqrt{6}}{6} \end{array}]]$

$\overset{combine into Matrix}{\to}$

$[\begin{array}{c} - \frac{4 \sqrt{53}}{53} & \frac{13 \sqrt{318}}{318} & \frac{\sqrt{6}}{6} \\ \frac{\sqrt{53}}{53} & \frac{5 \sqrt{318}}{159} & - \frac{\sqrt{6}}{3} \\ \frac{6 \sqrt{53}}{53} & \frac{7 \sqrt{318}}{318} & \frac{\sqrt{6}}{6} \end{array}]$

$\overset{assign to a name}{\to}$

$Q$

Verify that $Q$ is an orthogonal matrix

•	Write the name $Q$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Queries≻Orthogonal?

$Q$ = $[\begin{array}{c} - \frac{4 \sqrt{53}}{53} & \frac{13 \sqrt{318}}{318} & \frac{\sqrt{6}}{6} \\ \frac{\sqrt{53}}{53} & \frac{5 \sqrt{318}}{159} & - \frac{\sqrt{6}}{3} \\ \frac{6 \sqrt{53}}{53} & \frac{7 \sqrt{318}}{318} & \frac{\sqrt{6}}{6} \end{array}]$ $\overset{is orthogonal?}{\to}$ $true$

An alternative verification consists in showing that $Q^{T} Q = {QQ}^{T} = I$ , thereby confirming that the rows (and columns) of $Q$ are sets of orthonormal vectors.

$Q^{%T} &period; Q$ = $[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

$Q &period; Q^{%T}$ = $[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Vector Spaces

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Example 1: Four Fundamental Subspaces of a 5×3

Find the row space, column space, null space, and null space of the transpose for the matrix

$[\begin{array}{r} 32 & - 8 & - 4 \\ - 40 & 10 & 5 \\ - 32 & 8 & 4 \\ 8 & - 2 & - 1 \\ 72 & - 18 & - 9 \end{array}]$

(Gilbert Strang of MIT calls these the four fundamental subspaces of $A$ .)

The 5×3 matrix $A$ maps ${&reals;}^{3}$ to ${&reals;}^{5}$ . Maple provides bases for each of the four fundamental subspaces.

The row and null spaces of $A$ are orthogonal subspaces of ${&reals;}^{3}$ ; the column space of $A$ and the null space of $A^{T}$ are orthogonal subspaces in ${&reals;}^{5}$ . Figure 4 illustrates the relationships between these four subspaces.

Figure 4 The four fundamental subspaces of $A$

Data entry

•	Control-drag (or copy/paste) the given matrix. Context Panel: Assign to a Name≻ $A$

$[\begin{array}{r} 32 & - 8 & - 4 \\ - 40 & 10 & 5 \\ - 32 & 8 & 4 \\ 8 & - 2 & - 1 \\ 72 & - 18 & - 9 \end{array}]$ $\overset{assign to a name}{\to}$ $A$

Row space of $A$

•	Write the name $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Row Space

$A$ = $[\begin{array}{c} 32 & −8 & −4 \\ −40 & 10 & 5 \\ −32 & 8 & 4 \\ 8 & −2 & −1 \\ 72 & −18 & −9 \end{array}]$ $\overset{row space}{\to}$ $[[\begin{array}{c} 1 & - \frac{1}{4} & - \frac{1}{8} \end{array}]]$

Column space of $A$

•	Write the name $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Column Space

$A$ = $[\begin{array}{c} 32 & −8 & −4 \\ −40 & 10 & 5 \\ −32 & 8 & 4 \\ 8 & −2 & −1 \\ 72 & −18 & −9 \end{array}]$ $\overset{column space}{\to}$ $[[\begin{array}{c} 1 \\ - \frac{5}{4} \\ −1 \\ \frac{1}{4} \\ \frac{9}{4} \end{array}]]$

Null space of $A$

•	Write the name $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Null Space

$A$ = $[\begin{array}{c} 32 & −8 & −4 \\ −40 & 10 & 5 \\ −32 & 8 & 4 \\ 8 & −2 & −1 \\ 72 & −18 & −9 \end{array}]$ $\overset{null space}{\to}$ $\{[\begin{array}{c} \frac{1}{8} \\ 0 \\ 1 \end{array}], [\begin{array}{c} \frac{1}{4} \\ 1 \\ 0 \end{array}]\}$

Null space of $A^{T}$

•	Write the notation for the transpose of $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Null Space

$A^{%T}$ = $[\begin{array}{c} 32 & −40 & −32 & 8 & 72 \\ −8 & 10 & 8 & −2 & −18 \\ −4 & 5 & 4 & −1 & −9 \end{array}]$ $\overset{null space}{\to}$ $\{[\begin{array}{c} - \frac{9}{4} \\ 0 \\ 0 \\ 0 \\ 1 \end{array}], [\begin{array}{c} - \frac{1}{4} \\ 0 \\ 0 \\ 1 \\ 0 \end{array}], [\begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}], [\begin{array}{c} \frac{5}{4} \\ 1 \\ 0 \\ 0 \\ 0 \end{array}]\}$

Example 2: Four Fundamental Subspaces of a 4×5

Find the row space, column space, null space, and null space of the transpose for the matrix

$[\begin{array}{r} 50 & 42 & - 6 & - 20 & 20 \\ - 8 & - 1 & - 37 & 37 & - 24 \\ 47 & 43 & - 29 & 2 & 6 \\ - 25 & - 21 & 3 & 10 & - 10 \end{array}]$

(Gilbert Strang of MIT calls these the four fundamental subspaces of $A$ .)

The 4×5 matrix $A$ maps ${&reals;}^{5}$ to ${&reals;}^{4}$ . Maple provides bases for each of the four fundamental subspaces.

The row and null spaces of $A$ are orthogonal subspaces of ${&reals;}^{5}$ ; the column space of $A$ and the null space of $A^{T}$ are orthogonal subspaces in ${&reals;}^{4}$ . Figure 5 illustrates the relationships between these four subspaces.

Figure 5 The four fundamental subspaces of $A$

Data entry

•	Control-drag (or copy/paste) the given matrix. Context Panel: Assign to a Name≻ $A$

$[\begin{array}{r} 50 & 42 & - 6 & - 20 & 20 \\ - 8 & - 1 & - 37 & 37 & - 24 \\ 47 & 43 & - 29 & 2 & 6 \\ - 25 & - 21 & 3 & 10 & - 10 \end{array}]$ $\overset{assign to a name}{\to}$ $A$

Row space of $A$

•	Write the name $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Row Space

$A$ = $[\begin{array}{c} 50 & 42 & −6 & −20 & 20 \\ −8 & −1 & −37 & 37 & −24 \\ 47 & 43 & −29 & 2 & 6 \\ −25 & −21 & 3 & 10 & −10 \end{array}]$ $\overset{row space}{\to}$ $[[\begin{array}{c} 1 & 0 & \frac{60}{11} & - \frac{59}{11} & \frac{38}{11} \end{array}], [\begin{array}{c} 0 & 1 & - \frac{73}{11} & \frac{65}{11} & - \frac{40}{11} \end{array}]]$

Column space of $A$

•	Write the name $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Column Space

$A$ = $[\begin{array}{c} 50 & 42 & −6 & −20 & 20 \\ −8 & −1 & −37 & 37 & −24 \\ 47 & 43 & −29 & 2 & 6 \\ −25 & −21 & 3 & 10 & −10 \end{array}]$ $\overset{column space}{\to}$ $[[\begin{array}{c} 1 \\ 0 \\ \frac{27}{26} \\ - \frac{1}{2} \end{array}], [\begin{array}{c} 0 \\ 1 \\ \frac{8}{13} \\ 0 \end{array}]]$

Null space of $A$

•	Write the name $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Null Space

$A$ = $[\begin{array}{c} 50 & 42 & −6 & −20 & 20 \\ −8 & −1 & −37 & 37 & −24 \\ 47 & 43 & −29 & 2 & 6 \\ −25 & −21 & 3 & 10 & −10 \end{array}]$ $\overset{null space}{\to}$ $\{[\begin{array}{c} - \frac{38}{11} \\ \frac{40}{11} \\ 0 \\ 0 \\ 1 \end{array}], [\begin{array}{c} \frac{59}{11} \\ - \frac{65}{11} \\ 0 \\ 1 \\ 0 \end{array}], [\begin{array}{c} - \frac{60}{11} \\ \frac{73}{11} \\ 1 \\ 0 \\ 0 \end{array}]\}$

Null space of $A^{T}$

•	Write the notation for the transpose of $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Null Space

$A^{%T}$ = $[\begin{array}{c} 50 & −8 & 47 & −25 \\ 42 & −1 & 43 & −21 \\ −6 & −37 & −29 & 3 \\ −20 & 37 & 2 & 10 \\ 20 & −24 & 6 & −10 \end{array}]$ $\overset{null space}{\to}$ $\{[\begin{array}{c} \frac{1}{2} \\ 0 \\ 0 \\ 1 \end{array}], [\begin{array}{c} - \frac{27}{26} \\ - \frac{8}{13} \\ 1 \\ 0 \end{array}]\}$

Special Matrices

•	Tools≻Load Package: Student Linear Algebra

Loading Student:-LinearAlgebra

Example 1: Inverse by Adjoint

Divide the adjoint of $A = [\begin{array}{r} 9 & 2 & 1 \\ - 5 & - 4 & 1 \\ 4 & 6 & - 2 \end{array}]$ by the determinant of $A$ , and show that the resulting matrix is $A^{- 1}$ , the multiplicative inverse of $A$ .

Data entry

•	Control-drag the matrix $A$ . Context Panel: Assign to a Name≻ $A$

$[\begin{array}{r} 9 & 2 & 1 \\ - 5 & - 4 & 1 \\ 4 & 6 & - 2 \end{array}]$ $\overset{assign to a name}{\to}$ $A$

Obtain the determinant of $A$

•	Write the name $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻ Standard Operations≻Determinant

$A$ = $[\begin{array}{c} 9 & 2 & 1 \\ −5 & −4 & 1 \\ 4 & 6 & −2 \end{array}]$ $\overset{determinant}{\to}$ $−8$

Obtain the adjoint of $A$

•	Write the name $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Standard Operations≻Adjoint

•	Context Panel: Assign to a Name≻adjA

$A$ = $[\begin{array}{c} 9 & 2 & 1 \\ −5 & −4 & 1 \\ 4 & 6 & −2 \end{array}]$ $\overset{adjoint}{\to}$ $[\begin{array}{c} 2 & 10 & 6 \\ −6 & −22 & −14 \\ −14 & −46 & −26 \end{array}]$ $\overset{assign to a name}{\to}$ $adjA$

Divide the adjoint by the determinant

•	Context Panel: Evaluate and Display Inline

$\frac{adjA}{- 8}$ = $[\begin{array}{c} - \frac{1}{4} & - \frac{5}{4} & - \frac{3}{4} \\ \frac{3}{4} & \frac{11}{4} & \frac{7}{4} \\ \frac{7}{4} & \frac{23}{4} & \frac{13}{4} \end{array}]$

Obtain $A^{- 1}$ , the multiplicative inverse of $A$

•	Write the name $A$ Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻ Standard Operations≻Inverse

$A$ = $[\begin{array}{c} 9 & 2 & 1 \\ −5 & −4 & 1 \\ 4 & 6 & −2 \end{array}]$ $\overset{inverse}{\to}$ $[\begin{array}{c} - \frac{1}{4} & - \frac{5}{4} & - \frac{3}{4} \\ \frac{3}{4} & \frac{11}{4} & \frac{7}{4} \\ \frac{7}{4} & \frac{23}{4} & \frac{13}{4} \end{array}]$

Example 2: Reflection Matrix (across a Line)

Obtain a matrix that reflects vectors in ${&reals;}^{2}$ across the line $y = x / 3$ .

•	The red dashed line line in Figure 6 is the graph of $y = x / 3$ . The green vector, $3 i + j$ , is along this line.

•	The gold vector, $- i + 3 j$ , is orthogonal to the line $y = x / 3$ .

•	The black vector, $2 i + 2 j$ , is an arbitrary vector in ${&reals;}^{2}$ . Its reflection across the line $y = x / 3$ is the red vector.

•	The reflection matrix is constructed from the gold vector, that is, from a vector orthogonal to the "mirror" across which reflection is to take place.

Figure 6

Construct the rotation matrix

•	On a vector orthogonal to the line of reflection: Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Constructions≻Reflection Matrix

•	Context Panel: Assign to a Name≻ $R$

$⟨- 1, 3⟩$ = $[\begin{array}{c} −1 \\ 3 \end{array}]$ $\overset{reflection matrix}{\to}$ $[\begin{array}{c} \frac{4}{5} & \frac{3}{5} \\ \frac{3}{5} & - \frac{4}{5} \end{array}]$ $\overset{assign to a name}{\to}$ $R$

Test the rotation matrix

•	Write sequences of two vectors (black & green, red & green, in Figure 6); press the Enter key.

•	Context Panel: Student Linear Algebra≻Standard Operations≻Vector Angle

$⟨2, 2⟩, ⟨3, 1⟩$

$[\begin{array}{c} 2 \\ 2 \end{array}], [\begin{array}{c} 3 \\ 1 \end{array}]$

$\overset{angle between}{\to}$

$\arccos (\frac{\sqrt{2} \sqrt{10}}{5})$

$R &period; ⟨2, 2⟩, ⟨3, 1⟩$

$[\begin{array}{c} \frac{14}{5} \\ - \frac{2}{5} \end{array}], [\begin{array}{c} 3 \\ 1 \end{array}]$

$\overset{angle between}{\to}$

$\arccos (\frac{\sqrt{2} \sqrt{10}}{5})$

Example 3: Reflection Matrix (across a Plane)

Obtain a matrix that reflects vectors in ${&reals;}^{3}$ across the plane $x + 2 y + 3 z = 0$ .

•	Figure 7 shows the plane across which reflections are to take place. In addition, N, the black vector in the figure, is a normal to the plane.

•	The red vector, $V = i + j + k$ , is an arbitrary vector in ${&reals;}^{3}$ .

•	The green vector is $R V$ , the reflection of V across the given plane, where $R$ is the requisite reflection matrix.

•	The angles between V and N and RV and -N should be equal if RV is the reflection of V across the plane.

use plots, Student:-VectorCalculus, Student:-LinearAlgebra in
module()
local p1,p2,p3,R,N,V;
N:=<1,2,3>/2;
V:=<1,1,1>;
R:=ReflectionMatrix(N);
p1:=implicitplot3d(x+2*y+2*z=0,x=-1..1,y=-1..1,z=-2..2,style=wireframe);
p2:=PlotVector([N,V,R.V],color=[black,red,green],width=.2);
p3:=display(p1,p2,scaling=constrained,labels=[x,y,z],axes=frame,orientation=[-5,80,0],tickmarks=[3,4,6],lightmodel=none);
print(p3);
end module:
end use:

Figure 7

Construct the rotation matrix

•	On a vector orthogonal to the plane: Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Constructions≻Reflection Matrix

•	Context Panel: Assign to a Name≻ $R$

$⟨1, 2, 3⟩$ = $[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}]$ $\overset{reflection matrix}{\to}$ $[\begin{array}{c} \frac{6}{7} & - \frac{2}{7} & - \frac{3}{7} \\ - \frac{2}{7} & \frac{3}{7} & - \frac{6}{7} \\ - \frac{3}{7} & - \frac{6}{7} & - \frac{2}{7} \end{array}]$ $\overset{assign to a name}{\to}$ $R$

Test the rotation matrix

•	Write sequences of two vectors (V and N, RV and -N, in Figure 7); press the Enter key.

•	Context Panel: Student Linear Algebra≻Standard Operations≻Vector Angle

$⟨1, 1, 1⟩, ⟨1, 2, 3⟩$

$[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}], [\begin{array}{c} 1 \\ 2 \\ 3 \end{array}]$

$\overset{angle between}{\to}$

$\arccos (\frac{\sqrt{3} \sqrt{14}}{7})$

$R &period; ⟨1, 1, 1⟩, - ⟨1, 2, 3⟩$

$[\begin{array}{c} \frac{1}{7} \\ - \frac{5}{7} \\ - \frac{11}{7} \end{array}], [\begin{array}{c} −1 \\ −2 \\ −3 \end{array}]$

$\overset{angle between}{\to}$

$\arccos (\frac{\sqrt{3} \sqrt{14}}{7})$

Example 4: Rotation Matrix

Rotate the vector $V = i - j + 2 k$ through an angle of $π / 6$ radians about the line $x = t, y = 2 t, z = 3 t$ .

•	In Figure 8, the black vector, $i + 2 j + 3 k$ , is along the axis of rotation, shown as the dashed red line.

•	In Figure 8, the red vector is $V = i - j + 2 k$ ; its $π / 6$ rotation about the axis of rotation, is the green vector.

use plots, Student:-VectorCalculus, Student:-LinearAlgebra in
module()
local p1,p2,p3,V,N,R;
R:=RotationMatrix(Pi/6,<1,2,3>);
V:=<1,-1,2>;
N:=<1,2,3>;
p1:=spacecurve([t,2*t,3*t],t=-1/5..1.2,color=red,linestyle=dash);
p2:=PlotVector([V,R.V,N],color=[red,green,black],width=.2);
p3:=display(p1,p2,scaling=constrained,labels=[x,y,z],tickmarks=[3,3,5],orientation=[-65,85,0]);
print(p3);
end module:
end use:

Figure 8

Construct the requisite rotation matrix

•	Write a sequence of the rotation angle and a vector along the axis of rotation; press the Enter key.

•	Context Panel: Student Linear Algebra≻Constructions≻Rotation Matrix

$π / 6, ⟨1, 2, 3⟩$

$\frac{π}{6}, [\begin{array}{c} 1 \\ 2 \\ 3 \end{array}]$

$\overset{rotation matrix}{\to}$

$[\begin{array}{c} \frac{13 \sqrt{3}}{28} + \frac{1}{14} & - \frac{\sqrt{3}}{14} - \frac{3 \sqrt{14}}{28} + \frac{1}{7} & - \frac{3 \sqrt{3}}{28} + \frac{\sqrt{14}}{14} + \frac{3}{14} \\ - \frac{\sqrt{3}}{14} + \frac{3 \sqrt{14}}{28} + \frac{1}{7} & \frac{5 \sqrt{3}}{14} + \frac{2}{7} & - \frac{3 \sqrt{3}}{14} - \frac{\sqrt{14}}{28} + \frac{3}{7} \\ - \frac{3 \sqrt{3}}{28} - \frac{\sqrt{14}}{14} + \frac{3}{14} & - \frac{3 \sqrt{3}}{14} + \frac{\sqrt{14}}{28} + \frac{3}{7} & \frac{5 \sqrt{3}}{28} + \frac{9}{14} \end{array}]$

Matrix Operators

•	Tools≻Load Package: Student Linear Algebra

Loading Student:-LinearAlgebra

Example 1: Matrix Norm Subordinate to Vector Norm

Obtain the Euclidean norm of the matrix $A = [\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}]$ and show that it is the maximum value of the Euclidean norm of the vector $A v$ , where v is a unit vector.

Obtain the Euclidean norm of $A$

•	Control-drag the matrix $A$ Context Panel: Student Linear Algebra≻Standard Operations≻Norm≻Euclidean

•	Context Panel: Simplify≻Simplify

$[\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}]$ $\overset{Euclidean-norm}{\to}$ $\sqrt{\frac{15}{2} + \frac{\sqrt{29}}{2}}$ $\overset{simplify}{=}$ $\frac{\sqrt{29}}{2} + \frac{1}{2}$

Obtain the norm of $A x$ , where x is a unit vector

•	Write $A$ times a unit vector and press the Enter key.

•	Context Panel: Student Linear Algebra≻Standard Operations≻Norm≻Euclidean

•	Context Panel: Simplify≻Simplify

•	Context Panel: Assign to a Name≻ $f$

$[\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}] &period; ⟨x, \sqrt{1 - x^{2}}⟩$

$[\begin{array}{c} x + 2 \sqrt{- x^{2} + 1} \\ 3 x - \sqrt{- x^{2} + 1} \end{array}]$

$\overset{Euclidean-norm}{\to}$

$\sqrt{{(x + 2 \sqrt{- x^{2} + 1})}^{2} + {(3 x - \sqrt{- x^{2} + 1})}^{2}}$

$\overset{simplify}{=}$

$\sqrt{5 x^{2} + 5 - 2 x \sqrt{- x^{2} + 1}}$

$\overset{assign to a name}{\to}$

$f$

Maximize the norm of $A x$

•	Write $f$ and press the Enter key.

•	Context Panel: Differentiate≻With Respect To≻ $x$

•	Context Panel: Conversions≻Equate to 0 (This step is optional.)

•	Context Panel: Solve≻Solve

•	Context Panel: Assign to a Name≻ $S$

$f$

$\sqrt{5 x^{2} + 5 - 2 x \sqrt{- x^{2} + 1}}$

$\overset{differentiate w.r.t. x}{\to}$

$\frac{10 x - 2 \sqrt{- x^{2} + 1} + \frac{2 x^{2}}{\sqrt{- x^{2} + 1}}}{2 \sqrt{5 x^{2} + 5 - 2 x \sqrt{- x^{2} + 1}}}$

$\overset{equate to 0}{\to}$

$\frac{10 x - 2 \sqrt{- x^{2} + 1} + \frac{2 x^{2}}{\sqrt{- x^{2} + 1}}}{2 \sqrt{5 x^{2} + 5 - 2 x \sqrt{- x^{2} + 1}}} = 0$

$\overset{solve}{\to}$

$\{x = \frac{\sqrt{1682 - 290 \sqrt{29}}}{58}\}, \{x = - \frac{\sqrt{1682 + 290 \sqrt{29}}}{58}\}$

$\overset{assign to a name}{\to}$

$S$

Evaluate $f = ∥A x∥$ at each critical value of $x$

•	Expression palette: Evaluation template Evaluate at each of the two critical values.

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Simplify

$\binom{f}{} | \binom{}{S_{1}}$ = $\sqrt{\frac{15}{2} - \frac{25 \sqrt{29}}{58} - \frac{\sqrt{1682 - 290 \sqrt{29}} \sqrt{\frac{1}{2} + \frac{5 \sqrt{29}}{58}}}{29}}$ $\overset{simplify}{=}$ $- \frac{1}{2} + \frac{\sqrt{29}}{2}$

$\binom{f}{} | \binom{}{S_{2}}$ = $\sqrt{\frac{15}{2} + \frac{25 \sqrt{29}}{58} + \frac{\sqrt{1682 + 290 \sqrt{29}} \sqrt{\frac{1}{2} - \frac{5 \sqrt{29}}{58}}}{29}}$ $\overset{simplify}{=}$ $\frac{\sqrt{29}}{2} + \frac{1}{2}$

Example 2: Matrix Norm and Singular Values

Show that the Euclidean norm of the matrix $A = [\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}]$ is the largest singular value of $A$ , and the square root of the largest eigenvalue of $A^{T} A$ .

From Example 1:

Obtain the Euclidean norm of $A$

•	Control-drag the matrix $A$ Context Panel: Student Linear Algebra≻Standard Operations≻Norm≻Euclidean

•	Context Panel: Simplify≻Simplify

$[\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}]$ $\overset{Euclidean-norm}{\to}$ $\sqrt{\frac{15}{2} + \frac{\sqrt{29}}{2}}$ $\overset{simplify}{=}$ $\frac{\sqrt{29}}{2} + \frac{1}{2}$

Obtain the singular values of $A$

•	Control-drag the matrix $A$ Context Panel: Student Linear Algebra≻ Eigenvalues, etc≻Singular Values

$[\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}]$ $\overset{singular values}{\to}$ $[\begin{array}{c} \frac{\sqrt{29}}{2} + \frac{1}{2} \\ - \frac{1}{2} + \frac{\sqrt{29}}{2} \end{array}]$

Obtain the eigenvalues of $A^{T} A$

•	Write the product $A^{T} &period; A$ . Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Eigenvalues, etc≻Eigenvalues

${[\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}]}^{%T} &period; [\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}]$ = $[\begin{array}{c} 10 & −1 \\ −1 & 5 \end{array}]$ $\overset{eigenvalues}{\to}$ $[\begin{array}{c} \frac{15}{2} + \frac{\sqrt{29}}{2} \\ \frac{15}{2} - \frac{\sqrt{29}}{2} \end{array}]$

•	Control-drag the larger of the two eigenvalues.

•	Select, and click $\sqrt{a}$ in the Expression palette

•	Context Panel: Evaluate and Display Inline

$\sqrt{\frac{15}{2} + \frac{1}{2} \sqrt{29}}$ = $\frac{\sqrt{29}}{2} + \frac{1}{2}$

Vectors and Vector Operators

•	Tools≻Load Package: Student Linear Algebra

Loading Student:-LinearAlgebra

Example 1: Vector Angle, Dot and Cross Products

Determine the angle between the vectors $u = i + 2 j + 3 k$ and $v = 3 i - 7 j + 5 k$ , then obtain their dot and cross products.

Data entry

•	Context Panel: Assign to a Name≻ $u$ and $v$ , as appropriate

$⟨1, 2, 3⟩$ $\overset{assign to a name}{\to}$ $u$

$⟨3, - 7, 5⟩$ $\overset{assign to a name}{\to}$ $v$

Determine the angle between u and v

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Standard Operations≻Vector Angle

$u, v$ = $[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}], [\begin{array}{c} 3 \\ −7 \\ 5 \end{array}]$ $\overset{angle between}{\to}$ $\arccos (\frac{2 \sqrt{14} \sqrt{83}}{581})$

Dot product

Cross product

•	Common Symbols palette: Dot product operator

•	Context Panel: Evaluate and Display Inline

Common Symbols palette: Cross product operator

Context Panel: Evaluate and Display Inline

$u \cdot v$ = $4$

$u \times v$ = $[\begin{array}{c} 31 \\ 4 \\ −13 \end{array}]$

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Student Linear Algebra≻Standard Operations≻Dot Product (or Cross Product)

$u, v$ = $[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}], [\begin{array}{c} 3 \\ −7 \\ 5 \end{array}]$ $\overset{dot product}{\to}$ $4$

$u, v$ = $[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}], [\begin{array}{c} 3 \\ −7 \\ 5 \end{array}]$ $\overset{cross product}{\to}$ $[\begin{array}{c} 31 \\ 4 \\ −13 \end{array}]$

Example 2: Orthonormalization

Orthonormalize the columns of the matrix $A = [\begin{array}{r} - 8 & - 4 & 9 \\ - 6 & 4 & 3 \\ 3 & - 9 & 1 \end{array}]$ , then form $Q$ , a matrix with these orthonormalized vectors, and show that $Q$ is an orthogonal matrix.

•	Control-drag the matrix $A$ and press the Enter key.

•	Context Panel: Select Elements≻Split into Columns

•	Context Panel: Student Linear Algebra≻Vector Spaces≻Gram-Schmidt≻normalized

•	Context Panel: Assign to a Name≻ $Q$

$[\begin{array}{r} - 8 & - 4 & 9 \\ - 6 & 4 & 3 \\ 3 & - 9 & 1 \end{array}]$

$[\begin{array}{c} −8 & −4 & 9 \\ −6 & 4 & 3 \\ 3 & −9 & 1 \end{array}]$

$\overset{split into columns}{\to}$

$[[\begin{array}{c} −8 \\ −6 \\ 3 \end{array}], [\begin{array}{c} −4 \\ 4 \\ −9 \end{array}], [\begin{array}{c} 9 \\ 3 \\ 1 \end{array}]]$

$\overset{Gram-Schmidt (normalized)}{\to}$

$[[\begin{array}{c} - \frac{8 \sqrt{109}}{109} \\ - \frac{6 \sqrt{109}}{109} \\ \frac{3 \sqrt{109}}{109} \end{array}], [\begin{array}{c} - \frac{42 \sqrt{6649}}{6649} \\ \frac{23 \sqrt{6649}}{6649} \\ - \frac{66 \sqrt{6649}}{6649} \end{array}], [\begin{array}{c} \frac{3 \sqrt{61}}{61} \\ - \frac{6 \sqrt{61}}{61} \\ - \frac{4 \sqrt{61}}{61} \end{array}]]$

$\overset{combine into Matrix}{\to}$

$[\begin{array}{c} - \frac{8 \sqrt{109}}{109} & - \frac{42 \sqrt{6649}}{6649} & \frac{3 \sqrt{61}}{61} \\ - \frac{6 \sqrt{109}}{109} & \frac{23 \sqrt{6649}}{6649} & - \frac{6 \sqrt{61}}{61} \\ \frac{3 \sqrt{109}}{109} & - \frac{66 \sqrt{6649}}{6649} & - \frac{4 \sqrt{61}}{61} \end{array}]$

$\overset{assign to a name}{\to}$

$Q$

Verify that $Q$ is an orthogonal matrix

•	Context Panel: Evaluate and Display Inline

$Q^{%T} &period; Q$ = $[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

$Q &period; Q^{%T}$ = $[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Visualizing a Linear Transform

•	Tools≻Load Package: Student Linear Algebra

Loading Student:-LinearAlgebra

Example 1: Linear Transform Induced by a 2×2 Matrix

Visualize the effect of applying to unit vectors, the linear transformation determined by the matrix $A = [\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}]$ .

Access the Linear Transform Plot tutor through the Context Panel applied to the matrix $A$ . The result is Figure 9.

Context Panel: Student Linear Algebra≻Tutors≻Linear Transform Plot

Figure 9

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