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

Data entry

$\left[\begin{array}{rr}-1& -12\\ 1& 6\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$A$

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

$A$ = $\left[\begin{array}{cc}\mathrm{-1}& \mathrm{-12}\\ 1& 6\end{array}\right]$$\stackrel{\text{eigenvectors}}{\to }$$\left[\begin{array}{c}3\\ 2\end{array}\right],\left[\begin{array}{cc}\mathrm{-3}& \mathrm{-4}\\ 1& 1\end{array}\right]$$\stackrel{\text{select entry 2}}{\to }$$\left[\begin{array}{cc}\mathrm{-3}& \mathrm{-4}\\ 1& 1\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$P$

Diagonalize $A$ by applying $P$ 

$P^{-1}\.A\.P$ = $\left[\begin{array}{cc}3& 0\\ 0& 2\end{array}\right]$

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

Data entry

$\left[\begin{array}{rr}-1& -12\\ 1& 6\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$A$

Obtain the singular values

$A$ = $\left[\begin{array}{cc}\mathrm{-1}& \mathrm{-12}\\ 1& 6\end{array}\right]$$\stackrel{\text{singular values}}{\to }$$\left[\begin{array}{c}\frac{\sqrt{194}}{2}+\frac{\sqrt{170}}{2}\\ \frac{\sqrt{194}}{2}-\frac{\sqrt{170}}{2}\end{array}\right]$

From first principles

$A^{\mathrm{\%T}}\.A$ = $\left[\begin{array}{cc}2& 18\\ 18& 180\end{array}\right]$$\stackrel{\text{eigenvalues}}{\to }$$\left[\begin{array}{c}91+\sqrt{8245}\\ 91-\sqrt{8245}\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$V$

$\sqrt{{\mathbf{V}}_1}$ = $\frac{\sqrt{194}}{2}+\frac{\sqrt{170}}{2}$

$\sqrt{{\mathbf{V}}_2}$ = $\frac{\sqrt{194}}{2}-\frac{\sqrt{170}}{2}$

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

$\left[\begin{array}{rrr}5& -5& -4\\ -4& 8& 5\\ 7& -11& -7\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$A$

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

$\left[\begin{array}{rrr}5& -5& -4\\ -4& 8& 5\\ 7& -11& -7\end{array}\right]$$\stackrel{\text{Jordan form}}{\to }$$\left[\begin{array}{ccc}2& 1& 0\\ 0& 2& 1\\ 0& 0& 2\end{array}\right]$

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

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

$A-2$ = $\left[\begin{array}{ccc}3& \mathrm{-5}& \mathrm{-4}\\ \mathrm{-4}& 6& 5\\ 7& \mathrm{-11}& \mathrm{-9}\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$C$

$C$ = $\left[\begin{array}{ccc}3& \mathrm{-5}& \mathrm{-4}\\ \mathrm{-4}& 6& 5\\ 7& \mathrm{-11}& \mathrm{-9}\end{array}\right]$$\stackrel{\text{null space}}{\to }$$\left\{\left[\begin{array}{c}\frac{1}{2}\\ -\frac{1}{2}\\ 1\end{array}\right]\right\}$

$C^2$ = $\left[\begin{array}{ccc}1& \mathrm{-1}& \mathrm{-1}\\ \mathrm{-1}& 1& 1\\ 2& \mathrm{-2}& \mathrm{-2}\end{array}\right]$$\stackrel{\text{null space}}{\to }$$\left\{\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]\,\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]\right\}$

Select a vector in ${\mathrm{\ℝ}}^3$ that is not in the null space of $C^2$ and verify this choice

$⟨1\,0\,0⟩$$\stackrel{\text{assign to a name}}{\to }$$b_3$

(Non-vanishing of the determinant shows ${\mathbf{b}}_3$ is not a member of the null space of $C^2$)

$\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$$\stackrel{\text{determinant}}{\to }$$\mathrm{-1}$

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

$C\.{\mathbf{b}}_3$ = $\left[\begin{array}{c}3\\ \mathrm{-4}\\ 7\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$b_2$

(Note that ${\mathbf{b}}_1$ is an eigenvector.)

$C\.{\mathbf{b}}_2$ = $\left[\begin{array}{c}1\\ \mathrm{-1}\\ 2\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$b_1$

Construct the transition matrix whose columns are the vectors ${\mathbf{b}}_1\,{\mathbf{b}}_2\,{\mathbf{b}}_3$ 

$\left[{\mathbf{b}}_1\,{\mathbf{b}}_2\,{\mathbf{b}}_3\right]$ = $\left[\left[\begin{array}{c}1\\ \mathrm{-1}\\ 2\end{array}\right]\,\left[\begin{array}{c}3\\ \mathrm{-4}\\ 7\end{array}\right]\,\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\right]$$\stackrel{\text{combine into Matrix}}{\to }$$\left[\begin{array}{ccc}1& 3& 1\\ \mathrm{-1}& \mathrm{-4}& 0\\ 2& 7& 0\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$Q$

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

$Q^{-1}\.A\.Q$ = $\left[\begin{array}{ccc}2& 1& 0\\ 0& 2& 1\\ 0& 0& 2\end{array}\right]$

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

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

Convert to a linear system

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

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

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

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

$x+y+z=1,x-y-2z=3,5x+2y-7z=9,3x-7y+9z=\mathrm{-4}$

$\stackrel{\text{to Matrix form}}{\to }$

$\left[\begin{array}{ccc}1& 1& 1\\ 1& \mathrm{-1}& \mathrm{-2}\\ 5& 2& \mathrm{-7}\\ 3& \mathrm{-7}& 9\end{array}\right],\left[\begin{array}{c}1\\ 3\\ 9\\ \mathrm{-4}\end{array}\right]$

$\stackrel{\text{least squares}}{\to }$

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

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

Obtain the minimum-norm least-squares solution

$\left[\begin{array}{rrrr}7& 0& 5& -10\\ 1& -5& 1& 9\\ 10& -8& 11& -1\\ 4& -6& 9& 1\end{array}\right]\mathbf{\,}\left[\begin{array}{r}1\\ 2\\ 3\\ 4\end{array}\right]$$\stackrel{\text{least squares}}{\to }$$\left[\begin{array}{c}-\frac{10341}{75110}\\ -\frac{5434}{37555}\\ \frac{26981}{75110}\\ \frac{1847}{37555}\end{array}\right]$

Obtain the general solution

$\left[\begin{array}{rrrr}7& 0& 5& -10\\ 1& -5& 1& 9\\ 10& -8& 11& -1\\ 4& -6& 9& 1\end{array}\right]\mathbf{\,}\left[\begin{array}{r}1\\ 2\\ 3\\ 4\end{array}\right]$$\stackrel{\text{least squares}}{\to }$$\left[\begin{array}{c}{s}_1\\ 3s_1+\frac{139}{518}\\ \frac{7s_1}{5}+\frac{3574}{6475}\\ \frac{7s_1}{5}+\frac{3133}{12950}\end{array}\right]$$\stackrel{\text{evaluate at point}}{\to }$$\left[\begin{array}{c}s\\ 3s+\frac{139}{518}\\ \frac{7s}{5}+\frac{3574}{6475}\\ \frac{7s}{5}+\frac{3133}{12950}\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$X$

Obtain the norm and minimize it

$\mathbf{X}$

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

$\stackrel{\text{Euclidean-norm}}{\to }$

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

$\stackrel{\text{differentiate w.r.t. s}}{\to }$

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

$\stackrel{\text{equate to 0}}{\to }$

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

$\stackrel{\text{solve}}{\to }$

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

$\stackrel{\text{assign to a name}}{\to }$

$S$

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

 $\left[\begin{array}{rrrr}5& -6& -2& 3\\ -4& 3& -3& -1\\ 1& 1& 0& 2\end{array}\right]$$\stackrel{\text{row-reduced form}}{\to }$$\left[\begin{array}{cccc}1& 0& 0& \frac{59}{47}\\ 0& 1& 0& \frac{35}{47}\\ 0& 0& 1& -\frac{28}{47}\end{array}\right]$$\stackrel{\text{restrict columns}}{\to }$$\left[\begin{array}{c}\frac{59}{47}\\ \frac{35}{47}\\ -\frac{28}{47}\end{array}\right]$

Figure 3   Elementary row operations via the Context Panel system

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

 $\left[\begin{array}{rrr}1& 6& 5\\ -2& 2& 4\\ 5& 2& -6\end{array}\right]$$\stackrel{\text{LU decomposition}}{\to }$$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ \mathrm{-2}& 1& 0\\ 5& \mathrm{-2}& 1\end{array}\right],\left[\begin{array}{ccc}1& 6& 5\\ 0& 14& 14\\ 0& 0& \mathrm{-3}\end{array}\right]$

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

$\left[\begin{array}{rrr}1& 6& 5\\ -2& 2& 4\\ 5& 2& -6\end{array}\right]$$\stackrel{\text{QR decomposition}}{\to }$$\left[\begin{array}{ccc}\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}\right],\left[\begin{array}{ccc}\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}\right]$

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

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

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

$\left[\begin{array}{rrr}7& 4& 1\\ 4& 5& 3\\ 1& 3& 6\end{array}\right]$$\stackrel{\text{is positive definite?}}{\to }$$\mathrm{true}$

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

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

Data entry

$\left[\begin{array}{rrr}-5& -2& -2\\ 3& 4& -1\\ 4& -2& -8\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$A$

$\left[\begin{array}{rrr}-15& -1530& -2334\\ 11& 1433& 2191\\ -5& -931& -1427\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$B$

Test for similarity and find $C$ 

$A\,B$

$\left[\begin{array}{ccc}\mathrm{-5}& \mathrm{-2}& \mathrm{-2}\\ 3& 4& \mathrm{-1}\\ 4& \mathrm{-2}& \mathrm{-8}\end{array}\right],\left[\begin{array}{ccc}\mathrm{-15}& \mathrm{-1530}& \mathrm{-2334}\\ 11& 1433& 2191\\ \mathrm{-5}& \mathrm{-931}& \mathrm{-1427}\end{array}\right]$

$\stackrel{\text{is similar?}}{\to }$

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

$\stackrel{\text{select entry 2}}{\to }$

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

$\stackrel{\text{assign to a name}}{\to }$

$C$

Verify similarity

$C\.A$ = $\left[\begin{array}{ccc}\mathrm{-5}& \mathrm{-2}& \mathrm{-2}\\ \frac{2428078}{522111}& \frac{326169}{174037}& \frac{985481}{522111}\\ -\frac{1577354}{522111}& -\frac{212023}{174037}& -\frac{645562}{522111}\end{array}\right]$

$B\.C$ = $\left[\begin{array}{ccc}\mathrm{-5}& \mathrm{-2}& \mathrm{-2}\\ \frac{2428078}{522111}& \frac{326169}{174037}& \frac{985481}{522111}\\ -\frac{1577354}{522111}& -\frac{212023}{174037}& -\frac{645562}{522111}\end{array}\right]$

Construct a (nontrivial) 3×3 orthogonal matrix.

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

$\left[\left[\begin{array}{c}\mathrm{-4}\\ 1\\ 6\end{array}\right]\,\left[\begin{array}{c}5\\ 3\\ 1\end{array}\right]\,\left[\begin{array}{c}7\\ \mathrm{-8}\\ 9\end{array}\right]\right]$

$\stackrel{\text{Gram-Schmidt (normalized)}}{\to }$

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

$\stackrel{\text{combine into Matrix}}{\to }$

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

$\stackrel{\text{assign to a name}}{\to }$

$Q$

Verify that $Q$ is an orthogonal matrix

$Q$ = $\left[\begin{array}{ccc}-\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}\right]$$\stackrel{\text{is orthogonal?}}{\to }$$\mathrm{true}$

$Q^{\mathrm{\%T}}\.Q$ = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$Q\.Q^{\mathrm{\%T}}$ = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

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

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

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

Figure 4   The four fundamental subspaces of $A$

Data entry

 $\left[\begin{array}{rrr}32& -8& -4\\ -40& 10& 5\\ -32& 8& 4\\ 8& -2& -1\\ 72& -18& -9\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$A$

Row space of $A$

$A$ = $\left[\begin{array}{ccc}32& \mathrm{-8}& \mathrm{-4}\\ \mathrm{-40}& 10& 5\\ \mathrm{-32}& 8& 4\\ 8& \mathrm{-2}& \mathrm{-1}\\ 72& \mathrm{-18}& \mathrm{-9}\end{array}\right]$$\stackrel{\text{row space}}{\to }$$\left[\left[\begin{array}{ccc}1& -\frac{1}{4}& -\frac{1}{8}\end{array}\right]\right]$

Column space of $A$ 

$A$ = $\left[\begin{array}{ccc}32& \mathrm{-8}& \mathrm{-4}\\ \mathrm{-40}& 10& 5\\ \mathrm{-32}& 8& 4\\ 8& \mathrm{-2}& \mathrm{-1}\\ 72& \mathrm{-18}& \mathrm{-9}\end{array}\right]$$\stackrel{\text{column space}}{\to }$$\left[\left[\begin{array}{c}1\\ -\frac{5}{4}\\ \mathrm{-1}\\ \frac{1}{4}\\ \frac{9}{4}\end{array}\right]\right]$

Null space of $A$ 

$A$ = $\left[\begin{array}{ccc}32& \mathrm{-8}& \mathrm{-4}\\ \mathrm{-40}& 10& 5\\ \mathrm{-32}& 8& 4\\ 8& \mathrm{-2}& \mathrm{-1}\\ 72& \mathrm{-18}& \mathrm{-9}\end{array}\right]$$\stackrel{\text{null space}}{\to }$$\left\{\left[\begin{array}{c}\frac{1}{8}\\ 0\\ 1\end{array}\right]\,\left[\begin{array}{c}\frac{1}{4}\\ 1\\ 0\end{array}\right]\right\}$

Null space of $A^{\mathrm{T}}$ 

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

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

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

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

Figure 5   The four fundamental subspaces of $A$

Data entry

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