Solutions

1.
The projection matrix P onto the plane in $\Bbb$R3 spanned by the vectors u = $\left[\vphantom{
1,1,0}\right.$1, 1, 0$\left.\vphantom{
1,1,0}\right]$ and v = $\left[\vphantom{ 1,1,1}\right.$1, 1, 1$\left.\vphantom{ 1,1,1}\right]$ is the product P = A$\left(\vphantom{
A^{T}A}\right.$ATA$\left.\vphantom{
A^{T}A}\right)^{{-1}}_{}$AT, where u and v are the columns of A. BITMAPSETProbSolvHint0.2006in0.243in0inq1

Note that Pw is a linear combination of u and v for any vector w = $\left(\vphantom{ x,y,z}\right.$x, y, z$\left.\vphantom{ x,y,z}\right)$ in $\Bbb$R3, so P maps $\Bbb$R3 onto the plane spanned by u and v.

To find a nonzero vector b that is projected to zero, leave the insertion point in the matrix P, and from the Matrices submenu choose Nullspace Basis.

2.
The matrix $\left[\vphantom{
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & \;0 & -3 & \;2
\end{array}
}\right.$$\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & \;0 & -3 & \;2
\end{array}$$\left.\vphantom{
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & \;0 & -3 & \;2
\end{array}
}\right]$ has characteristic polynomial $\left(\vphantom{ X-2}\right.$X - 2$\left.\vphantom{ X-2}\right)^{{4}}_{}$, minimum polynomial 4 - 4X + X2 =  $\left(\vphantom{ X-2}\right.$X - 2$\left.\vphantom{ X-2}\right)^{{2}}_{}$, and eigenvalue and eigenvectors BITMAPSETProbSolvHint0.2006in0.243in0inq2

$\displaystyle \left\{\vphantom{ 2,4,\left[
\begin{array}{c}
0 \\
0 \\
...
...t[
\begin{array}{c}
0 \\
1 \\
0 \\
0
\end{array}
\right] }\right.$2, 4,$\displaystyle \left[\vphantom{
\begin{array}{c}
0 \\
0 \\
0 \\
1
\end{array}
}\right.$$\displaystyle \begin{array}{c}
0 \\
0 \\
0 \\
1
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
0 \\
0 \\
0 \\
1
\end{array}
}\right]$,$\displaystyle \left[\vphantom{
\begin{array}{c}
0 \\
1 \\
0 \\
0
\end{array}
}\right.$$\displaystyle \begin{array}{c}
0 \\
1 \\
0 \\
0
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
0 \\
1 \\
0 \\
0
\end{array}
}\right]$$\displaystyle \left.\vphantom{ 2,4,\left[
\begin{array}{c}
0 \\
0 \\
0...
...[
\begin{array}{c}
0 \\
1 \\
0 \\
0
\end{array}
\right] }\right\}$

The minimal polynomial is a factor of the characteristic polynomial. The eigenvalue 2 occurs with multiplicity 4 as a root of the characteristic polynomial $\left(\vphantom{ X-2}\right.$X - 2$\left.\vphantom{ X-2}\right)^{{4}}_{}$. The eigenvalue 2 has two linearly independent eigenvectors. Note that

 

3.
The solutions of this equation are in $\Bbb$R3, and the column space of A is a subset of $\Bbb$R2, so these solutions cannot be the column space of A. They do form the nullspace of A by the definition of nullspace; consequently, this set is a subspace of $\Bbb$R3. The product of A with the first row of A is $\left[\vphantom{
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 2
\end{array}
}\right.$$\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 2
\end{array}$$\left.\vphantom{
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 2
\end{array}
}\right]$$\left[\vphantom{
\begin{array}{c}
1 \\
1 \\
1
\end{array}
}\right.$$\begin{array}{c}
1 \\
1 \\
1
\end{array}$$\left.\vphantom{
\begin{array}{c}
1 \\
1 \\
1
\end{array}
}\right]$ =  $\left[\vphantom{
\begin{array}{c}
3 \\
3
\end{array}
}\right.$$\begin{array}{c}
3 \\
3
\end{array}$$\left.\vphantom{
\begin{array}{c}
3 \\
3
\end{array}
}\right]$, which is not $\left[\vphantom{
\begin{array}{c}
0 \\
0
\end{array}
}\right.$$\begin{array}{c}
0 \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
0 \\
0
\end{array}
}\right]$, so the solution set is not the row space of A.BITMAPSETProbSolvHint0.2006in0.243in0inq3

To determine whether this subspace of $\Bbb$R3 is a point, line, or plane, solve the system of equations: from the Solve submenu, choose Exact

$\displaystyle \left[\vphantom{
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 2
\end{array}
}\right.$$\displaystyle \begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 2
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 2
\end{array}
}\right]$$\displaystyle \left[\vphantom{
\begin{array}{c}
x_{1} \\
x_{2} \\
x_{3}
\end{array}
}\right.$$\displaystyle \begin{array}{c}
x_{1} \\
x_{2} \\
x_{3}
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
x_{1} \\
x_{2} \\
x_{3}
\end{array}
}\right]$ = $\displaystyle \left[\vphantom{
\begin{array}{c}
0 \\
0
\end{array}
}\right.$$\displaystyle \begin{array}{c}
0 \\
0
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
0 \\
0
\end{array}
}\right]$, Solution is: $\displaystyle \left[\vphantom{
\begin{array}{r}
-2t_{1} \\
t_{1} \\
t_{1}
\end{array}
}\right.$$\displaystyle \begin{array}{r}
-2t_{1} \\
t_{1} \\
t_{1}
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{r}
-2t_{1} \\
t_{1} \\
t_{1}
\end{array}
}\right]$

The subspace is a line. It is the line that passes through the origin and the point $\left[\vphantom{
\begin{array}{ccc}
-2 & 1 & 1
\end{array}
}\right.$$\begin{array}{ccc}
-2 & 1 & 1
\end{array}$$\left.\vphantom{
\begin{array}{ccc}
-2 & 1 & 1
\end{array}
}\right]$.

4.
Apply the following operations in turn.BITMAPSETProbSolvHint0.2006in0.243in0inq4

This result proves the first identity. For the second part, carry out the following steps.