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  {\bf Solutions: Practise Exam 2}
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\bigskip {\bf 2.}

\medskip 1. False: every set of 5 or more vectors in $\mathbb{R}^4$ is
linearly dependent since the dimension of $\mathbb{R}^4$ is 4.

\medskip 2. False: a subspace has smaller ($\le$) dimension.

\medskip 3. False: the identity matrix is a counter example.

\medskip 4. True: $P(I_m + A)P^{-1} = I_m + PAP^{-1} = I_m + B$.

\medskip 5. True: the eigenvectors form a subspace.

\medskip 6. True: if $v$ is an eigenvector for the eigenvalues
$\lambda$ of $A$ then it is an eigenvector for the eigenvalues
$\lambda^m$ of $A^m$

\medskip 7. True: by the Rank Theorem the dimension of null space = n
- rank = 3 - 3 = 0.

\medskip 8. True: it is clear the rank of the matrix:
$$\begin{pmatrix}
  3&0&0&0\\
  0&3&0&0\\
  0&0&0&0\\
  0&0&0&0
\end{pmatrix}$$is 2 (this problem is slightly too hard).

\bigskip {\bf 3.} It is clear that the matrix is row equivalent to the
matrix
$$\begin{pmatrix}
  1&0&1&0\\
  0&1&3&0\\
  0&0&0&1
\end{pmatrix}$$thus $x_3$ is free and the null space is
spanned by the vector $(-1, -3, 1, 0)$. The column space is spanned by
the 3 vectors $(1, 0, 0), (0, 1, 0), (2, 0, 1)$. The rank of $A = 3$.

\bigskip {\bf 4.} The rank of $A = 9 - 4 = 5$ so there are 5 pivots
and the dimension of the row space is 5.

\bigskip {\bf 5.} A basis is given by the vectors $\mathcal{B} =
\{\vec{a}_1 = (1, 0, 0), \vec{a}_2 = (2, 1, 0), \vec{a}_3 = (0, 3,
1)\}$. It is clear that $[\vec{a}_1]_{\mathcal{B}} = (1, 0, 0),
[\vec{a}_2]_{\mathcal{B}} = (0, 1, 0), [\vec{a}_3]_{\mathcal{B}} =
(0, 0, 1)$. The vector $(1, 1, 0) = - \vec{a}_1 + \vec{a}_2$ so
$[(1, 1, 0)]_{\mathcal{B}} = (-1, 1, 0)$.

\bigskip {\bf 6.} (a) It is clear that
$$\begin{bmatrix}
  1&1\\
  1&-2
\end{bmatrix} \sim \begin{bmatrix}
  1&1\\
  0&-3
\end{bmatrix}$$ so the two column vectors are linearly
independent. (b) $\vec{x} = 2 (1, 1) + (1, -2) = (3, 0)$. (c) $(2,
1) = 5/3(1, 1) + 1/3(1, -2)$ thus $[(2, 1)]_{_{\mathcal{B}}} =
(5/3, 1/3)$.

\bigskip {\bf 7.} (a) $[{\bf p}]_{\mathcal{E}} = [1, 1, 1]^T, [{\bf
  q}]_{\mathcal{E}} = [1, 2, 3]^T$ (b) the two vectors are linearly
independent because $[1, 1, 1]^T, [1, 2, 3]^T$ are not multiple of
each other.

\bigskip {\bf 8.} $A = PBP^{-1}, B = QCQ^{-1}$ so $A = PQCQ^{-1}P^{-1}
= (PQ)C(PQ)^{-1}$.

\bigskip {\bf 9.} This is a Theorem in the textbook.

\bigskip {\bf 10.} The characteristic polynomial is $(3 - \lambda)(2 -
\lambda) - 4 = \lambda^2 - 5 \lambda + 2$. The two eigenvalues are
$\lambda_1 = (5 + \sqrt{17})/2, \lambda_2 = (5 - \sqrt{17})/2$. An
eigenvector of $\lambda_1$ is obtained by taking a solution of the
matrix
$$\begin{bmatrix}\frac{1-\sqrt{17}}{2}&2\\
  2&\frac{-1-\sqrt{17}}{2}\end{bmatrix}\begin{bmatrix}x_1\\
  x_2\end{bmatrix} = \begin{bmatrix}0\\
  0\end{bmatrix}$$
for example,
$$\begin{bmatrix}x_1\\
  x_2\end{bmatrix} = \begin{bmatrix}\frac{1 + \sqrt{17}}{4}\\
  1\end{bmatrix}.$$
An eigenvector of $\lambda_2$ is obtained by
taking a solution of the matrix
$$\begin{bmatrix}\frac{1+\sqrt{17}}{2}&2\\
  2&\frac{-1+\sqrt{17}}{2}\end{bmatrix}\begin{bmatrix}x_1\\
  x_2\end{bmatrix} = \begin{bmatrix}0\\
  0\end{bmatrix}$$
for example,
$$\begin{bmatrix}x_1\\
  x_2\end{bmatrix} = \begin{bmatrix}1\\
  -\frac{1 + \sqrt{17}}{4}\end{bmatrix}.$$The matrix $P$ is formed by
taking these 2 eigenvectors as columns:
$$P = \begin{bmatrix}\frac{1 + \sqrt{17}}{4}&1\\1& -\frac{1 +
    \sqrt{17}}{4}\end{bmatrix}.$$  (The numbers here are fairly bad
and this problem would not OK for an exam.)

\bigskip {\bf 11.} (a) The condition
$$c_1 E_{11} + c_2E_{12} + c_3 E_{21} + c_4E_{22} = \vec{0}$$
is
equivalent to the condition
$$\begin{bmatrix}c_1&c_2\\c_3& c_4\end{bmatrix} =
\begin{bmatrix}0&0\\0& 0\end{bmatrix}$$i.e., $c_1 = c_2 = c_3 = c_4 =
0$ so $\dim V = 4$.

\medskip (b) $T(M_1 + M_2) = A(M_1 + M_2) - (M_1 + M_2) = AM_1 - M_1 +
AM_2 - M_2 = T(M_1) + T(M_2); T(cM) = A(cM) - cM = c(AM - M) = cT(M).$

\medskip (c) The matrix of $T$ is obtaind by consider the image of the
basis under $T$:
$$T(E_{11}) = \begin{bmatrix}0&-1\\0& -2\end{bmatrix} = - E_{12} -
2E_{22},$$
$$
T(E_{12}) =
\begin{bmatrix}1&0\\0& -2\end{bmatrix} = E_{11} - 2 E_{22}
$$
$$
T(E_{21}) =
\begin{bmatrix}0&-1\\2& -2\end{bmatrix} = -E_{12} + 2E_{21} -
2 E_{22}
$$
$$
T(E_{22}) =
\begin{bmatrix}-1&0\\0& 0\end{bmatrix} = - E_{11}.
$$Thus the matrix of $T$ is given by
$$\begin{bmatrix}0&-1&0&-1\\1&
  0&-1&0\\0&0&2&0\\-2&-2&-2&0\end{bmatrix}.
$$

\medskip (d) The matrix is invertible so the null space is trivial.

\bigskip {\bf 12.} (a) and (b) Obviously we have to exclude $h = 0$.
Next we find the null space of the matrix
$$A - I_4 = \begin{bmatrix}0&1&1/h&1\\0&
  1&h&1\\0&0&0&1\\0&0&0&2\end{bmatrix} \sim
\begin{bmatrix}0&1&1/h&0\\0&
  0&h-1/h&0\\0&0&0&1\\0&0&0&0\end{bmatrix}$$thus we see that the
dimension of the column space is 3 unless $h = 1$ and for $h = 1$ the
dimension of the column space is 2. By Rank Theorem we conclude that
the dimension of the null space is 2 if and only if $h = 1$ (and if $h
\ne 1$ then the dimension of the null space is 1). If $h = 1$ then the
matrix is diagonalizable because we then get 4 linearly independent
eigenvectors, 2 for the eigenvalue 1 and one each of the eigenvalues 2
and 3. (c) for $\lambda = 3$ we consider the null space of the matrix
$$A - 3 I_4 = \begin{bmatrix}-2&1&1/h&1\\0&
  -1&h&1\\0&0&-2&1\\0&0&0&0\end{bmatrix} \sim
\begin{bmatrix}-2&0&0&2+ (1/h + h)/2\\0&
  -1&0&1+h/2\\0&0&-2&1\\0&0&0&0\end{bmatrix}$$so $x_4$ is a free
variable and so $[1 + (1/h + h)/4, 1 + h/2, 1/2, 1]^T$ is an
eigenvector.


\bigskip {\bf 13.} (a) $A\vec{b}_1 = [\sqrt{5}, 5 + 2 \sqrt{5}]^T =
\sqrt{5} \vec{b}_1$ so $[A\vec{b}_1]_{\mathcal{B}} = [\sqrt{5}, 0]^T.$
$A\vec{b}_2 = [-\sqrt{5}, 5 - 2 \sqrt{5}]^T = - \sqrt{5} \vec{b}_2$ so
$[A\vec{b}_2]_{\mathcal{B}} = [0, -\sqrt{5}]^T.$ (b) the matrix
representing the linear transformation relative to the basis
$\mathcal{B}$ is given by
$$\begin{bmatrix}\sqrt{5}&0\\0&-\sqrt{5}\end{bmatrix}$$
(c) this means
we are solving the equation
$$\begin{bmatrix}\sqrt{5}&0\\0&-\sqrt{5}\end{bmatrix}
\begin{bmatrix}c_1\\c_2\end{bmatrix} =
\begin{bmatrix}1\\2\end{bmatrix}$$and the solution is
clearly $c_1 = 1/\sqrt{5}, c_2 = - 2/\sqrt{5}$.

\bigskip {\bf 14.} The 3 given vectors are linearly independent which
shows that the column space is 3 dimension, i.e., the rank of the
matrix is 3 hence the dimension of the null space is 0 because $\dim
nul(A) + 3 = 3$.



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