-
State the matrix product to obtain 3 linear combinations of vectors
\(\displaystyle \boldsymbol{u}= \left[ \begin{array}{l}
1\\
0\\
1
\end{array} \right], \boldsymbol{v}= \left[ \begin{array}{l}
-
1\\
0\\
1
\end{array} \right],\)
with scaling coefficients \((\alpha_1, \beta_1) = (1, 1)\),
\((\alpha_2, \beta_2) = (- 1, 1)\), \((\alpha_3, \beta_3) = (1, -
1)\).
Solution. The matrix product is
\(\boldsymbol{C}=\boldsymbol{A}\boldsymbol{B}\), \(\boldsymbol{C}
\in \mathbb{R}^{3 \times 3}\) (3 linear combinations, each with 3
components), \(\boldsymbol{A}= \left[ \begin{array}{ll}
\boldsymbol{u} & \boldsymbol{v}
\end{array} \right] \in
\mathbb{R}^{3 \times 2}\) (vectors entering into linear
combination), \(\boldsymbol{B} \in \mathbb{R}^{2 \times 3}\)
(scaling coefficients of each linear combination)
\(\displaystyle \boldsymbol{A}= \left[ \begin{array}{ll}
1 & -
1\\
0 & 0\\
1 & 1
\end{array} \right], \boldsymbol{B}= \left[
\begin{array}{lll}
1 & - 1 & 1\\
1 & 1 & - 1
\end{array}
\right] .\)
-
Orthonormalize the vectors
\(\displaystyle \boldsymbol{u}= \left[ \begin{array}{l}
1\\
0\\
1
\end{array} \right], \boldsymbol{v}= \left[ \begin{array}{l}
-
1\\
0\\
1
\end{array} \right], \boldsymbol{w}= \left[
\begin{array}{l}
1\\
1\\
- 1
\end{array} \right] .\)
Solution. Note that \(\boldsymbol{u}^T
\boldsymbol{v}=\boldsymbol{u}^T \boldsymbol{w}= 0\), such that these
vector pairs are already orthogonal. Also note that the vector
\(\boldsymbol{e}_2 = \left[ \begin{array}{lll}
0 & 1 &
0
\end{array} \right]^T\) is orthogonal to both \(\boldsymbol{u}\)
and \(\boldsymbol{v}\). Scale vectors to have unit norm
\(\displaystyle \boldsymbol{Q}= \left[ \begin{array}{lll}
\boldsymbol{q}_1 & \boldsymbol{q}_2 & \boldsymbol{q}_3
\end{array}
\right] = \left[ \begin{array}{lll}
\boldsymbol{u}/ \sqrt{2} &
\boldsymbol{v}/ \sqrt{2} & \boldsymbol{e}_2
\end{array} \right]
.\)
-
For \(x, y \in \mathbb{R}\), expansion of \((x - y)^3\) leads to
\((x - y)^3 = x^3 - 3 x^2 y + 3 x y^2 - y^3\). Find the
corresponding expansion of \((\boldsymbol{A}-\boldsymbol{B})^3\) for
\(\boldsymbol{A}, \boldsymbol{B} \in \mathbb{R}^{m \times m}\).
Solution. Compute, recalling that matrix
multiplication is not commutative
\(\displaystyle (\boldsymbol{A}-\boldsymbol{B})^3 =
(\boldsymbol{A}-\boldsymbol{B})
(\boldsymbol{A}^2
-\boldsymbol{A}\boldsymbol{B}-\boldsymbol{B}\boldsymbol{A}+\boldsymbol{B}^2)
=\boldsymbol{A}^3
-\boldsymbol{A}^2
\boldsymbol{B}-\boldsymbol{A}\boldsymbol{B}\boldsymbol{A}+\boldsymbol{A}\boldsymbol{B}^2
-\boldsymbol{B}\boldsymbol{A}^2
+\boldsymbol{B}\boldsymbol{A}\boldsymbol{B}+\boldsymbol{B}^2
-\boldsymbol{B}^3
.\)
-
Find the projection of \(\boldsymbol{b}\) onto \(C
(\boldsymbol{A})\) for
\(\displaystyle \boldsymbol{b}= \left[ \begin{array}{l}
1\\
2\\
3
\end{array} \right], \boldsymbol{A}= \left[ \begin{array}{ll}
1
& - 1\\
1 & 0\\
1 & 1
\end{array} \right] .\)
Solution. With \(\boldsymbol{A}= \left[
\begin{array}{ll}
\boldsymbol{a}_1 & \boldsymbol{a}_2
\end{array}
\right]\) note that \(\boldsymbol{b}= 2\boldsymbol{a}_1
+\boldsymbol{a}_2\), hence \(\boldsymbol{b} \in C
(\boldsymbol{A})\), and the projection of \(\boldsymbol{b}\) onto
\(C (\boldsymbol{A})\) is \(\boldsymbol{b}\) itself.
Alternatively, orthonormalize \(\boldsymbol{A}= \left[
\begin{array}{ll}
\boldsymbol{a}_1 & \boldsymbol{a}_2
\end{array}
\right]\) to obtain
\(\displaystyle \boldsymbol{Q}= \left[ \begin{array}{ll}
\boldsymbol{q}_1 & \boldsymbol{q}_2
\end{array} \right] = \left[
\begin{array}{ll}
\boldsymbol{a}_1 / \sqrt{3} & \boldsymbol{a}_2 /
\sqrt{2}
\end{array} \right] .\)
The projection onto \(C (\boldsymbol{A})\) is
\(\displaystyle \boldsymbol{c}=\boldsymbol{Q}\boldsymbol{Q}^T
\boldsymbol{b}=\boldsymbol{Q}
(\boldsymbol{Q}^T \boldsymbol{b}) =
\left[ \begin{array}{ll}
1 / \sqrt{3} & - 1 / \sqrt{2}\\
1 /
\sqrt{3} & 0\\
1 / \sqrt{3} & 1 / \sqrt{2}
\end{array} \right]
\left( \left[ \begin{array}{lll}
1 / \sqrt{3} & 1 / \sqrt{3} & 1 /
\sqrt{3}\\
- 1 / \sqrt{2} & 0 & 1 / \sqrt{2}
\end{array} \right]
\left[ \begin{array}{l}
1\\
2\\
3
\end{array} \right] \right)
\Rightarrow\)
\(\displaystyle \boldsymbol{c}= \left[ \begin{array}{ll}
1 /
\sqrt{3} & - 1 / \sqrt{2}\\
1 / \sqrt{3} & 0\\
1 / \sqrt{3} &
1 / \sqrt{2}
\end{array} \right] \left[ \begin{array}{l}
2
\sqrt{3}\\
\sqrt{2}
\end{array} \right] = \left[
\begin{array}{l}
1\\
2\\
3
\end{array} \right] .\)
-
Find the \(L U\) decomposition of
\(\displaystyle \boldsymbol{A}= \left[ \begin{array}{lll}
1 & 1 &
1\\
2 & 3 & 3\\
3 & 5 & 6
\end{array} \right] .\)
Solution. Carry out reduction to upper triangular
form, noting multipliers used in the process
\(\displaystyle \boldsymbol{L}_1 \boldsymbol{A}= \left[
\begin{array}{lll}
1 & 0 & 0\\
- 2 & 1 & 0\\
- 3 & 0 &
1
\end{array} \right] \left[ \begin{array}{lll}
1 & 1 & 1\\
2 &
3 & 3\\
3 & 5 & 6
\end{array} \right] = \left[ \begin{array}{lll}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 2 & 3
\end{array} \right] .\)
\(\displaystyle \boldsymbol{L}_2 \boldsymbol{L}_1 \boldsymbol{A}=
\left[ \begin{array}{lll}
1 & 0 & 0\\
0 & 1 & 0\\
0 & - 2 &
1
\end{array} \right] \left[ \begin{array}{lll}
1 & 1 & 1\\
0 &
1 & 1\\
0 & 2 & 3
\end{array} \right] = \left[ \begin{array}{lll}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1
\end{array} \right]
=\boldsymbol{U}.\)
Find \(\boldsymbol{A}=\boldsymbol{L}_1^{- 1} \boldsymbol{L}_2^{-
1}
\boldsymbol{U}=\boldsymbol{L}\boldsymbol{U}\). Compute
\(\displaystyle \boldsymbol{L}=\boldsymbol{L}_1^{- 1}
\boldsymbol{L}_2^{- 1} = \left[
\begin{array}{lll}
1 & 0 & 0\\
2
& 1 & 0\\
3 & 0 & 1
\end{array} \right] \left[ \begin{array}{lll}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 2 & 1
\end{array} \right] = \left[
\begin{array}{lll}
1 & 0 & 0\\
2 & 1 & 0\\
3 & 2 &
1
\end{array} \right] .\)
Verify
\(\displaystyle \boldsymbol{L}\boldsymbol{U}= \left[
\begin{array}{lll}
1 & 0 & 0\\
2 & 1 & 0\\
3 & 2 &
1
\end{array} \right] \left[ \begin{array}{lll}
1 & 1 & 1\\
0 &
1 & 1\\
0 & 0 & 1
\end{array} \right] = \left[ \begin{array}{lll}
1 & 1 & 1\\
2 & 3 & 3\\
3 & 5 & 6
\end{array} \right] .
\checked\)
-
State the eigenvalues and eigenvectors of \(\boldsymbol{R} \in
\mathbb{R}^{2 \times 2}\), the matrix describing reflection across
the vector \(\boldsymbol{w}= \left[ \begin{array}{ll}
1 &
2
\end{array} \right]^T\).
Solution. From eigenvalue relation
\(\boldsymbol{R}\boldsymbol{x}= \lambda \boldsymbol{x}\) note that
directions not changed by reflection are along \(\boldsymbol{w}\)
and orthogonal to \(\boldsymbol{w}\). Deduce
\(\displaystyle \boldsymbol{x}_1 =\boldsymbol{w}= \left[
\begin{array}{l}
1\\
2
\end{array} \right], \lambda_1 = 1\)
\(\displaystyle \boldsymbol{x}_2 =\boldsymbol{w}= \left[
\begin{array}{l}
- 2\\
1
\end{array} \right], \lambda_2 = -
1.\)
-
Compute the eigendecomposition of
\(\displaystyle \boldsymbol{A}= \left[ \begin{array}{lll}
5 / 2 &
0 & 1 / 2\\
0 & 1 & 0\\
1 / 2 & 0 & 5 / 2
\end{array} \right]
.\)
Solution. Eigenvalue problem is
\(\boldsymbol{A}\boldsymbol{x}= \lambda \boldsymbol{x}\). Observe
that \(\boldsymbol{x}_2 =\boldsymbol{e}_2 = \left[
\begin{array}{lll}
0 & 1 & 0
\end{array} \right]^T\), \(\lambda_2
= 1\) is an eigenvector, value pair. Compute characteristic
polynomial by
\(\displaystyle p (\lambda) = \det (\lambda
\boldsymbol{I}-\boldsymbol{A}) = \left|
\begin{array}{lll}
\lambda
- 5 / 2 & 0 & - 1 / 2\\
0 & \lambda - 1 & 0\\
- 1 / 2 & 0 &
\lambda - 5 / 2
\end{array} \right| = (\lambda - 1) \left|
\begin{array}{ll}
\lambda - 5 / 2 & - 1 / 2\\
- 1 / 2 & \lambda
- 5 / 2
\end{array} \right| \Rightarrow\)
\(\displaystyle p (\lambda) = (\lambda - 1) \left( \lambda^2 - 5
\lambda + \left( \frac{5}{2}
\right)^2 - \left( \frac{1}{2}
\right)^2 \right) = (\lambda - 1) (\lambda^2 -
5 \lambda + 6) =
(\lambda - 1) (\lambda - 2) (\lambda - 3) .\)
The other eigevalues are \(\lambda_1 = 2\), \(\lambda_3 = 3\). Find
eigenvectors by computing bases for eigenspaces \(N (\boldsymbol{A}-
\lambda_1 \boldsymbol{I})\) and \(N (\boldsymbol{A}- \lambda_3
\boldsymbol{I})\).
\(\displaystyle \boldsymbol{A}- \lambda_1 \boldsymbol{I}= \left[
\begin{array}{lll}
1 / 2 & 0 & 1 / 2\\
0 & 1 & 0\\
1 / 2 & 0 &
1 / 2
\end{array} \right] \sim \left[ \begin{array}{lll}
1 / 2 & 0
& 1 / 2\\
0 & 1 & 0\\
0 & 0 & 0
\end{array} \right] \Rightarrow
\boldsymbol{x}_1 = \left[ \begin{array}{l}
1\\
0\\
-
1
\end{array} \right]\)
\(\displaystyle \boldsymbol{A}- \lambda_3 \boldsymbol{I}= \left[
\begin{array}{lll}
- 1 / 2 & 0 & 1 / 2\\
0 & 1 & 0\\
1 / 2 &
0 & - 1 / 2
\end{array} \right] \sim \left[ \begin{array}{lll}
-
1 / 2 & 0 & 1 / 2\\
0 & 1 & 0\\
0 & 0 & 0
\end{array} \right]
\Rightarrow \boldsymbol{x}_3 = \left[ \begin{array}{l}
1\\
0\\
1
\end{array} \right] .\)
-
Find the SVD of
\(\displaystyle \boldsymbol{A}= \left[ \begin{array}{ll}
1 & -
1\\
0 & 0\\
1 & 1
\end{array} \right] .\)
Solution. Matrix has orthogonal columns that are
not of unit norm. Construct SVD as
\(\displaystyle \boldsymbol{A}= \left[ \begin{array}{ll}
1 /
\sqrt{2} & - 1 / \sqrt{2}\\
0 & 0\\
1 / \sqrt{2} & 1 /
\sqrt{2}
\end{array} \right] \left[ \begin{array}{ll}
\sqrt{2} &
0\\
0 & \sqrt{2}\\
0 & 0
\end{array} \right] \left[
\begin{array}{ll}
1 & 0\\
0 & 1
\end{array} \right] .\)
-
Find the matrix of the reflection of \(\mathbb{R}^2\) vectors across
the vector \(\boldsymbol{u}= \left[ \begin{array}{ll}
1 &
2
\end{array} \right]^T\).
Solution. Let \(\boldsymbol{w}\) be the reflection
of \(\boldsymbol{v}\) across \(\boldsymbol{u}\),
\(\boldsymbol{w}=\boldsymbol{R}\boldsymbol{v}\). Let
\(\displaystyle \boldsymbol{q}= \frac{1}{\| \boldsymbol{u} \|}
\boldsymbol{u}=
\frac{1}{\sqrt{5}} \left[ \begin{array}{l}
1\\
0
\end{array} \right] .\)
The projection of \(\boldsymbol{v}\) onto the direction of
\(\boldsymbol{u}\) is
\(\boldsymbol{z}=\boldsymbol{q}\boldsymbol{q}^T \boldsymbol{v}\).
The travel from \(\boldsymbol{v}\) to \(\boldsymbol{z}\) is
\(\boldsymbol{z}-\boldsymbol{v}\)
\(\displaystyle \boldsymbol{z}=\boldsymbol{v}+
(\boldsymbol{z}-\boldsymbol{v}) .\)
The reflection is obtained by doubling the travel distance
\(\displaystyle \boldsymbol{w}=\boldsymbol{v}+ 2
(\boldsymbol{z}-\boldsymbol{v}) =
2\boldsymbol{z}-\boldsymbol{v}=
(2\boldsymbol{q}\boldsymbol{q}^T
-\boldsymbol{I}) \boldsymbol{v}.\)
Deduce that the reflection matrix is
\(\displaystyle \boldsymbol{R}= 2\boldsymbol{q}\boldsymbol{q}^T
-\boldsymbol{I}.\)
-
Find bases for the four fundamental spaces of
\(\displaystyle \boldsymbol{A}= \left[ \begin{array}{ll}
1 & -
1\\
0 & 0\\
1 & 1
\end{array} \right] .\)
Solution. With \(\boldsymbol{A}= \left[
\begin{array}{ll}
\boldsymbol{a}_1 & \boldsymbol{a}_2
\end{array}
\right] \in \mathbb{R}^{3 \times 2}\), the bases are:
\(C (\boldsymbol{A})\): \(\{ \boldsymbol{a}_1, \boldsymbol{a}_2 \}\)
\(N (\boldsymbol{A}^T)\): \(\{ \boldsymbol{e}_2 \}\) since
\(\boldsymbol{e}_2\) is orthogonal to both \(\boldsymbol{a}_1\) and
\(\boldsymbol{a}_2\)
\(C (\boldsymbol{A}^T)\): \(\{ \boldsymbol{a}^T_1,
\boldsymbol{a}^T_2 \}\)
\(N (\boldsymbol{A})\): \(\{ \boldsymbol{0} \}\)
