0.1 Solutions for Note 2

1.

(a)

Consider the following computation. \begin{align*} A - 2B &= \begin {bmatrix} 1 & 2 \\ 2 & 0 \end {bmatrix} - 2\begin {bmatrix} -1 & 2 \\ 0 & 3 \end {bmatrix} = \begin {bmatrix} 1 & 2 \\ 2 & 0 \end {bmatrix} + \begin {bmatrix} 2 & -4 \\ 0 & -6 \end {bmatrix} \\ &= \begin {bmatrix} 3 & -2 \\ 2 & -6 \end {bmatrix} \end{align*}

(b)

Consider the following equation. \[ AB = \begin {bmatrix} 1 & 2 \\ 2 & 0 \end {bmatrix} \begin {bmatrix} -1 & 2 \\ 0 & 3 \end {bmatrix} = \begin {bmatrix} -1 & 8 \\ -2 & 7 \end {bmatrix} \]

(c)

Because \(AB\) is found above, \(BA\) could be found. \[ BA = \begin {bmatrix} -1 & 2 \\ 0 & 3 \end {bmatrix} \begin {bmatrix} 1 & 2 \\ 2 & 0 \end {bmatrix} = \begin {bmatrix} 3 & -2 \\ 6 & 0 \end {bmatrix} \] Therefore, the following equation is true. \begin{align*} (AB - BA)^\intercal &= \left ( \begin {bmatrix} -1 & 8 \\ -2 & 7 \end {bmatrix} - \begin {bmatrix} 3 & -2 \\ 6 & 0 \end {bmatrix} \right )^\intercal = \left ( \begin {bmatrix} -4 & 10 \\ -8 & 7 \end {bmatrix} \right )^\intercal \\ &= \begin {bmatrix} -4 & -8 \\ 10 & 7 \end {bmatrix} \end{align*}

2.

By the definition of matrix multiplication, \(a_2B = [4, 5, 6]\).

3.

Notice that \(A\) is a \(2 \times 2\) square matrix. Therefore, define \(A\) as the following. \[ A = \begin {bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end {bmatrix} \] Therefore, the following equation is obtained. \begin{align*} 2A - 3A^\intercal &= \begin {bmatrix} 2a_{11} & 2a_{12} \\ 2a_{21} & 2a_{22} \end {bmatrix} - \begin {bmatrix} 3a_{11} & 3a_{21} \\ 3a_{12} & 3a_{22} \end {bmatrix} \\ &= \begin {bmatrix} 2a_{11} - 3a_{11} & 2a_{12} - 3a_{21} \\ 2a_{21} - 3a_{12} & 2a_{22} - 3a_{22} \end {bmatrix} = \begin {bmatrix} -1 & -5 \\ 0 & -4 \end {bmatrix} \end{align*}

Continuing, the elements of \(A\) could be found. \begin{align*} 2a_{11} - 3a_{11} &= -1 \\ 2a_{12} - 3a_{21} &= -5 \\ 2a_{21} - 3a_{12} &= 0 \\ 2a_{22} - 3a_{22} &= -4 \end{align*}

Therefore, \(a_{11} = 1\), \(a_{12} = 2\), \(a_{21} = 3\), and \(a_{22} = 4\). Thus, \(A\) can be defined as the following. \[ A = \begin {bmatrix} 1 & 2 \\ 3 & 4 \end {bmatrix} \]

4.

For any square matrix \(S\), define \(A\) and \(B\) as the following. \[ A = \frac {S + S^\intercal }{2}, \quad B = \frac {S - S^\intercal }{2} \] Notice that \(S = A + B\). Therefore, it suffices to show that \(A\) is symmetric and \(B\) is skew-symmetric. By Theorem ??, the following equations hold. \[ A^\intercal = \frac {S + S^\intercal }{2}, \quad B^\intercal = \frac {S^\intercal - S}{2} = -B \] Thus, all square matrices can be represented as the sum of a symmetric and skew-symmetric matrices.

5.

By definition, there exist \(n, m \in \mathbb {Z}\) such that \(A^n = 0\) and \(B^m = 0\). Because \(AB = BA\), \(A^{nm} B^{nm} = 0\) can be written as \((AB)^{nm} = 0\) and \(AB\) is nilpotent.

6.

By Theorem ??, \((AB + (AB)^\intercal )^\intercal = (AB)^\intercal + AB = AB + (AB)^\intercal \). Therefore, \(AB + (AB)^\intercal \) is symmetric.

7.

An easy trap for this problem is that you might be tempted to do \(\det (A^2 - B^2) = \det (A + B) \det (A - B)\). However, notice that this does not hold in general as \(AB \neq BA\) in general.

Consider the following matrix \(P\) and its inverse \(P^{-1}\). \[ P = \begin {bmatrix} I & I \\ I & -I \end {bmatrix}, \quad P^{-1} = \frac {1}{2} \begin {bmatrix} I & I \\ I & -I \end {bmatrix} \] Continuing, \begin{align*} P^{-1} \begin {bmatrix} A & B \\ B & A \end {bmatrix} P &= \frac {1}{2} \begin {bmatrix} A + B & B + A \\ A - B & B - A \end {bmatrix} \begin {bmatrix} I & I \\ I & -I \end {bmatrix} \\ &= \begin {bmatrix} A + B & 0 \\ 0 & A - B \end {bmatrix} \end{align*}

holds and \(\begin {bmatrix} A & B \\ B & A \end {bmatrix} \cong \begin {bmatrix} A + B & 0 \\ 0 & A - B \end {bmatrix}\). Therefore, the determinant of the original matrix is \(\det (A + B) \det (A - B)\).

8.

By definition, \(A\) is skew-symmetric if \(A^\intercal = -A\). Notice that \((A^3)^\intercal = (A^\intercal )^3\) by Theorem ??. Therefore, \((A^3)^\intercal = (A^\intercal )^3 = -A^3\). Thus, if \(A\) is skew-symmetric, then \(A^3\) is also skew-symmetric.