2.3 Matrix Algebra

The sum (difference) of two \(m\) by \(n\) matrices \(A=(a_{ij})\) and \(B=(b_{ij})\) is the \(m\) by \(n\) matrix obtained by adding (subtracting) their corresponding entries.

That is \(A+B=(a_{ij}+b_{ij})\)

Let \(A= \begin {pmatrix} 1&4\\0&-2\\6&5\\ \end {pmatrix} \) and \(B= \begin {pmatrix} 0&3\\-1&1\\-3&2\\ \end {pmatrix} \).

Then \(A+B= \begin {pmatrix} 1&7\\-1&-1\\3&7\\ \end {pmatrix} \) and \( A-B= \begin {pmatrix} 1&1\\1&-3\\9&3\\ \end {pmatrix} \).

Let \(A=(a_{ij})\) be an \(m\) by \(n\) matrix and let \(c\) be any scalar. Then \(cA\) is the matrix obtained by multiplying each entry on \(A\) by \(c\).

Example 2.3.1. If \(A= \begin {pmatrix} 1&4&6\\2&-1&0\\ \end {pmatrix} \) and \(c=-3\), then \(cA= \begin {pmatrix} -3&-12&-18\\-6&3&0\\ \end {pmatrix} \).

If \(A=(a_{ij})\) is an \(m\) by \(n\) matrix and \(B=(b_{ij})\) is an \(n\) by \(r\) matrix, then the product \(C=AB\) is defined at it is an \(m\) by \(r\) matrix. Then \((i,j)\) entry of the matrix is calculated as follows:
\(c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots +a_{in}b_{nj}\).

Example 2.3.2. \(A= \begin {pmatrix} -1&-3&3\\1&-1&3\\ \end {pmatrix} \) and \(B= \begin {pmatrix} 4&0&-3&1\\-5&2&1&-1\\1&-2&0&6\\ \end {pmatrix} \) then \[AB= \begin {pmatrix} -1&-3&3\\1&-1&3\\ \end {pmatrix} \begin {pmatrix} 4&0&-3&1\\-5&2&1&-1\\1&-2&0&6\\ \end {pmatrix} = \begin {pmatrix} 14&-10&0&14\\12&-8&-4&20\\ \end {pmatrix} \]

Example 2.3.3. Let \(A\) and \(B\) be any \(m\) by \(n\) matrices. Then

1.: \((A+B)^t=A^t+B^t\)
2.: \((AB)^t=A^tB^t\)
3.: \((A^t)^t=A\)
4.: \((A^t)^n=(A^n)^t\) for any positive integer \(n\).

Proof.

1.: Let \(A=(a_{ij})=A^t\), \(B=(b_{ij})=B^t\) \begin {align*} (A+B) &=(a_{ij}+b_{ij})\\ (A+B)^t &=(a_{ij}+b_{ij})\\ A^t+B^t &=(a_{ij}+b_{ij})\\\\ \therefore \hspace {0.5cm} (A+B)^t &=A^t+B^t\\ \end {align*}
2.: \begin {align*} (AB)' &=(a_{ij}(b_{ij})=(a_{ij}\hspace {0.1cm}.\hspace {0.1cm}b_{ij})\\\\ (AB)^t &=(a_{ij}\hspace {0.1cm}\cdot \hspace {0.1cm}b_{ij})^t\\ A^t\hspace {0.1cm}\cdot \hspace {0.1cm}B^t &=(a_{ij})^t\hspace {0.1cm}\cdot \hspace {0.1cm}(b_{ij})^t\\ \end {align*}

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Example 2.3.4.

1.: If \(A\) is a square matrix, then \(1/2(A+A^t)\) is symmetric and \(1/2(A-A^t)\) is skew-symmetric.
2.: For any matrix \(A\), both \(AA^t\) and \(A^tA\) are symmetric.

Proof.

1.: \begin {align*} \Big (\frac {1}{2}(A+A^t)\Big )^t &=\frac {1}{2}\Big (A^t+(A^t)^t\Big )\\ &=\frac {1}{2}(A^t+A)\\ &=\frac {1}{2}(A+A^t)\hspace {2cm}\text {and}\\ \end {align*}

\begin {align*} \Big (\frac {1}{2}(A-A^t)\Big )^t &=\frac {1}{2}\Big (A^t-(A^t)^t\Big )\\ &=\frac {1}{2}(A^t-A)\\ &=-\frac {1}{2}(A-A^t)\\ \end {align*}
2.: \((AA^t)^t =(A^t)^tA^t =AA^t\qquad \text {and}\qquad (A^tA)^t = A^t(A^t)^t=A^tA\)

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