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--- lecture_8 [2015/02/12 10:28] – rupert
+++ lecture_8 [2016/02/18 10:20] (current) – rupert
@@ Line 1: / Line 1: @@
-===== Operations on matrices =====
+=== Definition of matrix multiplication ===
-We want to define operations on matrices: some (useful) ways of taking two matrices and making a new matrix.
+{{page>matrix multiplication}}
+=== Examples ===
-Before we begin, a remark about $1\times 1$ matrices. These are of the form $[a_{11}]$ where $a_{11}$ is just a number. The square brackets $[\,]$ don't really matter here; they just keep the inside of a matrix in one place. So really: a $1\times 1$ matrix is just a number. This means that special cases of the operations we define will be operations on ordinary numbers. You should check that in the special case when all the matrices involved are $1\times 1$ matrices, the operations become the ordinary operations on numbers, so we are //generalising// the familiar operations (addition, subtraction, multiplication and so on) from numbers to matrices.
+  - If $\newcommand{\mat}[1]{\begin{bmatrix}#1\end{bmatrix}} A=\mat{1&0&5\\2&-1&3}$ and  $B=\mat{1&2\\3&4\\5&6}$, then $AB=\mat{26&32\\14&18}$ and $BA=\mat{5&-2&11\\11&-4&27\\17&-6&43}$. Note that $AB$ and $BA$ are both defined, but $AB\ne BA$ since $AB$ and $BA$ don't even have the [[same size]].
+  - If $A=\mat{1&2\\3&4\\5&6}$, $B=\mat{2&1&1\\1&2&0\\1&0&2\\2&2&1}$ and $C=\mat{1&3&0&7\\0&4&6&8}$, then $A$ is $3\times 2$, $B$ is $4\times 3$ and $C$ is $2\times 4$, so
+    * $AB$, $CA$ and $BC$ don't exist (i.e., they are undefined);
+    * $AC$ exists and is $3\times 4$;
+    * $BA$ exists and is $4\times 2$; and
+    * $CB$ exists and is $2\times 2$.
+    * In particular, $AB\ne BA$ and $AC\ne CA$ and $BC\ne CB$, since in each case one of the matrices doesn't exist.
+  - If $A=\mat{0&1\\0&0}$ and $B=\mat{0&0\\1&0}$, then $AB=\mat{1&0\\0&0}$ and $BA=\mat{0&0\\0&1}$. So $AB$ and $BA$ are both defined and have the [[same size]], but they are not [[equal matrices]]: $AB\ne BA$.
+  - If $A=\mat{1&0\\0&0}$ and $B=\mat{0&0\\0&1}$, then $AB=\mat{0&0\\0&0}$ and $BA=\mat{0&0\\0&0}$. So $AB=BA$ in this case.
+  - If $A=0_{n\times n}$ is the $n\times n$ zero matrix and $B$ is any $n\times n$ matrix, then $AB=0_{n\times n}$ and $BA=0_{n\times n}$. So in this case, we do have $AB=BA$.
+  - If $A=\mat{1&2\\3&4}$ and $B=\mat{7&10\\15&22}$, then $AB=\mat{37&54\\81&118}=BA$, so $AB=BA$ for these particular matrices $A$ and $B$.
+  - If $A=\mat{1&2\\3&4}$ and $B=\mat{6&10\\15&22}$, then $AB=\mat{36&54\\78&118}$ and $BA= \mat{36&52\\81&118}$, so $AB\ne BA$.
-==== Matrix addition and subtraction ====
+=== Commuting matrices ===
-=== Definition of matrix addition ===
+{{page>commute}}
-{{page>matrix addition}}
+Which matrices commute? Suppose $A$ is an $n\times m$ matrix and $B$ is an $\ell\times k$ matrix, and $A$ and $B$ commute, i.e., $AB=BA$.
-=== Example ===
+  * $AB$ must be defined, so $m=\ell$
-\[ \begin{bmatrix}1&2&-2\\3&0&5\end{bmatrix}+\begin{bmatrix}-2&2&0\\1&1&1\end{bmatrix}=\begin{bmatrix}-1&4&-2\\4&1&6\end{bmatrix}.\]
+  * $BA$ must be defined, so $k=n$
+  * $AB$ is an $n\times k$ matrix and $BA$ is an $\ell\times n$ matrix. Since $AB$ has the same size as $BA$, we must have $n=\ell$ and $k=m$.
+Putting this together: we see that if $A$ and $B$ commute, then $A$ and $B$ must both be $n\times n$ matrices for some number $n$. In other words, they must be //square matrices of the same size//.
-=== Example ===
+Examples 4 and 5 above show that for some square matrices $A$ and $B$ of the same size, it is true that $A$ and $B$ commute. On the other hand, examples 3 and 6 show that it's not true that square matrices of the same size must always commute.
-\[ \begin{bmatrix}1&2&-2\\3&0&5\end{bmatrix}+\begin{bmatrix}-2&2\\1&1\end{bmatrix}\text{ is undefined.}\]
-=== Definition of matrix subtraction ===
+Because it's not true in general that $AB=BA$, we say that **matrix multiplication is not commutative**.
-{{page>matrix subtraction}}
+=== Definition of the $n\times n$ identity matrix ===
-=== Example ===
+{{page>identity matrix}}
-\[ \begin{bmatrix}1&2&-2\\3&0&5\end{bmatrix}-\begin{bmatrix}-2&2&0\\1&1&1\end{bmatrix}=\begin{bmatrix}3&0&-2\\2&-1&4\end{bmatrix}.\]
-=== Example ===
+=== Examples ===
-\[ \begin{bmatrix}1&2&-2\\3&0&5\end{bmatrix}-\begin{bmatrix}-2&2\\1&1\end{bmatrix}\text{ is undefined.}\]
+  - $I_1=[1]$
+  - $I_2=\mat{1&0\\0&1}$
+  - $I_3=\mat{1&0&0\\0&1&0\\0&0&1}$
+  - $I_4=\mat{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$, and so on!
+=== Proposition: properties of $I_n$ ===
+  - $I_nA=A$ for any $n\times m$ matrix $A$;
+  - $AI_m=A$ for any $n\times m$ matrix $A$; and
+  - $I_nB=B=BI_n$ for any $n\times n$ matrix $B$. In particular, $I_n$ commutes with every other square $n\times n$ matrix $B$.
+=== Proof of the proposition ===
+. We want to show that $I_nA=A$ for any $n\times m$ matrix $A$. These matrices the [[same size]], since $I_n$ has size $n\times n$ and $A$ has size $n\times m$, so $I_n A$ has size $n\times m$ by the definition of [[matrix multiplication]], which is the same as the size of $A$.
+ Note that $\text{row}_i(I_n)=[0~0~\dots~0~1~0~\dots~0]$, where the $1$ is in the $i$th place, by definition of the [[identity matrix]] $I_n$; and the $j$th column of $A$ is $\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\end{bmatrix}$. The (i,j) entry of $I_nA$ is
+$\text{row}_i(I_n)\cdot \text{col}_j(A)$, by the definition of [[matrix multiplication]], which is therefore
+\begin{align*} [0~0~\dots~0~1~0~\dots~0]\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\end{bmatrix} &= 0a_{1j}+0a_{2j}+\dots+0a_{i-1,j}+1a_{ij}+0a_{i+1,j}+\dots+0a_{nj} \\&= a_{ij}.\end{align*}
+So the matrices $I_nA$ and $A$ have the same size, and the same $(i,j)$ entries, for any $(i,j)$. So $I_nA=A$.