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--- lecture_9 [2015/02/17 10:58] – rupert
+++ lecture_9 [2017/02/21 10:02] (current) – rupert
@@ Line 1: / Line 1: @@
-=== Definition of matrix multiplication ===
+=== Proof of the proposition, continued ===
-If $A$ is an $n\times m$ matrix and $B$ is an $m\times k$ matrix, then the product $AB$ is the $n\times k$ matrix whose $(i,j)$ entry is the row-column product of the $i$th row of $A$ with the $j$th column of $B$. That is:
+. To show that $AI_m=A$ for any $n\times m$ matrix $A$ is similar to the first part of the proof; the details are left as an exercise.
-\[ (AB)_{i,j} = \text{row}_i(A)\cdot \text{col}_j(B).\]
-If $A$ is an $n\times m$ matrix and $B$ is an $\ell\times k$ matrix with $m\ne \ell$, then the matrix product $AB$ is undefined.
-=== Examples ===
-  - 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 $B$ is any $n\times n$ matrix, then $I_nB=B$ by part 1 and $BI_n=B$ by part 2, so $I_nB=B=BI_n$. In particular, $I_nB=BI_n$ so $I_n$ commutes with $B$, for every square $n\times n$ matrix $B$. ■
-  - If $A=\mat{1&2\\3&4\\5&6}$, $B=\mat{2&1&1\\1&2&0\\1&0&2\\1&0&2\\2&2&1}$ 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 $A=0_{n\times n}$ 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 $AB\ne BA$.
-=== Commuting matrices ===
+===== Algebraic properties of matrix multiplication =====
+==== The associative law ====
+=== Proposition: associativity of matrix multiplication ===
+Matrix multiplication is //associative//. This means that $(AB)C=A(BC)$ whenever $A,B,C$ are matrices which can be multiplied together in this order.
-{{page>commute}}
+We omit the proof, but this is not terribly difficult; it is a calculation in which you write down two formulae for the $(i,j)$ entries of $(AB)C$ and $A(BC)$, and carefully check they are equal using the fact that if $a,b,c$ are real numbers, then $(ab)c=a(bc)$.
-Which matrices commute? Suppose $A$ is an $n\times m$ matrix and $B$ is an $\ell\times k$ matrix, and $A$ adn $B$ commute, i.e., $AB=BA$.
+=== Example ===
+We saw above that $\newcommand{\m}[1]{\begin{bmatrix}#1\end{bmatrix}}A=\m{1&2\\3&4}$ commutes with $B=\m{7&10\\15&22}$. We can explain why this is so using associativity. You can check that $B=AA$ (which we usually write as $B=A^2$). Hence, using associativity at $\stackrel*=$,
+\[ AB=A(AA)\stackrel*=(AA)A=BA.\]
+The same argument for any square matrix $A$ gives a proof of:
+=== Proposition ===
+If $A$ is any square matrix, then $A$ commutes with $A^2$.■
-  * $AB$ must be defined, so $m=\ell$
+The powers of a square matrix $A$ are defined by $A^1=A$, and $A^{k+1}=A(A^k)$ for $k\in \mathbb{N}$. Using [[wp>mathematical induction]], you can prove the following more general proposition.
-  * $BA$ must be defined, so $k=n$
+===Proposition: a square matrix commutes with its powers===
-  * $AB$ 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 $k=m$.
+If $A$ is any square matrix and $k\in\mathbb{N}$, then $A$ commutes with $A^k$.■
-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//.
-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.
-Because it's not true in general that $AB=BA$, we say that **matrix multiplication is not commutative**.
+====The distributive laws====
-=== Definition of the $n\times n$ identity matrix ===
+=== Lemma: the distributive laws for row-column multiplication ===
-{{page>identity matrix}}
+  - If $a$ is a $1\times m$ row vector and $b$ and $c$ are $m\times 1$ column vectors, then $a\cdot (b+c)=a\cdot b+a\cdot c$.
+  - If $b$ and $c$ are $1\times m$ row vectors and $a$ is an $m\times 1$ column vector, then $(b+c)\cdot a=b\cdot a+c\cdot a$.
+The proof is an exercise (see tutorial worksheet 5).
+=== Proposition: the distributive laws for matrix multiplication ===
+If $A$ is an $n\times m$ matrix and $k\in\mathbb{N}$, then:
+  - $A(B+C)=AB+AC$ for any $m\times k$ matrices $B$ and $C$; and
+  - $(B+C)A=BA+CA$ for any $k\times n$ matrices $B$ and $C$.
+In other words, $A(B+C)=AB+AC$ whenever the matrix products make sense, and similarly $(B+C)A=BA+CA$ whenever this makes sense.
+===Proof===
+. First note that
+  * $B$ and $C$ are both $m\times k$, so $B+C$ is $m\times k$ by the definition of [[matrix addition]];
+  * $A$ is $n\times m$ and $B+C$ is $m\times k$, so $A(B+C)$ is $m\times k$ by the definition of [[matrix multiplication]];
+  * $AB$ and $AC$ are both $n\times k$ by the definition of matrix multiplication
+  * so $AB+AC$ is $n\times k$ by the definition of matrix addition.
+So we have (rather long-windedly) checked that $A(B+C)$ and $AB+AC$ have the [[same size]].
+By the Lemma above, the [[row-column product]] has the property that \[a\cdot (b+c)=a\cdot b+a\cdot c.\]
+So the $(i,j)$ entry of $A(B+C)$ is
+\begin{align*}\def\row{\text{row}}\def\col{\text{col}}
+\text{row}_i(A)\cdot \col_j(B+C) &= \text{row}_i(A)\cdot \big(\col_j(B)+\col_j(C)\big)
+\\ &= \text{row}_i(A)\cdot \col_j(B)+\row_i(A)\cdot\col_j(C).\end{align*}
+On the other hand,
+  * the $(i,j)$ entry of $AB$ is $\text{row}_i(A)\cdot \col_j(B)$; and
+  * the $(i,j)$ entry of $AC$ is $\row_i(A)\cdot\col_j(C)$;
+  * so the $(i,j)$  entry of $AB+AC$ is also $\text{row}_i(A)\cdot \col_j(B)+\row_i(A)\cdot\col_j(C)$.
+So the entries of $A(B+C)$ and $AB+AC$ are all equal, so $A(B+C)=AB+AC$.
+. The proof is similar, and is left as an exercise.■
-=== Examples ===
-  - $I_1=[1]$
-  - $I_2=\mat{1&0\\0&1}$
-  - $I_3=\mat{1&0&0\\0&1&0\\0&0&1}$
-  - $I_4=\mat{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$, and so on!
-=== Proposition ===
-  - $I_nA=A$ 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 $B$.