Differences

This shows you the differences between two versions of the page.

--- lecture_9_slides [2016/02/22 17:59] – rupert
+++ lecture_9_slides [2017/02/21 10:03] (current) – rupert
@@ Line 1: / Line 1: @@
 ~~REVEAL~~
+==== Recap: matrix multiplication and the identity matrix ====
+  * $AB$ defined if $A:n\times m$ and $B:m\times k$
+    * then $AB$ is $n\times k$ ...
+    * with $(i,j)$ entry is $\text{row}_i(A).\text{col}_j(B)$
+  * Sometimes $AB=BA$ (need $A,B$ to both be $n\times n$)
+    * say $A$ and $B$ commute
+  * But often $AB\ne BA$ (even if $A,B$ both $n\times n$)
+  * $n\times n$ identity matrix $I_n$: $1$s on diagonal, zeros elsewhere
+  * $I_n$ commutes with every $n\times n$ matrix, in a nice way...
 ==== ====
   - Last time: proved that $I_nA=A$ for any $n\times m$ matrix A.
@@ Line 68: / Line 79: @@
 ==== $A(B+C)=AB+AC$ continued ====
-  * In tutorial 4: $a\cdot (b+c)=a\cdot b+a\cdot c$ (row-col product) whenever $a$: $1\times m$ and $b,c$: $m\times 1$.
+  * In tutorial 4: $a\cdot (b+c)=a\cdot b+a\cdot c$ (row-col product)
-  * So the $(i,j)$ entry of $A(B+C)$ is\begin{align*}\def\xx{\!\!\!\!}\def\xxx{\xx\xx\xx\xx}\xxx\xxx\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*}
+  * (for $a$: $1\times m$ and $b,c$: $m\times 1$)
-  * the $(i,j)$ entry of $AB$ is $\text{row}_i(A)\cdot \col_j(B)$; and
+  * Write $\def\row{\text{row}}\def\col{\text{col}}a_i=\row_i(A)$, $b_j=\col_j(B)$, $c_j=\col_j(C)$.
-  * the $(i,j)$ entry of $AC$ is $\row_i(A)\cdot\col_j(C)$;
+  * $(i,j)$ entry of $A(B+C)$ is:\begin{align*}\def\xx{\!\!\!\!}\def\xxx{\xx\xx\xx\xx}\xxx\xxx \row_i(A)\cdot \col_j(B+C) &= a_i\cdot \big(b_j+c_j\big)\\ &= a_i\cdot b_j+a_i\cdot c_j.\end{align*}
-  * so the $(i,j)$  entry of $AB+AC$ is also $\text{row}_i(A)\cdot \col_j(B)+\row_i(A)\cdot\col_j(C)$.
+  * $(i,j)$ entry of $AB$ is $a_i\cdot b_j$; and
-  * Same entries; so $A(B+C)=AB+AC$.
+  * $(i,j)$ entry of $AC$ is $a_i\cdot c_j$;
+  * so $(i,j)$ entry of $AB+AC$ is also $a_i\cdot b_j+a_i\cdot c_j$
+  * Same sizes and same entries, so $A(B+C)=AB+AC$.
 ==== Proof that $(B+C)A=BA+CA$ ====
   * This is very similar, and is left as an exercise.■
-===== Matrix equations =====
-  * A linear equation can be written using [[row-column multiplication]].
-  * e.g. $ 2x-3y+z=8$ is same as $ \m{2&-3&1}\m{x\\y\\z}=8$
-  * or $ a\vec x=8$ where $a=\m{2&-3&1}$ and $\vec x=\m{x\\y\\z}$.
-==== ====
-  * We can write a whole [[system of linear equations]] in a similar way, as a matrix equation using [[matrix multiplication]].
-  * e.g. the linear system $\begin{align*} 2x-3y+z&=8\\ y-z&=4\\x+y+z&=0\end{align*}$
-  * is same as $\m{2&-3&1\\0&1&-1\\1&1&1}\m{x\\y\\z}=\m{8\\4\\0}$
-  * or $ A\vec x=\vec b$ where $A=\m{2&-3&1\\0&1&-1\\1&1&1}$, $\vec x=\m{x\\y\\z}$ and $\vec b=\m{8\\4\\0}$.
-==== ====
-In a similar way, any linear system \begin{align*} a_{11}x_1+a_{12}x_2+\dots+a_{1m}x_m&=b_1\\ a_{21}x_1+a_{22}x_2+\dots+a_{2m}x_m&=b_2\\ \hphantom{a_{11}}\vdots \hphantom{x_1+a_{22}}\vdots\hphantom{x_2+\dots+{}a_{nn}} \vdots\ & \hphantom{{}={}\!} \vdots\\ a_{n1}x_1+a_{n2}x_2+\dots+a_{nm}x_m&=b_n \end{align*}
-can be written in the form
-\[ A\vec x=\vec b\]
-where $A$ is the $n\times m $ matrix, called the **coefficient matrix** of the linear system, whose $(i,j)$ entry is $a_{ij}$ (the number in front of $x_j$ in the $i$th equation of the system) and $\vec x=\m{x_1\\x_2\\\vdots\\x_m}$, and $\vec b=\m{b_1\\b_2\\\vdots\\b_n}$.
-==== Solutions of matrix equations ====
-  * More generally, might want to solve a matrix equation like \[AX=B\] where $A$, $X$ and $B$ are matrices of any size, with $A$ and $B$ fixed matrices and $X$ a matrix of unknown variables.
-  * If $A$ is $n\times m$, we need $B$ to be $n\times k$ for some $k$, and then $X$ must be $m\times k$
-    * so we know the size of any solution $X$.
-  *  But which $m\times k$ matrices $X$ are solutions?
-==== Example ====
-If $A=\m{1&0\\0&0}$ and $B=0_{2\times 3}$, then any solution $X$ to $AX=B$ must be $2\times 3$.
-  * One solution is $X=0_{2\times 3}$
-    *  because then we have $AX=A0_{2\times 3}=0_{2\times 3}$.
-  * This is not the only solution!
-  * For example, $X=\m{0&0&0\\1&2&3}$ is another solution
-    * because then we have $AX=\m{1&0\\0&0}\m{0&0&0\\1&2&3}=\m{0&0&0\\0&0&0}=0_{2\times 3}.$
-  * From this example, we see that a matrix equation can have many solutions.