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--- lecture_11_slides [2016/03/01 11:02] – [Proof] rupert
+++ lecture_11_slides [2017/02/27 12:35] (current) – [Proof] rupert
@@ Line 2: / Line 2: @@
 ==== Last time ====
+  * An $n\times n $ matrix  $A$ is [[invertible]] if there is a matrix $C$ solving the matrix equations $AC=I_n$ and $CA=I_n$. We then say $C$ is an [[inverse]] of $A$.
+  * e.g. $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\mat{2&4\\0&1}$ is invertible, with inverse $C=\mat{0.5&-2\\0&1}$
+  * if $[a]$ is $1\times 1$ and $a\ne 0$, then $C=[\tfrac 1a]$ is an inverse
+  * $I_n$ is its own inverse for any $n$
+  * $0_{n\times n}$, $\mat{1&0\\0&0}$ and $\mat{1&2\\-3&-6}$ aren't invertible
+==== Proposition: uniqueness of the inverse ====
+If $A$ is an invertible $n\times n$ matrix, then $A$ has a //unique// inverse.
+=== Proof ===
+  * $A$ invertible, so $A$ has at least one inverse.
+  * Suppose it has two inverses, say $C$ and $D$.
+  * Then $AC=I_n=CA$ and $AD=I_n=DA$.
+  * So $C=CI_n=C(AD)=(CA)D=I_nD=D$
+  * So $C=D$.
+  * So any two inverses of $A$ are equal.
+  * So $A$ has a unique inverse.■
+==== Definition/notation: $A^{-1}$ ====
 {{page>the inverse}}
-  * So $A^{-1}$ is the matrix with $AA^{-1}=I_n=A^{-1}A$, if such a matrix exists
+==== Examples again ====
-  * e.g. $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\mat{2&4\\0&1}^{-1}=\mat{0.5&-2\\0&1}$
+  * $A=\mat{2&4\\0&1}$ is invertible, and $A^{-1}=\mat{\tfrac12&-2\\0&1}$.
+    * i.e. $\mat{2&4\\0&1}^{-1}=\mat{\tfrac12&-2\\0&1}$
   * if $a$ is a non-zero scalar, then $[a]^{-1}=[\tfrac 1a]$
   * $I_n^{-1}=I_n$ for any $n$
   * $0_{n\times n}^{-1}$, $\mat{1&0\\0&0}^{-1}$ and $\mat{1&2\\-3&-6}^{-1}$ **do not exist**
+    * because these matrices aren't invertible
 ==== Warning ====
-If $A$, $B$ are matrices, then $\frac AB$ doesn't make sense!
+If $A$, $B$ are matrices, never write down $\frac AB$. It doesn't make (unambiguous) sense!
-  * Never write this down as it will almost always lead to mistakes.
+  * $\frac AB$ is ambiguous, or not well defined:
+    * it could mean $B^{-1}A$
+    * or equally well it could mean $AB^{-1}$
+    * and these are usually different because matrix multiplication isn't commutative!
   * In particular, $A^{-1}$ definitely doesn't mean $\frac 1A$. Never write this either!
 ==== Proposition: solving $AX=B$ when $A$ is invertible ====
@@ Line 79: / Line 103: @@
 If $A=\mat{a&b\\c&d}$ and $J=\mat{d&-b\\-c&a}$, then we have
-\[ AJ=\delta I_2=JA\]
+\[ AJ=(ad-bc) I_2=JA.\]
-where $\delta=ad-bc$.
-  * Proof is a calculation!
+  * Proof is an easy calculation!
+  * Note that $(ad-bc) I_2=(ad-bc)\mat{1&0\\0&1}=\mat{ad-bc&0\\0&ad-bc}$
+  * Now just show that $AJ$ and $JA$ both give the same matrix (exercise).
 ==== Definition: the determinant of a $2\times 2$ matrix ====
@@ Line 97: / Line 122: @@
 ==== Proof ====
-  * Let $J=\mat{d&-b\\-c&a}$ and write $\delta=\det(A)$.
+  * Let $J=\mat{d&-b\\-c&a}$, so $AJ=\det(A) I_2=JA$ (Lemma).
-  * By the previous lemma, $AJ=\delta I_2=JA$.
-If $\delta\ne 0$:
+If $\det(A)\ne 0$:
-  * Multiply by $\frac1{\delta}$: $\quad A(\tfrac1{\delta}J)=I_2=(\tfrac1{\delta}J) A$
+  * Multiply by $\frac1{\det(A)}$: $\quad A(\tfrac1{\det(A)}J)=I_2=(\tfrac1{\det(A)}J) A$
-  * So $ AB=I_2=BA$, where $B=\tfrac1{\delta}J$
+  * So $A$ invertible with $A^{-1}=\frac1{\det(A)}J=\frac1{\det(A)}\mat{d&-b\\-c&a}$.
-  * So $A$ invertible with $A^{-1}=B=\frac1{\delta}J=\frac1{\det(A)}\mat{d&-b\\-c&a}$.
-If $\delta=0$:
+If $\det(A)=0$:
   * $AJ=0_{2\times 2}$
-  * If $J=0_{2\times 2}$ then $A=0_{2\times 2}$ so $A$ isn't invertible
+  * If $J\ne 0_{2\times 2}$ then (by corollary) $A$ isn't invertible.
-  * If $J\ne 0_{2\times 2}$ then by the previous corollary, $A$ isn't invertible.  ■
+  * If $J=0_{2\times 2}$ then $A=0_{2\times 2}$, which isn't invertible.■
 ==== Using the inverse to solve a matrix equation ====