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--- lecture_11_slides [2017/02/27 11:57] – [Proposition: solving $AX=B$ when $A$ is invertible] rupert
+++ lecture_11_slides [2017/02/27 12:35] (current) – [Proof] rupert
@@ Line 6: / Line 6: @@
   * 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 a non-zero scalar, then $C=[\tfrac 1a]$ is an inverse
+  * 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
@@ Line 43: / Line 43: @@
     * 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 ====
-If $A$ is an invertible $n\times n$ matrix and $B$ is an $n\times k$ matrix, then $ AX=B$ has a unique solution: $X=A^{-1}B$.
-=== Proof ===
-  - Existence: check that $X=A^{-1}B$ really is a solution:
-    * LHS is $AX=A(A^{-1}B)=(AA^{-1})B=I_nB=B$
-    * same as RHS. So $X=A^{-1}B$ is indeed a solution.
-  - Uniqueness: check there are no other solutions:
-    * Suppose that $X$ is any solution. Then $AX=B$.
-    * Multiply both sides on the left by $A^{-1}$ to get $A^{-1}AX=B$
-    * Since $A^{-1}A=I_n$ and $I_nX=X$, we get $X=B$
-    * Conclusion: if $X$ is any solution, we must have $X=A^{-1}B$■
-==== Corollary ====
-If $A$ is an $n\times n$ matrix and there is a non-zero $n\times m$ matrix $K$ so that $AK=0_{n\times m}$, then $A$ is not invertible.
-=== Proof ===
-  * $AX=0_{n\times m}$ has (at least) two solutions:
-    * $X=K$,
-    * and $X=0_{n\times m}$
-  * If $A$ was invertible, this would contradict the Proposition.
-  * So $A$ cannot be invertible. ■
-==== Example ====
-  * Let $A=\mat{1&2\\-3&-6}$. $A$ isn't invertible... why?
-  * One column of $A$ is $2$ times the other... exploit this.
-  * Let $K=\mat{-2\\1}$
-  * $K$ is non-zero but $AK=\mat{1&2\\-3&-6}\mat{-2\\1}=\mat{0\\0}=0_{2\times 1}$
-  * So $A$ is not invertible, by the Corollary.
-==== Warning ====
-If $A$, $B$ are matrices, then $\frac AB$ doesn't make sense!
-  * Never write this down as it will almost always lead to mistakes.
-  * In particular, $A^{-1}$ definitely doesn't mean $\frac 1A$. Never write this either!
 ==== Proposition: solving $AX=B$ when $A$ is invertible ====
@@ Line 145: / 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 163: / 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 ====