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--- lecture_11 [2015/02/24 10:44] – [Proposition: solving $AX=B$ when $A$ is invertible] rupert
+++ lecture_11 [2017/02/28 12:07] (current) – rupert
@@ Line 1: / Line 1: @@
-===== Invertibility =====
+==== Proposition: solving $AX=B$ when $A$ is invertible ====
-We've seen that solving matrix equations $AX=B$ is useful, since they generalise systems of linear equations.
+If $A$ is an invertible $n\times n$ matrix and $B$ is an $n\times k$ matrix, then the matrix equation \[ AX=B\] has a unique solution: $X=A^{-1}B$.
-How can we solve them?
+=== Proof ===
-==== Example ====
+First we check that $X=A^{-1}B$ really is a solution to $AX=B$. To see this, note that if $X=A^{-1}B$, then
+\begin{align*}
+ AX&=A(A^{-1}B)\\&=(AA^{-1})B\\&=I_n B \\&= B.
+\end{align*}
+Now we check that the solution is unique. If $X$ and $Y$ are both solutions, then $AX=B$ and $AY=B$, so \[AX=AY.\] Multiplying both sides on the left by $A^{-1}$, we get
+\[ A^{-1}AX=A^{-1}AY\implies I_nX=I_nY\implies X=Y.\]
+So any two solutions are equal, so $AX=B$ has a unique solution. ■
-Take $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\mat{2&4\\0&1}$ and $B=\mat{3&4\\5&6}$, so we want to find all matrices $X$ so that $AX=B$, or \[ \mat{2&4\\0&1}X=\mat{3&4\\5&6}.\] Note that $X$ must be a $2\times 2$ matrix for this to work, by the definition of [[matrix multiplication]]. So one way to solve this is to write $X=\mat{x_{11}&x_{12}\\x_{21}&x_{22}}$ and plug it in:
+==== Corollary ====
-\[\mat{2&4\\0&1}\mat{x_{11}&x_{12}\\x_{21}&x_{22}}=\mat{3&4\\5&6}\iff \mat{2x_{11}+4x_{21}&2 x_{12}+4x_{22}\\x_{21}&x_{22}}=\mat{3&4\\5&6}\]
-and then equate entries to get four linear equations:
-\begin{align*}2x_{11}+4x_{21}&=3\\2 x_{12}+4x_{22}&=4\\x_{21}&=5\\x_{22}&=6\end{align*}
-which we can solve in the usual way.
-But this is a bit tedious! We will develop a slicker method by first thinking about solving ordinary equations $ax=b$ where $a,x,b$ are all numbers, or  if you like, $1\times 1$ matrices.
+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.
-==== Solving $ax=b$ and $AX=B$ ====
+=== Proof ===
-If $a\ne0$, then solving $ax=b$ where $a,b,x$ are numbers is easy. We just divide both sides by $a$, or equivalently, we multiply both sides by $\tfrac1a$, to get the solution: $x=\tfrac1a\cdot b$.
+Since $A0_{n\times m}=0_{n\times m}$ and $AK=0_{n\times m}$, the equation $AX=0_{n\times m}$ has (at least) two solutions: $X=0_{n\times m}$ and $X=K$. Since $K$ is non-zero, these two solutions are different.
-Why does this work? If $x=\tfrac1a\cdot b$, then
+So there is not a unique solution to $AX=B$, for $B$ the zero matrix. If $A$ was invertible, this would contradict the uniqueness statement of the last Proposition. So $A$ cannot be invertible. ■
-\begin{align*} ax&=a(\tfrac1a\cdot b)\\&=(a\cdot \tfrac1a)b\\&=1b\\&=b\end{align*}
-so $ax$ really is equal to $b$, and we do have a solution to $ax=b$.
-What is special about $\tfrac1a$ which made this all work?
+==== Examples ====
-  - we have $\tfrac1a \cdot a = 1$,
+  * We can now see why the matrix $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\mat{1&2\\-3&-6}$ is not invertible. If $X=\mat{-2\\1}$ and $K=\mat{2\\-1}$, then $K$ is non-zero, but $AK=0_{2\times 1}$. So $A$ is not invertible, by the Corollary.
-  - and $1b = b$.
+  * $A=\mat{1&4&5\\2&5&7\\3&6&9}$ is not invertible, since $K=\mat{1\\1\\-1}$ is non-zero and $AK=0_{3\times 1}$.
-Now for an $n\times k$ matrix $B$, we know that the identity matrix $I_n$ does the same sort of thing as $1$ is doing in the relation $1b=b$: we have $I_nB=B$ for any $n\times k$ matrix $B$. So instead of $\tfrac1a$, we want to find a matrix $A'$ with the property: $A'\cdot A=I_n$. In fact, because matrix multiplication is not commutative, we also require that $A\cdot A'=I_n$. It's then easy to argue that $X=A'\cdot B$ is a solution to $AX=B$, since
+===== $2\times 2$ matrices: determinants and invertibility =====
-\begin{align*} AX&=A(A'\cdot B)\\&=(A\cdot A')B\\&=I_nB\\&=B.\end{align*}
-==== Example revisited ====
+==== Question ====
-If $A=\mat{2&4\\0&1}$, then the matrix $A'=\mat{\tfrac12&-2\\0&1}$ does have the property that
-\[ A\cdot A'=I_2=A'\cdot A.\]
-(You should check this!). So a solution to $AX=B$ where $B=\mat{3&4\\5&6}$ is $X=A'B=\mat{\tfrac12&-2\\0&1}\mat{3&4\\5&6} = \mat{-8.5&-10\\5&6}$.
-Notice that having found the matrix $A'$, then we can solve $AX=C$ easily for any $2\times 2$ matrix $C$: the answer is $X=A'C$. This is quicker than having to solve four new linear equations using our more tedious method above.
+Which $2\times 2$ matrices are invertible? For the invertible matrices, can we find their inverse?
-==== Definition: invertible ====
+==== Lemma ====
-{{page>invertible}}
+If $A=\mat{a&b\\c&d}$ and $J=\mat{d&-b\\-c&a}$, then we have
+\[ AJ=\delta I_2=JA\]
-==== Examples ====
+where $\delta=ad-bc$.
-  * $A=\mat{2&4\\0&1}$ is invertible, and the matrix $A'=\mat{\tfrac12&-2\\0&1}$ is an inverse of $A$
-  * a $1\times 1$ matrix $A=[a]$ is invertible if and only if $a\ne0$, and if $a\ne0$ then an inverse of $A=[a]$ is $A'=[\tfrac1a]$.
-  * $I_n$ is invertible for any $n$, since $I_n\cdot I_n=I_n=I_n\cdot I_n$, so an inverse of $I_n$ is $I_n$.
-  * $0_{n\times n}$ is not invertible for any $n$, since $0_{n\times n}\cdot A'=0_{n\times n}$ for any $n\times n$ matrix $A'$, so $0_{n\times n}\cdot A'\ne I_n$.
-  * $A=\mat{1&0\\0&0}$ is not invertible, since for any $2\times 2$ matrix $A'=\mat{a&b\\c&d}$ we have $AA'=\mat{a&b\\0&0}$ which is not equal to $I_2=\mat{1&0\\0&1}$ since the $(2,2)$ entries are not equal.
-  * $A=\mat{1&2\\-3&-6}$ is not invertible. We'll see why later!
-==== Proposition: uniqueness of the inverse ====
-If $A$ is an invertible $n\times n$ matrix, then $A$ has a //unique// inverse.
 === Proof ===
-Suppose $A'$ and $A''$ are both inverses of $A$. Then
-$AA'=I_n=A'A$ and $AA''=I_n=A''A$. So
-\begin{align*} A'&=A'I_n\\&=A'(AA'')\quad\mbox{because }AA''=I_n\\&=(A'A)A''\quad\mbox{because matrix multiplication is associative}\\&=I_nA''\quad\mbox{because }A'A=I_n\\&=A''.\end{align*}
-So $A'=A''$, whenever $A'$ and $A''$ are inverses of $A$. So $A$ has a unique inverse. ■
-==== Definition/notation: $A^{-1}$ ====
+This is a calculation (done in the lectures; you should also check it yourself). ■
-{{page>the inverse}}
+==== Definition: the determinant of a $2\times 2$ matrix ====
-=== Warning ===
+{{page>determinant of a 2x2 matrix}}
-If $A$ is a matrix then $\frac 1A$ doesn't make sense! You should never write this down. In particular, $A^{-1}$ definitely doesn't mean $\frac 1A$.
+==== Theorem: the determinant determines the invertibility (and inverse) of a $2\times 2$ matrix ====
-==== Proposition: solving $AX=B$ when $A$ is invertible ====
+Let $A=\mat{a&b\\c&d}$ be a $2\times 2$ matrix.
-If $A$ is an invertible $n\times n$ matrix and $B$ is an $n\times k$ matrix, then the matrix equation \[ AX=B\] has a unique solution: $X=A^{-1}B$.
+  - $A$ is invertible if and only if $\det(A)\ne0$.
+  - If $A$ is invertible, then $A^{-1}=\frac{1}{\det(A)}\mat{d&-b\\-c&a}$.
 === Proof ===
-First we check that $X=A^{-1}B$ really is a solution to $AX=B$. To see this, note that if $X=A^{-1}B$, then
+If $A=0_{2\times 2}$, then $\det(A)=0$ and $A$ is not invertible. So the statement is true is this special case.
-\begin{align*}
- AX&=A(A^{-1}B)\\&=(AA^{-1})B\\&=I_n B \\&= B.
-\end{align*}
-Now we check that the solution is unique. If $X$ and $Y$ are both solutions, then $AX=B$ and $AY=B$, so \[AX=AY.\] Multiplying both sides on the left by $A^{-1}$, we get
-\[ A^{-1}AX=A^{-1}AY\implies I_nX=I_nY\implies X=Y.\]
-So any two solutions are equal, so $AX=B$ has a unique solution. ■
-==== Corollary ====
+Now assume that $A\ne0_{2\times 2}$ and let $J=\mat{d&-b\\-c&a}$.
-If $A$ is an $n\times n$ matrix and there exist different matrices $X$ and $Y$ so that $AX=AY$, then $A$ is not invertible.
+By the previous lemma, we have \[AJ=(\det(A))I_2=JA.\]
-=== Proof ===
+If $\det(A)\ne0$, then multiplying this equation through by the scalar $\frac1{\det(A)}$, we get
+\[ A\left(\frac1{\det(A)}J\right)=I_2=\left(\frac1{\det(A)}J\right) A,\]
+so if we write $B=\frac1{\det(A)}J$ to make this look simpler, then  we obtain
+\[ AB=I_2=BA,\]
+so in this case $A$ is invertible with inverse $B=\frac1{\det(A)}J=\frac1{\det(A)}\mat{d&-b\\-c&a}$.
-Write $B=AX$. The matrix equation $AX=B$ has two different solutions, $X$ and $Y$. If $A$ was invertible, this would contradict the Proposition. So $A$ cannot be invertible. ■
+If $\det(A)=0$, then $AJ=0_{2\times 2}$  and $J\ne 0_{2\times2}$ (since $A\ne0_{2\times2}$, and $J$ is obtained from $A$ by swapping two entries and multiplying the others by $-1$). Hence by the previous corollary, $A$ is not invertible in this case.  ■
-==== Example ====
-We can now see why the matrix $A=\mat{1&2\\-3&-6}$ is not invertible. If $X=\mat{-2\\1}$ and $Y=\mat{2\\-1}$, then $X\ne Y$ but $AX=0_{2\times 1}$ and $AY=0_{2\times 1}$, so $AX=AY$. So $A$ is not invertible, by the Corollary.