MST10030 notes wiki

Last time

• $A\vec{x}=\vec{b}$ is equivalent to a system of linear equations
• More generally, we might want to solve \[AX=B\] where
  • $A$, $B$ are fixed matrices
  • $X$ is an unknown matrix we want to find.
• Can figure out the size of $X$ from the size of $A$ and $B$
• To solve: find the matrices $X$. How?
• Can't “divide by $A$” (can't divide by a matrix…)

Example

• Consider $AX=B$, where
  • $A=\def\m#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\m{1&0\\0&0}$
  • $B=0_{2\times 3}$,
• i.e. $\m{1&0\\0&0}X=\m{0&0&0\\0&0&0}$
• then any solution $X$ to $AX=B$ must be $2\times 3$.
• One solution is $X=0_{2\times 3}$
• Are there any more?
• Yes! e.g. $X=\m{0&0&0\\1&2&3}$
• So a matrix equation can have more than one solution.

Invertibility

Example

• Take $A=\def\mat#1{\m{#1}}\mat{2&4\\0&1}$ and $B=\mat{3&4\\5&6}$, consider $AX=B$
• $\mat{2&4\\0&1}X=\mat{3&4\\5&6}$
• $X$ must be $2\times 2$ matrix
• One way to solve: write $X=\mat{x_{11}&x_{12}\\x_{21}&x_{22}}$
• Plug in and do matrix multiplication: $\mat{2x_{11}+4x_{21}&2 x_{12}+4x_{22}\\x_{21}&x_{22}}=\mat{3&4\\5&6}$
• Get four linear equations: $\begin{align*}2x_{11}+4x_{21}&=3\\2 x_{12}+4x_{22}&=4\\x_{21}&=5\\x_{22}&=6\end{align*}$
• Can solve in the usual way….
• Tedious! Can we do better?

Simpler: $1\times 1$ matrix equations

• Let $a,b$ be ordinary numbers ($1\times 1$ matrices) with $a\ne0$. How do we solve $ax=b$?
• Answer: multiply both sides by $a^{-1}$
  • (for numbers, $a^{-1}$ is same as $\tfrac1a$)
• Solution: $x=a^{-1}b$.

• Why does this work?
• If $x=\tfrac1a\cdot b$, then $ax=a(\tfrac1a\cdot b)=(a\cdot \tfrac1a)b=1b=b$
• so $ax$ really is equal to $b$
• We do have a solution to $ax=b$.

Thinking about $a^{-1}$ for $a$ a number

• What is special about $a^{-1}$ which made this all work?
• Write $c=a^{-1}$

• $1b=b$, and $a c = 1$
• so $ x=cb$ has $ax=acb=1b=b$

What about matrices?

• Can we do something similar for an $n\times n$ matrix $A$?
• i.e. find a matrix $C$ with similar properties?
  • Replace $1$ with $I_n$
  • Then $I_nB=B$
  • So if we had a matrix $C$ with $CA=I_n$…
  • Then $X=CB$ would have $AX=ACB=I_nB=B$
• The key property of $C$ is that $AC=I_n$.
• Turns out we also want $CA=I_n$

Example revisited

• Let $A=\mat{2&4\\0&1}$
• The matrix $C=\mat{\tfrac12&-2\\0&1}$ does have the property\[ C A =I_2= AC.\]
• Check this!
• Multiply both sides of $AX=B$ on the left by $C$ and use $CA=I_2$:
• Get $X=CB=\mat{\tfrac12&-2\\0&1}\mat{3&4\\5&6} = \mat{-8.5&-10\\5&6}$.
• This is quicker than solving lots of equations
  • although we don't yet know how to find $C$ from $A$
• Also, if we change $B$ we can easily find solutions to the new equation

Definition: invertible

Examples

• $A=\mat{2&4\\0&1}$ is invertible, and $C=\mat{\tfrac12&-2\\0&1}$ is an inverse
• 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 $C=[\tfrac1a]$.
• $I_n$ is invertible for any $n$, with inverse $I_n$
• $0_{n\times n}$ is not invertible for any $n$… why?
• $A=\mat{1&0\\0&0}$ is not invertible… why?
• $A=\mat{1&2\\-3&-6}$ 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

• $A$ 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: $A^{-1}$

Examples again

• $A=\mat{2&4\\0&1}$ 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!

• 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

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

• First check that $X=A^{-1}B$ really is a solution:
  • $AX=A(A^{-1}B)=(AA^{-1})B=I_nB=B$
• Uniqueness: suppose $X$ and $Y$ are both solutions
• Then $AX=B$ and $AY=B$, so $AX=AY$.
• Multiply both sides on the left by $A^{-1}$:
  • $A^{-1}AX=A^{-1}AY$, or $I_nX=I_n Y$, or $X=Y$.
• So any two solutions are equal.
• So $AX=B$ has a unique solution.■

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 $A=\mat{1&2\\-3&-6}$. We can now explain why $A$ isn't invertible.
• Notice that 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.

• Next time: a more systematic way to determine when a matrix is invertible: determinants