08 - Matrix Inverses

Inverses

$$B\text{ is an inverse of }A⟺BA=AB=I$$

Invertible vs Singular ( $A$)

• Invertible: Exists matrix $B$ s.t. $B$ is the inverse of $A$
• Singular: There doesn’t exist an inverse of $A$

Properties of Inverses

1. Uniqueness: If a square matrix $A$ has an inverse then it is unique
   → Notation: $A^{-1}$

2. Product Rule: If $A,B$ are invertible $n\times n$ matrices then $AB$ is invertible with $(AB{)}^{-1}=B^{-1}{A}^{-1}$

3. Involution Rule: If $A$ is invertible then $A^{-1}$ is invertible with $(A^{-1}{)}^{-1}=A$

Proof

1. Let $B,C$ be inverses of $A$ then

$$C=IC=(BA)C=B(AC)=BI=B$$

2. Let $B$ be the inverse of $A$ then

$$(AB{)}^{-1}(AB)=I⟹(AB{)}^{-1}(AB)(B^{-1}{A}^{-1})=I{B}^{-1}{A}^{-1}⟹(AB{)}^{-1}=B^{-1}{A}^{-1}$$

3. Consider

$$(A^{-1})A=A(A^{-1})=I$$

So by definition $(A^{-1}{)}^{-1}=A$ by 1)

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Left and Right Inverses

• Left Inverse $B$ of $A$: $BA=I$
• Right Inverse $C$ of $A$: $AC=I$

If $A$ is non-square then $A$ cannot have both left and right inverses

If $A$ is square then any left or right inverse is also the other inverse!

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Inverses of $2\times 2$ matrices

Let

$$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$$

Iff $ad-bc\ne 0$ then

$$A^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$$

$ad-bc$ is commonly referred to as the determinant of $A$ ($detA$)

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Determining Invertibiilty

Let $A$ be a $n\times n$ matrix

Suppose there exists a sequence of EROs such that $A$ is taken to it’s RRE form $R$
Applying the same sequence of EROs to the augmented matrix $(A\mid I_n)$ to take it to $(R\mid P)$ for some matrix $P$

1. $R=I_n$ then $A$ is invertible and $P=A^{-1}$

2. $R\ne I_n$ then $A$ is singular

Proof

Suppose the finite sequence of EROs that reduce $A$ to $R$ are $E_1,E_2,\cdots ,E_k$
Applying it to $(A|I_n)$ s.t.

$$(E_k{E}_{k-1}\cdots E_{1}A\mid E_k{E}_{k-1}\cdots E_1)=(R\mid P)$$

With $R=PA$ and $P=E_K{E}_{k-1}\cdots E_1$

If $R=I_n$ then as elementary matrices are invertible then

$$PA=(E_k{E}_{k-1}\cdots E_1)A=I_n⟹A^{-1}=E_k{E}_{k-1}\cdots E_1=P$$

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If $R\ne I_n$ then $R$ contains at least one zero row
As in RRE form the zero row is at the bottom then

$$(0,0,\cdots ,1)(PA)=0\text{(selects the last row)}$$

Then as $P$ is invertible and if we assume $A$ is also invertible (i.e. $A^{-1}$) exists then

$$(0,0,\cdots ,1)=0\cdot A^{-1}{P}^{-1}=0$$

This is a contradiction so $A$ is not invertible! Therefore $A$ is singular

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Invertibility of Matrices from Inversible Product corollary

Let $A,B$ be square matrices of the same size
If $AB$ is invertible then

$$A\text{ is invertible and }B\text{ is invertible}$$

Criteria for Invertibility (Equivalent Statements) corollary

Let $A$ be a $n\times n$ matrix

1. $A$ is invertible
2. $A$ has a left inverse
3. $A$ has a right inverse
4. $\mathrm{Row}(A)={\mathbb{R}}^n$
5. The columns of $A$ are linearly independent
6. The rows of $A$ are linearly independent
7. The only solution $\mathbf{x}$ in ${\mathbb{R}}_{\mathrm{col}}^n$ to the system $A\mathbf{x}=0$ is $\mathbf{x}=0$
8. The row rank of $A$ is $n$
9. $\mathrm{RRE}(A)=I_n$

^57bfa7

Invertible Linear Maps and Matrices corollary

Let $V$ be a finite-dimensional vector space
Let $T:V\to V$ be an invertible linear transformation
Let $A$ be a matrix of $T$ with respect to an ordered basis (for both domain and codomain)

Then $A$ is invertible, and $A^{-1}$ is the matrix of $T^{-1}$ with respect to the same basis