Invertible matrices

Equivalent Statement to Invertible

Let $A$ be an invertible $m\times m$ matrix:

1. $T_A:{\mathbb{R}}^m\to {\mathbb{R}}^m$ is bijective
2. There is an $m\times m$ matrix $B$ such that $BA=I$
3. The columns of $A$ are linearly independent

Proof $S_2⟹S_3$ Assume $A\mathbf{x}=0$ $\mathbf{x}=(BA)\mathbf{x}=B(A\mathbf{x})=B0=0$ Since $A\mathbf{x}=0$ has only the trivial solution $\mathbf{x}=0$, the columns of $A$ are linearly independent (See Equivalent Statements to linear independence Statement 2)

NOTE

If $A$ is not invertible, it is called singular

Lemma 2.54

Let $A,B$ be $m\times m$ matrices such that $BA=I$. Then also $AB=I$ According to $S_1$ in Equivalent Statement to Invertible, we know $T_A$ is bijective. So for every $\mathbf{y}$ we have some $\mathbf{x}$ with $A\mathbf{x}=\mathbf{y}$ $AB\mathbf{y}=AB(A\mathbf{x})=A(BA)\mathbf{x}=A\mathbf{x}=\mathbf{y}$ Thus, we must have $AB=I$

Observation 2.56

$A$ is invertible if and only if there is an $B$ satisfying one of the following conditions (and therefore all): i)$AB=I$ ii)$BA=I$ iii)$AB=BA=I$ Proof of uniqueness of $B$

$$B=IB\stackrel{(ii)}{=}(B^{\prime }A)B=B^{\prime }(AB)\stackrel{i}{=}{B}^{\prime }I=B^{\prime }$$

Example

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

Lemma 5.2.4

Let $A\in {\mathbb{R}}^{m\times n}$ then $A^{\top }A$ is invertible if and only if columns of $A$ are linear independent

Proof: $N(A^{\top }A)=N(A)$ $A^{\top }A\in {\mathbb{R}}^{n\times n}$ $A^{\top }A$ invertible $⟺N(A^{\top }A)=\{0\}$ $⟺N(A)=\{0\}$ $⟹$ columns of $A$ are linear independent