Unit 3

Inverse Matrix

• Only for square matrix

Let A be $n\times n$. The inverse of A is a matrix B satisfying the following properties

1. $AB=I_N$
2. $BA=I_N$

We denote it $B=A^{-1}$

Example

So $A^{-1}=\left[\begin{array}{cc}2& -5\\ -1& 3\end{array}\right]$

General Formula

For $A2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, the inverse is $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

The inverse exists provided that $ad-bc\ne 0$

Suppose $A^{-1}$ exists, where A is $n\times n$.

• Consider $A\vec{x}=\vec{b}$
• Multiply by $A^{-1}$: $A^{-1}(A\vec{x})=A^{-1}\vec{b}$ = $I_N\vec{x}=A^{-1}\vec{b}$
• $\vec{x}=A^{-1}\vec{b}$
• Therefore, if $A^{-1}$ exists then the system $A\vec{x}=\vec{b}$ has a unique solution which $\vec{x}=A^{-1}\vec{b}$ for any $\vec{b}$

Theorem

The following statements are equivalent (all true or false) for $A$ that is $n\times n$:

• $A^{-1}$ exists
• $rank(A)=N$
• $A\vec{x}=\vec{b}$ Has a solution for all $\vec{b}$
• $A\vec{x}=\vec{0}$ Has only the trivial solution $\vec{x}=\vec{0}$
• The RREF of A is $I_N$

Suppose $A\vec{x}=\vec{b}$ has a solution for all $\vec{b}$. Let ${\vec{b}}_i$ be the $i-th$ column of the identity matrix. Let $\overrightarrow{c_i}$ be the solution of $A\vec{x}=\overrightarrow{b_i}$

To find $A^{-1}$:

Reduce it to:

Example

Recall $(AB{)}^T=B^T{A}^T$. Suppose $A^{-1}$ exists: $A{A}^{-1}=I_N$.

Then $(A{A}^{-1}{)}^T=(I_N{)}^T=I_N$.

Then $(A^{-1}{)}^T{A}^T=I_N$

Therefore, $(A^T{)}^{-1}=(A^{-1}{)}^T$

Note, if $A=A^T$ (i,e, $A$ is symmetric)

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

If A is symmetric, then $A^{-1}$

Subspace

A subspace $W$ a set of vectors of ${\mathbb{R}}^N$ satisfying:

• $W$ Is closed under addition: if $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ are in $W$. Then $\overrightarrow{v_1}+\overrightarrow{v_2}$ is in $W$.
• $W$ Is closed under scalar products: if $\vec{v}$ is in $W$ and a is a scalar, then $a\vec{v}$ is in $W$

Note: $W$ is closed under linear combinations

Note: $\vec{0}$ is in every subspace

Subspaces of ${\mathbb{R}}^2$

1. Each line through the origin is a subspace
2. $\{\vec{0}\}$ Is a subspace of ${\mathbb{R}}^N$ - trivial subspace
3. The whole space, ${\mathbb{R}}^2$, is also a subspace

Theorem

$span(\overrightarrow{v_1},\overrightarrow{v_2},...,\overrightarrow{v_k})$ is a subspace