# Day 11:

Inverse of a matrix

Abstract vector spaces

### Computing $$B$$ such that $$BA=\text{rref}(A)$$.

Theorem. Let $$A$$ be an $$m\times n$$ matrix. Let $$I$$ be an $$m\times m$$ identity matrix. If

$\text{rref}([A\ |\ I]) = [D\ |\ B]$

then $$BA=\text{rref}(A)$$.

Proof. Note that $\text{rref}([A\ |\ I]) = [\text{rref}(A)\ |\ B].$ Assume $$C$$ is the matrix such that

$C\cdot [A\ \vert\ I] = \text{rref}([A\ \vert\ I])= [\text{rref}(A)\ \vert\ B].$

Finally, note that $C\cdot [A\ \vert\ I] = [CA\ \vert\ CI] = [CA\ \vert\ C],$ and hence $$C=B$$ and $$BA=\text{rref}(A)$$. $$\Box$$

### Computing $$B$$ such that $$BA=\text{rref}(A)$$.

Example.

$\text{rref}\left(\left[\begin{array}{rrr|rr} 2 & 3 & 1 & 1 & 0\\ 2 & 3 & -2 & 0 & 1 \end{array}\right]\right) = \left[\begin{array}{rrr|rr} 1 & 3/2 & 0 & 1/3 & 1/6\\ 0 & 0 & 1 & 1/3 & -1/3 \end{array}\right]$

$\left[\begin{array}{rrr} 1/3 & 1/6\\ 1/3 & -1/3 \end{array}\right]\left[\begin{array}{rrr} 2 & 3 & 1\\ 2 & 3 & -2 \end{array}\right] = \left[\begin{array}{rrr} 1 & 3/2 & 0\\ 0 & 0 & 1\end{array}\right]$

### The Inverse of a matrix

Definition. Given a square matrix $$A$$, a square matrix $$B$$ such that $$AB=BA=I$$ is called the inverse of $$A$$. The inverse of $$A$$ is denoted $$A^{-1}$$. If a matrix $$A$$ has an inverse, then we say that $$A$$ is invertible.

Examples.

• If $$A=\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}$$, then $$A^{-1} = \begin{bmatrix} 1 & -1\\ 0 & 1\end{bmatrix}$$

• If $$A = \begin{bmatrix} 1 & 0 & -2\\ 3 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$, then $$A^{-1} = \begin{bmatrix} 1 & 0 & 2\\ -3 & 1 & -6\\ 0 & 0 & 1\end{bmatrix}$$

Theorem. A matrix $$A$$ is invertible if and only if $$A$$ is square and $$\text{rref}(A)=I$$.

Proof. Suppose $$\text{rref}(A)=I$$. There exists a matrix $$B$$ so that $$BA=\text{rref}(A)$$. From this we deduce that $$BA=I$$. For each elementary matrix $$E$$ there is an elementary matrix $$F$$ such that $$FE=I$$. Since $$B$$ is a product of elementary matrices $B= E_{k}\cdot E_{k-1}\cdots E_{1}$

We take $$F_{i}$$ such that $$F_{i}E_{i}=I$$ for each $$i$$, and set  $C = F_{1}\cdot F_{2}\cdots F_{k}$

and we see that $$CB=I$$. Finally, we have $AB = CBAB =C I B = CB = I.$

Now, suppose $$A$$ is invertible. By definition $$A$$ is square. If $$x$$ is a vector such that $$Ax=0$$, then $$A^{-1}Ax=A^{-1}0$$. Thus, $$x=0$$ is the only solution to $$Ax=0$$. Hence, $$\{0\}=N(A) = N(\operatorname{rref}(A))$$. This implies that each row of $$\operatorname{rref}(A)$$ must have a pivot. Since it is square, every column of $$\operatorname{rref}(A)$$ contains a pivot. This implies $$\operatorname{rref}(A)=I$$. $$\Box$$

Example. Let $A = \begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 0\\ 5 & 7 & 4\end{bmatrix}$

Is $$A$$ invertible? If it is, find $$A^{-1}$$.

Note that $\text{rref}(A)\neq I$ and hence $$A$$ is not invertible.

Note that

$\text{rref}\left(\left[\begin{array}{ccc|ccc} 2 & 3 & 4 & 1 & 0 & 0\\ 3 & 4 & 0 & 0 & 1 & 0\\ 5 & 7 & 4 & 0 & 0 & 1\end{array}\right]\right) = \left[\begin{array}{ccc|ccc} 1 & 0 &-16 & 0 & 7 & -4\\ 0 & 1 & 12 & 0 & -5 & 3\\ 0 & 0 & 0 & 1 & 1 & -1\end{array}\right]$

Example.  Let $A = \begin{bmatrix} 0 & -1 & -2\\ 1 & 1 & 1\\ -1 & -1 & 0\end{bmatrix}$

Is $$A$$ invertible? If it is, find $$A^{-1}$$.

Since $$\text{rref}(A)= I$$ we see that $$A$$ is invertible. Moreover, $A^{-1} = \begin{bmatrix} 1 & 2 & 1\\ -1 & -2 & -2\\ 0 & 1 & 1\end{bmatrix}$

Note that

$\text{rref}\left(\left[\begin{array}{ccc|ccc} 0 & -1 & -2 & 1 & 0 & 0\\ 1 & 1 & 1 & 0 & 1 & 0\\ -1 & -1 & 0& 0 & 0 & 1\end{array}\right]\right) = \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 2 & 1\\ 0 & 1 & 0 & -1 & -2 & -2\\ 0 & 0 & 1 & 0 & 1 & 1\end{array}\right]$

### Vector Spaces

Definition. A vector space (over $$\mathbb{R}$$) is a set $$V$$ together with two operations:

1. vector addition: a map from $$V\times V\to V$$. Given two inputs $$x,y\in V$$, the output of this map is denoted $$x+y$$
2. scalar multiplication: a map from $$\mathbb{R}\times V\to V$$. Given inputs $$a\in\mathbb{R}$$ and $$x\in V$$, the output of this map is denoted $$ax$$

and an element $$0_{V}\in V$$ called the zero vector such that for each $$x,y,z\in V$$ and $$a,b\in\mathbb{R}$$, the following hold:

(a) $$x+y=y+x$$

(b) $$(x+y)+z = x + (y+z)$$

(c) $$0x=0_{V}$$

(d) $$1x=x$$

(e) $$(ab)x = a(bx)$$

(f) $$a(x+y)=ax+ay$$

(g) $$(a+b)x = ax+bx$$

### Examples of Vector Spaces

• $$\R^{n}$$ is a vector space. Vector addition and scalar multiplication are defined as follows: $\begin{bmatrix} x_{1}\\ x_{2}\\ \vdots\\ x_{n}\end{bmatrix} + \begin{bmatrix} y_{1}\\ y_{2}\\ \vdots\\ y_{n}\end{bmatrix} = \begin{bmatrix} x_{1}+y_{1}\\ x_{2}+y_{2}\\ \vdots\\ x_{n}+y_{n}\end{bmatrix}\quad\text{and}\quad a\begin{bmatrix} x_{1}\\ x_{2}\\ \vdots\\ x_{n}\end{bmatrix} = \begin{bmatrix} ax_{1}\\ ax_{2}\\ \vdots\\ ax_{n}\end{bmatrix}$ and $0_{\mathbb{R}^{n}} := [0\ \ 0\ \ \cdots\ \ 0]^{\top}.$
• If $$W\subset\mathbb{R}^{n}$$ is a subspace, then $$W$$ is a vector space with the same vector addition, scalar multiplication, and zero vector as $$\mathbb{R}^{n}$$.

### Examples of Vector Spaces

• For any set $$S$$, consider the set of functions $\mathbb{R}^{S}:= \{f:S\to\mathbb{R}\}.$ Then $$\mathbb{R}^{S}$$ is a vector space with vector addition and scalar multiplication given by $(f+g)(s) := f(s)+g(s), \quad (af)(s):=a\cdot f(s),\quad s\in S$ and with the zero vector given by $$0_{\mathbb{R}^{S}}(s)=0$$ for all $$s\in S$$.
• A particular example of the above vector space: Consider the set $$\mathbb{R}^{\mathbb{N}}$$. This is the set of all functions with domain $$\mathbb{N}$$ and codomain $$\mathbb{R}.$$ For example, if we set $$f(n) := \frac{1}{n}$$ and $$g(n) := 2^{n}$$ for all $$n\in\mathbb{N}$$, then $$f$$ and $$g$$ are vectors in the vector space $$\mathbb{R}^{\mathbb{N}}$$.

### Examples of Vector Spaces

• For $$n\in\N\cup\{0\}$$, set $\mathbb{P}_{n} = \left\{a_{0}+a_{1}x+a_{2}x^2+\cdots+a_{n}x^{n} : a_{0},a_{1},a_{2},\ldots,a_{n}\in\mathbb{R}\right\},$ that is, $$\mathbb{P}_{n}$$ is the set of polynomials of degree $$\leq n$$. Define vector addition and scalar multiplication to be the usual sum of polynomials and multiplication of a polynomial by a scalar, respectively. Let $$0_{\mathbb{P}_{n}}$$ be the zero polynomial. With all of this $$\mathbb{P}_{n}$$ is a vector space.

### Familiar definitions in a new setting

Definition. Let $$V$$ be a vector space.

• A nonempty subset $$W\subset V$$ is a subspace if for every $$x,y\in W$$ and $$a\in\mathbb{R}$$ it follows that $$x+y\in W$$ and $$ax\in W$$.
• A linear combination of $$x_{1},x_{2},\ldots,x_{k}\in V$$, is any vector of the form $a_{1}x_{1}+a_{2}x_{2} + \cdots + a_{k}x_{k}$ where $$a_{1},a_{2},\ldots,a_{k}\in\mathbb{R}.$$
• The span of $$x_{1},x_{2},\ldots,x_{k}\in V$$ is the set of all linear combinations. This set is denoted $$\operatorname{span}\{x_{i}\}_{i=1}^{k}.$$ If $$\operatorname{span}\{x_{i}\}_{i=1}^{k}=V$$, then we say that $$\{x_{i}\}_{i=1}^{k}$$ spans $$V$$.
• The vectors $$\{x_{i}\}_{i=1}^{k}$$ are linearly dependent if there are scalars $$a_{1},a_{2},\ldots,a_{k}\in\mathbb{R}$$, at least one of which is not zero, such that $a_{1}x_{1}+a_{2}x_{2}+\cdots + a_{k}x_{k} = 0_{V}.$ Otherwise, $$\{x_{i}\}_{i=1}^{k}$$ is linearly independent

### Familiar definitions in a new setting

Definition continued. Let $$V$$ be a vector space.

• A linearly independent spanning set for $$V$$ is known as a basis for $$V$$.
• The dimension of $$V$$, denoted $$\operatorname{dim}V$$, is the size of any basis for $$V.$$ If $$V$$ has a finite basis, then we say that $$V$$ is finite-dimensional.

Note that if $$W\subset V$$ is a subspace, then $$W$$ is a vector space with the same vector addition, scalar multiplication, and zero vector as $$V$$. Hence, it also makes sense to talk about a spanning set and a basis for $$W$$.

Note that if $$x_{1},x_{2},\ldots,x_{k}\in V$$, then the set $$\operatorname{span}\{x_{i}\}_{i=1}^{k}$$ is a subspace of $$V$$.

### Linear maps

Definition. Suppose $$V$$ and $$W$$ are vector spaces. A function $$L:V\to W$$ is called linear if for every $$x,y\in V$$ and $$a\in\mathbb{R}$$ the following two properties hold:

1. $$L(x+y) = L(x)+L(y)$$
2. $$L(ax)=aL(x)$$.

Example 1. Let  $A = \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 0\end{bmatrix}.$ Define the function $$L:\mathbb{R}^{3}\to\mathbb{R}^{2}$$ by $$L(x) = Ax.$$ Using the properties of matrix multiplication we proved in Week 1, for any $$x,y\in\mathbb{R}^{3}$$ and $$a\in\mathbb{R}$$ we have $L(x+y) = A(x+y) = Ax+Ay = L(x) + L(y)$ and $L(ax) = A(ax) = aAx = aL(x).$ Thus, $$L$$ is a linear function.

### Linear maps

Example 2. Define the function $$D:\mathbb{P}_{2}\to\mathbb{P}_{1}$$ by $$Df(x) = f'(x)$$ for $$f(x)\in\mathbb{P}_{n}$$. For example,

$D(2x^2+x-1) = 4x+1.$

Using some properties of the derivative from calculus, if $$f(x),g(x)\in\mathbb{P}_{2}$$ and $$c\in\mathbb{R}$$, then

$D(f(x) + g(x)) = f'(x) + g'(x) = D(f(x)) + D(g(x))$

and

$D(cf(x)) = cf'(x) = cD(f(x)).$

Therefore, $$D$$ is linear.

Example 3.  The function $$F:\mathbb{R}\to\mathbb{R}$$ given by $$F(x) = 2x+1$$ is not linear, since

$F(1+1)=F(2) = 2(2)+1=5\neq 6 = F(1)+F(1).$

### Image and Kernel

Definition. Suppose $$V$$ and $$W$$ are vector spaces, and $$L:V\to W$$ is linear. The image of $$L$$ is the set

$\operatorname{im}(L) := \{Lv : v\in V\}.$

The kernel of $$L$$ is the set

$\operatorname{ker}(L):= \{v : L(v) = 0\}.$

If $$A\in\mathbb{R}^{m\times n}$$, and we set $$L(x) = Ax$$ for $$x\in\mathbb{R}^{n}$$, then $$L:\mathbb{R}^{n}\to\mathbb{R}^{m}$$ is linear, moreover,

$\operatorname{im}(L) = C(A)\quad\text{and}\quad \operatorname{ker}(L) = N(A).$

By John Jasper

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