- 2.3.1: Multiplicative Inverse
- 2.3.2: Non-singular Matrices
- 2.3.3: Computing Inverses
- Video Lecture 1 - Multiplication and Inverses
- Theorem 3 - Inverses are Well-Defined
- Proof - Inverses are Well-Defined
- Theorem 4 - Product Inverse
- Proof - Product Inverse
- Corollary 5 - Inverse of Finite Products
- Exercise 5 - Prove Corollary 5
- Theorem 6 - Inverse Transpose
- Exercise 6 - Prove Theorem 6
- 2.3.4: Computing Inverses by Row Reduction
- 2.3.5: Exercises
- copyleft
As we learned in the last section, the identity matrix $I_n$ represents the identity function $\mathfrak{I}$ in $\mathbb{R}^{n\times n}$. For every vector $\mathbf{x}\in\mathbb{R}^n$, the domain of function $\mathfrak{I}$, $\mathfrak{I}(\mathbf{x}) = \mathbf{x}$. Similarly,
$$ I_n\mathbf{x} = \mathbf{x} $$
Suppose that $L$ is a linear transformation on $\mathbb{R}^n$. That means $L\,:\,\mathbb{R}^n\rightarrow\mathbb{R}^n$, and there is a matrix $A\in\mathbb{R}^{n\times n}$ so that
$$ L(\mathbf{x}) = A\mathbf{x} $$
If we compose $L$ with the identity function $\mathfrak{I}$, then $L\circ\mathfrak{I} = \mathfrak{I}\circ L = L$. That is, for every vector $\mathbf{x}\in\mathbb{R}^n$, $(L\circ\mathfrak{I})(\mathbf{x}) = (\mathfrak{I}\circ L)(\mathbf{x}) = L(\mathbf{x})$. For the $n\times n$ identity matrix
$$ AI_n = I_nA = A. $$
We say that $I_n = [\delta_{ij}]$ is the multiplicative identity matrix. In scalar algebra every nonzero scalar $a\neq 0$ has a multiplicative inverse $a^{-1} = \frac{1}{a}$ so that
$$ a\cdot a^{-1} = a\cdot\frac{1}{a} = 1 = \frac{1}{a}\cdot a = a^{-1}\cdot a $$
Question¶
Does every nonzero $n\times n$ matrix have a multiplicative inverse?
The answer is no. Matrices and the linear transformations (functions) they represent are more complicated than scalars. Unlike scalars, there are nonzero functions that have no inverse, and there are nonzero $n\times n$ matrices with no multiplicative inverse.
Example 1 - No Multiplicative Inverse¶
$$ A = \begin{bmatrix}\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ \end{bmatrix} $$
This matrix has no multiplicative inverse. It is already in reduced row echelon form and it has a free column. Let us try to find a multiplicative inverse say
$$ B = \begin{bmatrix}\ \ a\ &\ \ b\ \\ \ \ c\ &\ \ d\ \end{bmatrix} $$
If $AB = I_2$, then
$$ \begin{bmatrix}\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ \end{bmatrix}\begin{bmatrix}\ \ a\ &\ \ b\ \\ \ \ c\ &\ \ d\ \end{bmatrix} = \begin{bmatrix}\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 1\ \end{bmatrix} $$
This gives us a linear system of four equations and four unknowns
$$ \begin{array}{rrrrrrrrcl} a & & & & + & 0c & & & = & 1 \\ 0a & & & & + & 0c & & & = & 0 \\ & & b & & + & & + & 0d & = & 0 \\ & & 0b & & + & & + & 0d & = & 1 \end{array} $$
The augmented matrix is given by
$$ \begin{align*} \begin{bmatrix} 1 & 0 & 0 & 0 & | & 1 \\ 0 & 0 & 0 & 0 &| & 0 \\ 0 & 1 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & | & 1 \end{bmatrix} &\rightarrow\begin{bmatrix} 1 & 0 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & | & 1 \\ 0 & 0 & 0 & 0 &| & 0 \end{bmatrix} \end{align*} $$
This linear system is inconsistent due to the third row so there is no matrix $B$ so that $AB = I_2$.
Definition¶
Invertible
If an $n\times n$ matrix $A\in\mathbb{R}^{n\times n}$ has a multiplicative inverse $B$ so that
$$ AB = BA = I_n $$
then we say that matrix $A$ is invertible.
If matrix $A\in\mathbb{R}^{n\times n}$ has no multiplicative inverse we say that matrix $A$ is non-invertible.
We never put a matrix in the denominator of an algebraic expression.
Now we must develop some tools for determining when a square matrix is non-singular, and computing the inverse if it exists
Question¶
Why doesn't matrix $A = \begin{bmatrix}\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ \end{bmatrix}$ in Example 1 have a multiplicative inverse?
There are two viewpoints that we need to consider about this matrix to understand why this matrix is non-invertible.
1. First let us define linear transformation $F\,:\,\mathbb{R}^n\rightarrow\mathbb{R}^n$ by $F(\mathbf{x})=A\mathbf{x}$ for every $\mathbf{x}\in\mathbb{R}^n$. Does $F$ have an inverse function? Is there a function $F^{-1}\,:\,\mathbb{R}^n\rightarrow\mathbb{R}^n$ so that $F\circ F^{-1} = F^{-1}\circ F = \mathfrak{I}$? Recall that in our pre-calculus algebra, a function needs to be __one-to-one__ and __onto__ in order to have a functional inverse.
The easiest way to show that $F$ is not one-to-one is to demonstrate it with an example. If we take any vector on the $x_2$-axis, such as $\langle 0, y \rangle$, then
$$ F(\langle 0, y \rangle) = A\begin{bmatrix}\ 0\ \\ \ y\ \end{bmatrix} = \begin{bmatrix}\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ \end{bmatrix}\begin{bmatrix}\ 0\ \\ \ y\ \end{bmatrix} = \begin{bmatrix}\ 0\ \\ \ 0\ \end{bmatrix} $$
Therefore, $F(\langle 0, 1 \rangle) = \mathbf{0} = F(\langle 0, 2 \rangle)$ and $F$ is not one-to-one. This means that there is no matrix $A^{-1}$ that represents $F^{-1}$. Neither inverse exists.
2. The second viewpoint is to look at the matrix. In light of our previous observations, multiplication by matrix $A$ results in a vector on the $x_1$-axis (It is the projection of the vector onto the $x_1$-axis). If we let $\langle z, y \rangle$ be any vector in $\mathbb{R}^2$. Then $A\langle z, y \rangle = \langle z, 0 \rangle$. Multiplying by matrix $A$ _squeezes_ two dimensional space onto the horizontal axis. In chapter one, we associated this geometrical phenomenon to the existence of a free column. The _loss_ of information when we squeeze space; the existence of a free column also informs us the operation is not one-to-one. The free column notifies us that the matrix has no multiplicative inverse.
Exercise 1 - Invertible Matrix¶
For matrices
$$ B = \begin{bmatrix}\ \ 5\ & -6\ \\ \ \ 7\ &\ \ 7\ \end{bmatrix},\qquad C = \begin{bmatrix}\ \ 2\ &\ \ 1\ & -6\ \\ -7\ &\ \ 8\ &\ \ 8\ \\ -4\ &\ \ 8\ &\ \ 8\ \end{bmatrix},\qquad\text{and }D = \begin{bmatrix}\ \ 0\ &\ \ 7\ & -8\ &\ \ 1\ \\ \ \ 5\ &\ \ 5\ &\ \ 6\ &\ \ 4\ \\ -6\ &\ \ 8\ &\ \ 7\ & -2\ \\ -1\ &\ \ 3\ &\ \ 3\ &\ \ 3\ \end{bmatrix}, $$
determine which matrices are invertible and which are non-invertible.
View Solution
$$ \begin{align*} B = \begin{bmatrix}\ \ 5\ & -6\ \\ \ \ 7\ &\ \ 7\ \end{bmatrix}\begin{array}{c} \\ R_2-R_1 \end{array} &\longrightarrow \begin{bmatrix}\ \ 5\ & -6\ \\ \ \ 2\ &\ \ 13\ \end{bmatrix}\begin{array}{c} R_1-2R_2 \\ \\ \end{array} \longrightarrow \begin{bmatrix}\ \ 1\ & -32\ \\ \ \ 2\ &\ \ 13\ \end{bmatrix}\begin{array}{c} \\ R_2-2R_1 \end{array} \\ &\longrightarrow \begin{bmatrix}\ \ 1\ & -32\ \\ \ \ 0\ &\ \ 77\ \end{bmatrix} \end{align*} $$Both columns of matrix $B$ are pivot columns, so the matrix is invertible.
$$ \begin{align*} C = \begin{bmatrix}\ \ 2\ &\ \ 1\ & -6\ \\ -7\ &\ \ 8\ &\ \ 8\ \\ -4\ &\ \ 8\ &\ \ 8\ \end{bmatrix}\begin{array}{c} \\ R_2+3R_1 \\ R_3+2R_1 \end{array} &\longrightarrow \begin{bmatrix}\ \ 2\ &\ \ 1\ & -6\ \\ -1\ &\ 11\ & -10\ \\ \ \ 0\ &\ 10\ & -4\ \end{bmatrix}\begin{array}{c} R_1+R_2 \\ \\ \\ \end{array} \longrightarrow \begin{bmatrix}\ \ 1\ &\ 12\ & -16\ \\ -1\ &\ 11\ & -10\ \\ \ \ 0\ &\ 10\ & -4\ \end{bmatrix}\begin{array}{c} \\ R_2+R_1 \\ \\ \end{array} \\ &\longrightarrow \begin{bmatrix}\ \ 1\ &\ 12\ & -16\ \\ \ \ 0\ &\ 23\ & -26\ \\ \ \ 0\ &\ 10\ & -4\ \end{bmatrix}\begin{array}{c} \\ R_2-2R_3 \\ \\ \end{array} \longrightarrow \begin{bmatrix}\ \ 1\ &\ 12\ & -16\ \\ \ \ 0\ &\ \ 3\ & -18\ \\ \ \ 0\ &\ 10\ & -4\ \end{bmatrix}\begin{array}{c} \\ \\ R_3-3R_2 \end{array} \\ &\longrightarrow \begin{bmatrix}\ \ 1\ &\ 12\ & -16\ \\ \ \ 0\ &\ \ 3\ & -18\ \\ \ \ 0\ &\ 1\ &\ 50\ \end{bmatrix}\begin{array}{c} \\ R_3 \\ R_2 \\ \end{array}\qquad\quad \longrightarrow \begin{bmatrix}\ \ 1\ &\ 12\ & -16\ \\ \ \ 0\ &\ 1\ &\ 50\ \\ \ \ 0\ &\ \ 3\ & -18\ \end{bmatrix}\begin{array}{c} \\ \\ R_3-3R_2 \end{array} \\ &\longrightarrow \begin{bmatrix}\ \ 1\ &\ 12\ & -16\ \\ \ \ 0\ &\ 1\ &\ 50\ \\ \ \ 0\ &\ \ 0\ & -168\ \end{bmatrix} \end{align*} $$
All three columns of matrix $C$ are pivot columns so matrix $C$ is invertible
$$ \begin{align*} D = \begin{bmatrix}\ \ 0\ &\ \ 7\ & -8\ &\ \ 1\ \\ \ \ 5\ &\ \ 5\ &\ \ 6\ &\ \ 4\ \\ -6\ &\ \ 8\ &\ \ 7\ & -2\ \\ -1\ &\ \ 3\ &\ \ 3\ &\ \ 3\ \end{bmatrix}\begin{array}{c} -R_4 \\ \\ \\ R_1 \end{array} &\longrightarrow \begin{bmatrix}\ \ 1\ & -3\ & -3\ & -3\ \\ \ \ 5\ &\ \ 5\ &\ \ 6\ &\ \ 4\ \\ -6\ &\ \ 8\ &\ \ 7\ & -2\ \\ \ \ 0\ &\ \ 7\ & -8\ &\ \ 1\ \end{bmatrix}\begin{array}{c} \\ R_2-5R_1 \\ R_3+6R_1\\ \\ \end{array} \longrightarrow \begin{bmatrix}\ \ 1\ & -3\ & -3\ & -3\ \\ \ \ 0\ &\ 20\ &\ 21\ &\ 19\ \\ \ \ 0\ & -10\ & -11\ & -20\ \\ \ \ 0\ &\ \ 7\ & -8\ &\ \ 1\ \end{bmatrix}\begin{array}{c} \\ R_2+R_3 \\ \\ \\ \end{array} \\ &\longrightarrow \begin{bmatrix}\ \ 1\ & -3\ & -3\ & -3\ \\ \ \ 0\ &\ 10\ &\ 10\ & -1\ \\ \ \ 0\ & -10\ & -11\ & -20\ \\ \ \ 0\ &\ \ 7\ & -8\ &\ \ 1\ \end{bmatrix}\begin{array}{c} \\ \\ R_3+R_2 \\ \\ \end{array} \longrightarrow \begin{bmatrix}\ \ 1\ & -3\ & -3\ & -3\ \\ \ \ 0\ &\ 10\ &\ 10\ & -1\ \\ \ \ 0\ &\ \ 0\ & -1\ & -21\ \\ \ \ 0\ &\ \ 7\ & -8\ &\ \ 1\ \end{bmatrix}\begin{array}{c} \\ R_2-R_4 \\ -R_3 \\ \\ \end{array} \\ &\longrightarrow \begin{bmatrix}\ \ 1\ & -3\ & -3\ & -3\ \\ \ \ 0\ &\ \ 3\ &\ 18\ & -2\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ 21\ \\ \ \ 0\ &\ \ 7\ & -8\ &\ \ 1\ \end{bmatrix}\begin{array}{c} \\ \\ \\ R_4-2R_2 \end{array} \longrightarrow \begin{bmatrix}\ \ 1\ & -3\ & -3\ & -3\ \\ \ \ 0\ &\ \ 3\ &\ 18\ & -2\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ 21\ \\ \ \ 0\ &\ \ 1\ & -44\ &\ \ 5\ \end{bmatrix}\begin{array}{c} \\ R_2-3R_4 \\ \\ \\ \end{array} \\ &\longrightarrow \begin{bmatrix}\ \ 1\ & -3\ & -3\ & -3\ \\ \ \ 0\ &\ \ 0\ &\ 150\ & -17\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ 21\ \\ \ \ 0\ &\ \ 1\ & -44\ &\ \ 5\ \end{bmatrix}\begin{array}{c} \\ R_4 \\ \\ R_2 \end{array} \longrightarrow \begin{bmatrix}\ \ 1\ & -3\ & -3\ & -3\ \\ \ \ 0\ &\ \ 1\ & -44\ &\ \ 5\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ 21\ \\ \ \ 0\ &\ \ 0\ &\ 150\ & -17\ \end{bmatrix}\begin{array}{c} \\ \\ \\ R_4-150R_3 \end{array} \\ &\longrightarrow \begin{bmatrix}\ \ 1\ & -3\ & -3\ & -3\ \\ \ \ 0\ &\ \ 1\ & -44\ &\ \ 5\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ 21\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ & -3167\ \end{bmatrix} \end{align*} $$
All four columns of matrix $D$ are pivot columns, so matrix $D$ is invertible.
We have seen another application of reducing a matrix using row operations. This is an important skill that you must practice throughout the course.
Definition¶
Non-Singular Matrices
If all of the columns of a square matrix are pivot columns, the matrix is called non-singular or a non-singular matrix. Non-singular matrices are also called non-degenerate.
If a square matrix has a free column, the matrix is called singular or a singular matrix. Singular matrices are also called degenerate.
Notice that a rectangular matrix, $m\neq n$, cannot be described as singular, invertible, or non-degenerate. Although one may find non-singular spelled without a hyphen in some articles and texts, the Oxford-English dictionary says it has a hyphen.
Throughout this course we will learn new vocabulary for describing $m\times n$ matrices, and the special class of square matrices. There will be many instances where matrices will share properties. The list of properties that are equivalent to non-singular is substantial. This is mainly due to different researchers developing vocabulary for their particular application in different fields of STEM, in different parts of the world, and at different times. It is up to us to understand when these properties describe the same kinds of matrices or linear systems.
Theorem 1¶
Properties Equivalent to Non-Singular
Given a square matrix $A\in\mathbb{R}^{n\times n}$ all of the following properties are equivalent:
- $A$ is a non-singular matrix (non-degenerate)
- all of the columns of $A$ are pivot columns
- The homogeneous linear system $A\mathbf{x} = 0$ has a unique solution
- $A$ is an invertible matrix
Proof of Theorem 1¶
In a theorem like this, every statement is equivalent to the other statements means that the logical implication, "Statement 1 implies Statement 2" is true. So are all of the $12$ different possible logical statements "Statement $j$ implies $k$". One of the simplest ways to prove all 12 statements is to prove that "1 implies 2", "2 implies 3", "3 implies 4", and "4 implies 1". Proving these 4 statements convinces one that the other 8 are true also; one just follows the circle of implications.
Suppose matrix $A$ is an $n\times n$ matrix.
$1\implies 2$¶
If matrix $A$ is non-singular, then by definition all of the columns of matrix $A$ are pivot columns.
$2\implies 3$¶
If all of the columns of matrix $A$ are pivot columns, then consider the linear system $A\mathbf{x}=\mathbf{0}$. This linear system has the solution $\mathbf{x}=\mathbf{0}$ because $A\mathbf{0}$ is a sum of the scalar zero times each column of matrix $A$. Since matrix $A$ has no free columns, this is the only solution.
$3\implies 4$¶
Suppose that the linear system $A\mathbf{x}=\mathbf{0}$ has a unique solution. If $\mathbf{b}\in\mathbb{R}^n$ is a nonzero vector, then either $A\mathbf{x}=\mathbf{b}$ is consistent or inconsistent. Assume that $A\mathbf{x}=\mathbf{b}$ is inconsistent. Then the reduced row echelon form of the augmented matrix $\left[\,A\,|\,\mathbf{b}\,\right]$ has a row $k$ with $n$ zeros followed by a nonzero entry in the $k^{\text{th}}$ row of the reduced form of $\mathbf{b}$. Thus matrix $A$ has a free column, and $A\mathbf{x}=\mathbf{0}$ has infinitely many solutions. This contradicts the fact $A\mathbf{x}=\mathbf{0}$ has only one solution. Thus for any $\mathbf{b}\in\mathbb{R}^n$, $A\mathbf{x}=\mathbf{b}$ is consistent.
Suppose that $A\mathbf{x}=\mathbf{b}$ has infinitely many solutions. Then there are nonzero vector $\mathbf{u}$ and $\mathbf{v}$ so that $A\mathbf{u}=\mathbf{b}=A\mathbf{v}$, and $\mathbf{u}\neq\mathbf{v}$. This implies that $A\left(\mathbf{u}-\mathbf{v}\right) = \mathbf{0}$ and $\mathbf{u}-\mathbf{v}\neq\mathbf{0}$. However, this contradicts the fact that $A\mathbf{x}=\mathbf{0}$ has a unique solution again. Thus for every $\mathbf{b}\in\mathbb{R}^n$, the linear system $A\mathbf{x}=\mathbf{b}$ has a unique solution.
For each elementary basis vector $\mathbf{e}_k$, $1\le k\le n$, let $\mathbf{c}_k$ be the unique solution for $A\mathbf{x}=\mathbf{e}_k$, and construct $C = \begin{bmatrix} \mathbf{c}_1 & \cdots & \mathbf{c}_n \end{bmatrix}$. We have that
$$ AC = \begin{bmatrix} A\mathbf{c}_1 & \cdots & A\mathbf{c}_n \end{bmatrix} = \begin{bmatrix} \mathbf{e}_1 & \cdots & \mathbf{e}_n \end{bmatrix} = I_n. $$
Hence $AC = I_n$ and matrix $A$ is invertible.
$4 \implies 1$¶
Suppose that matrix $A$ is invertible, and $B$ is an $n\times n$ matrix so that $BA = I_n$. Each row of $BA$ is a linear combination of the rows of matrix $A$, and $BA = I$. This means that the reduced row echelon form of matrix $A$ is the identity matrix $I_n$, and all of the columns of matrix $A$ are pivot columns. Hence matrix $A$ is non-singular.
This completes the proof. $\blacksquare$
A corollary is a theorem that is easily derived from a previous theorem. We can state the properties of Theorem 1 in the reverse direction.
Corollary 2¶
Singular Matrices
Given a square matrix $A\in\mathbb{R}^{n\times n}$ all of the following properties are equivalent:
- $A$ is a singular matrix (degenerate)
- $A$ has a free column
- The homogeneous linear system $A\mathbf{x} = 0$ has infinitely many solutions
- $A$ is not an invertible matrix
Proof of Corollary 2¶
The contrapositives of each of the statements proven in Theorem 1 are $2\implies 1$, $3\implies 2$, $4\implies 3$, and $1\implies 4$. $\blacksquare$
Exercise 2 - Inverses are Non-Singular¶
Prove that if matrix $A$ is non-singular, then matrix $A^{-1}$ is also non-singular.
View Solution
Proof: If matrix $A$ is non-singular, then it is invertible with unique inverse $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. This computation shows that $A^{-1}$ is invertible as well. By Theorem 1, $A^{-1}$ is non-singular. $\blacksquare$Exercise 3 - Non-singular Matrices are Consistent Linear Systems¶
Prove that if $A$ is a non-singular matrix and $\mathbf{b}\in\mathbb{R}^n$, then the linear system $A\mathbf{x}=\mathbf{b}$ has a unique solution.
View Solution
Proof: The proof is embedded in the proof of Theorem 1. If $A$ is a non-singular matrix, then for $1\le k\le n$ and each elementary basis vector of $\mathbb{R}^n$, the linear system $A\mathbf{x}=\mathbf{0}$ has a unique solution. Hence the linear system $A\mathbf{x}=\mathbf{b}$ also has a unique solution. If we assume otherwise, then we obtain a contradiction as in the proof of Theorem 1. $\blacksquare$Exercise 4 - Non-singular Matrices and Identity¶
Prove that the reduced row echelon form of a non-singular matrix is an identity matrix.
View Solution
Proof: If $A$ is a non-singular matrix, then for $1\le k\le n$ let $\mathbf{c}_k$ be the unique solution to $A\mathbf{x}=\mathbf{e}_k$ for each elementary basis vector $\mathbf{e}_k$. Construct $n\times n$ matrix $C = \begin{bmatrix} \mathbf{c}_1 & \cdots & \mathbf{c}_n \end{bmatrix}$. Then$$ AC = A\begin{bmatrix} \mathbf{c}_1 & \cdots & \mathbf{c}_n \end{bmatrix} = \begin{bmatrix} A\mathbf{c}_1 & \cdots & A\mathbf{c}_n \end{bmatrix} = \begin{bmatrix} \mathbf{e}_1 & \cdots & \mathbf{e}_n \end{bmatrix} = I_n $$ Thus non-singular matrix $C$ is the unique inverse of $A$, and $CA = I$. Each row of matrix $CA$ is a linear combination of the rows of $A$. This shows that the reduced row echelon form of $A$ is $I_n$. $\blacksquare$
Video Lecture 1: Multiplication and Inverses¶
Now you can watch this video by Dr. Strang starting at 21:20 minutes if you skipped inverses the first time you watched the video for matrix multiplication.
Matrix MultiplicationTheorem 3¶
Inverses of Square Matrices are Well-Defined
Let $A$ be an $n\times n$ non-singular matrix, and $B$ be an $n\times n$ matrix so that $AB = I_n$. Then $B$ is invertible, $BA = I_n$, and $B$ is the unique matrix with the property that $AB = BA = I_n$.
Proof of Theorem 3¶
Suppose that $A$ is an $n\times n$ non-singular matrix, and $B$ is an $n\times n$ matrix so that $AB = I_n$.
1. Given this we have
$$ A(BA) = (AB)A = (I_n)A = A $$
Thus $A(BA) - A = A(BA - I_n) = O$ ($O$ is the zero matrix). If we define $C = BA-I$, then
$$ AC = A\begin{bmatrix} \mathbf{c}_1 & \cdots & \mathbf{c}_n \end{bmatrix} = \begin{bmatrix} \mathbf{0} & \cdots & \mathbf{0} \end{bmatrix} $$
Using Theorem 1, for each $1\le k\le n$, the linear system $A\mathbf{c}_k = \mathbf{0}$ has a unique solution $\mathbf{c}_k=\mathbf{0}$. Hence $C=O$. Thus $BA-I_n=O$, or $BA = I$. This proves that the right inverse of a square matrix is also the left inverse.
2. Now that we have $AB = BA = I_n$, we also have that $B$ is an invertible matrix with inverse $A$.
3. Finally, suppose that there are two matrices $B$ and $D$ so that $AB = AD = I_n$. Then $AB - AD = A(B - D) = I_n - I_n = O$. Using the same argument as in part 1, $B-D=O$ and $B=D$. Hence the inverse of a square matrix is unique.
This establishes that the multiplicative inverse of a non-singular matrix is unique and unlike most matrix multiplication, the non-singular matrix and its inverse commute. $\blacksquare$
As a result, we denote the unique multiplicative inverse of a non-singular matrix $A$ by $A^{-1}$. This notation is unambiguous for square matrices.
Theorem 4¶
If $A$ and $B$ are non-singular $n\times n$ matrices, then the product $AB$ is also non-singular.
Proof of Theorem 4¶
Proof:¶
Since $A$ and $B$ are non-singular they have multiplicative inverses, $A^{-1}$ and $B^{-1}$.
$$ \begin{align*} (AB)(B^{-1}A^{-1}) &= A(BB^{-1})A^{-1} & &\text{Matrix product is associative} \\ &= A(I)A^{-1} & &\text{$B$ and $B^{-1}$ are inverses} \\ &= (AI)A^{-1} & &\text{Matrix product is associative} \\ &= AA^{-1} & &\text{I is the matrix product identity} \\ &= I & &\text{$A$ and $A^{-1}$ are inverses} \end{align*} $$
Furthermore,
$$ \begin{align*} (B^{-1}A^{-1})(AB) &= B^{-1}(A^{-1}A)B & &\text{Matrix product is associative} \\ &= B^{-1}(I)B & &\text{$A^{-1}$ and $A$ are inverses} \\ &= B^{-1}(IB) & &\text{Matrix product is associative} \\ &= B^{-1}B & &\text{I is the matrix product identity} \\ &= I & &\text{$B^{-1}$ and $B$ are inverses} \end{align*} $$
So there is an $n\times n$ matrix $B^{-1}A^{-1}$ that is the multiplicative inverse of product matrix $AB$. Thus the product $AB$ is non-singular and
$$ (AB)^{-1} = B^{-1}A^{-1}.\ \blacksquare $$
Notice that we show that matrix $AB$ is non-singular by producing its multiplicative inverse.
Corollary 5¶
If $A_1$, $A_2$, $\dots$, $A_k$ are all non-singular $n\times n$ matrices, then the product $A_1A_2\cdots A_k$ is a non-singular matrix and
$$ \left(A_1A_2\cdots A_k\right)^{-1} = A_k^{-1}\cdots A_2^{-1}A_1^{-1}. $$
Exercise 5 - Prove Corollary 5¶
Prove Corollary 5 using mathematical induction.
View Solution
1. Step one is to prove Corollary 5 for $k=2$. Theorem 4 proves that if $A_1$ and $A_2$ are non-singular matrices, then $(A_1A_2)^{-1} = A_2^{-1}A_1^{-1}$.2. Show that if Corollary 5 is true for $k=m$, then it is also true for $k=m+1$. Suppose that for positive integer $m\ge2$ Corollary 5 is true. Let $A_1$, $A_2$, $\dots$, $A_{m+1}$ be $m+1$ non-singular $n\times n$ matrices. Then by hypothesis, $A_1A_2\cdots A_m$ is non-singular, and $(A_1A_2\cdots A_m)^{-1} = A_m^{-1}\cdots A_2^{-1}A_1^{-1}$. Furthermore,
$$ (A_{m+1}^{-1}A_m^{-1}\cdots A_1^{-1})(A_1\cdots A_mA_{m+1}) = A_{m+1}^{-1}(A_m^{-1}\cdots A_1^{-1})(A_1\cdots A_m)A_{m+1} = A_{m+1}^{-1}I_nA_{m+1} = I_n $$
Thus $A_1\cdots A_mA_{m+1}$ is non-singular, and $(A_1\cdots A_mA_{m+1})^{-1} = A_{m+1}^{-1}A_m^{-1}\cdots A_1^{-1}$.
3. Invoke the Principle of Mathematical Induction. Therefore, for all positive integers $k\ge 2$, the Corollary is true. $\blacksquare$
Theorem 6¶
If matrix $A$ is non-singular, then $A^T$ is non-singular. Furthermore, $\left(A^T\right)^{-1} = \left(A^{-1}\right)^T$.
Exercise 6 - Prove Theorem 6¶
Prove Theorem 6 by producing the inverse.
View Solution
If matrix $A\in\mathbb{R}^{n\times n}$ is non-singular, then it is invertible and there is an inverse matrix $A^{-1}$ so that $A^{-1}A = AA^{-1} = I_n$. Consider the transpose of $A^{-1}$, $\left(A^{-1}\right)^T$.We learned in section 2.2 that $\left(BC\right)^T = C^TB^T$. So we let $C = A$ and $B = A^{-1}$. That gives us
$$ \begin{align*} A^T\left(A^{-1}\right)^T &= \left(A^{-1}A\right)^T & &\text{Property of the Algebra of the Transpose} \\ &= \left(I_n\right)^T & &\text{$A^{-1}$ and $A$ are inverses} \\ &= I_n & &\text{$I$ is a symmetric matrix} \end{align*} $$ $\blacksquare$
Using Row Reduction to Compute Inverses¶
Video Lecture 2: Computing Inverses by Row Reduction.¶
Computing Inverses by Row ReductionTo compute the inverse of a non-singular matrix $A\in\mathbb{R}^{n\times n}$, form the partitioned matrix $\left[\,A\,|\,I_n\,\right]$. Multiplying on the left by $A^{-1}$ gives
$$ A^{-1}\left[\,A\,|\,I_n\,\right] = \left[\,A^{-1}A\,|\,A^{-1}I_n\,\right] = \left[\,I_n\,|\,A^{-1}\,\right]. $$
Left-multiplication by $A^{-1}$ is the same as performing a sequence of elementary row operations, so we may reduce $\left[\,A\,|\,I_n\,\right]$ directly: apply row operations until the left partition becomes $I_n$, and the right partition becomes $A^{-1}$. If the left partition cannot be reduced to $I_n$ because a free column appears, then $A$ is singular and has no inverse.
For example, let $A = \begin{bmatrix}\ 2\ &\ 1\ \\ \ 1\ &\ 1\ \end{bmatrix}$. Then
$$ \begin{align*} \left[\,A\,|\,I_2\,\right] = \left[\begin{array}{cc|cc} 2 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array}\right]\begin{array}{c} R_1\leftrightarrow R_2 \\ \\ \end{array} &\rightarrow \left[\begin{array}{cc|cc} 1 & 1 & 0 & 1 \\ 2 & 1 & 1 & 0 \end{array}\right]\begin{array}{c} \\ R_2-2R_1 \end{array} \\ \\ &\rightarrow \left[\begin{array}{cc|cc} 1 & 1 & 0 & 1 \\ 0 & -1 & 1 & -2 \end{array}\right]\begin{array}{c} \\ -R_2 \end{array} \rightarrow \left[\begin{array}{cc|cc} 1 & 1 & 0 & 1 \\ 0 & 1 & -1 & 2 \end{array}\right]\begin{array}{c} R_1-R_2 \\ \\ \end{array} \\ \\ &\rightarrow \left[\begin{array}{cc|cc} 1 & 0 & 1 & -1 \\ 0 & 1 & -1 & 2 \end{array}\right] \end{align*} $$
The left partition is $I_2$, so
$$ A^{-1} = \begin{bmatrix}\ \ 1\ & -1\ \\ -1\ &\ \ 2\ \end{bmatrix}. $$
You can check that $AA^{-1} = A^{-1}A = I_2$.
Practice Computing Inverses¶
Exercise 7 - Reduce the Partitioned Matrix¶
Compute $B^{-1}$ when
$$
B = \begin{bmatrix}\ \ 1\ & -2\ &\ \ 2\ \\ \ \ 1\ &\ \ 1\ & -1\ \\ -2\ & -2\ &\ \ 1\ \end{bmatrix}
$$
Solution
We apply row operations to reduce matrix $B$ in the left partition to reduced row echelon form, the identity. The result in the right partition will be $B^{-1}$.$$ \begin{align*} \left[\,B\,|\,I_3\,\right] &= \left[\begin{array}{ccc|ccc} \ \ 1\ & -2\ &\ \ 2\ &\ \ 1\ &\ \ 0\ &\ \ 0\ \\ \ \ 1\ &\ \ 1\ & -1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ \\ -2\ & -2\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \end{array}\right]\begin{array}{c} \ \\ R_2 - R_1 \\ R_3 + 2R_1 \end{array} \rightarrow \left[\begin{array}{ccc|ccc} \ \ 1\ & -2\ &\ \ 2\ &\ \ 1\ &\ \ 0\ &\ \ 0\ \\ \ \ 0\ &\ \ 3\ & -3\ & -1\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ & -6\ &\ \ 5\ &\ \ 2\ &\ \ 0\ &\ \ 1\ \end{array}\right]\begin{array}{c} \ \\ \\ R_3 + 2R_2 \end{array} \\ \\ &\rightarrow \left[\begin{array}{ccc|ccc} \ \ 1\ & -2\ &\ \ 2\ &\ \ 1\ &\ \ 0\ &\ \ 0\ \\ \ \ 0\ &\ \ 3\ & -3\ & -1\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ & -1\ &\ \ 0\ &\ \ 2\ &\ \ 1\ \end{array}\right]\begin{array}{c} \ \\ \\ -R_3 \end{array} \rightarrow \left[\begin{array}{ccc|ccc} \ \ 1\ & -2\ &\ \ 2\ &\ \ 1\ &\ \ 0\ &\ \ 0\ \\ \ \ 0\ &\ \ 3\ & -3\ & -1\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ & -2\ & -1\ \end{array}\right]\begin{array}{c} R_1-2R_3 \\ R_2+3R_3 \\ \\ \end{array} \\ \\ &\rightarrow \left[\begin{array}{ccc|ccc} \ \ 1\ & -2\ &\ \ 0\ &\ \ 1\ &\ \ 4\ &\ \ 2\ \\ \ \ 0\ &\ \ 3\ &\ \ 0\ & -1\ & -5\ & -3\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ & -2\ & -1\ \end{array}\right]\begin{array}{c} R_1+R_2 \\ \\ \\ \end{array} \rightarrow \left[\begin{array}{ccc|ccc} \ \ 1\ &\ \ 1\ &\ \ 0\ &\ \ 0\ & -1\ & -1\ \\ \ \ 0\ &\ \ 3\ &\ \ 0\ & -1\ & -5\ & -3\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ & -2\ & -1\ \end{array}\right]\begin{array}{c} \\ \frac{1}{3}R_2 \\ \\ \end{array} \\ \\ &\rightarrow \left[\begin{array}{ccc|ccc} \ \ 1\ &\ \ 1\ &\ \ 0\ &\ \ 0\ & -1\ & -1\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ & -\frac{1}{3}\ & -\frac{5}{3}\ & -1\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ & -2\ & -1\ \end{array}\right]\begin{array}{c} R_1-R_2 \\ \\ \\ \end{array} \rightarrow \left[\begin{array}{ccc|ccc} \ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ \frac{1}{3}\ & \frac{2}{3}\ &\ \ 0\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ & -\frac{1}{3}\ & -\frac{5}{3}\ & -1\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ & -2\ & -1\ \end{array}\right] \end{align*} $$
From reducing the left partition to reduced row echelon form we have
$$ B^{-1} = \begin{bmatrix}\ \ \frac{1}{3}\ &\ \ \frac{2}{3}\ &\ \ 0\ \\ -\frac{1}{3}\ & -\frac{5}{3}\ & -1\ \\ \ \ 0\ & -2\ & -1\ \end{bmatrix} $$
You can verify this by mutliplying $B^{-1}$ times $B$ to obtain the identity matrix $I_3$.
Exercise 8 - Reduce the Partitioned Matrix¶
Find the inverse of $A = \begin{bmatrix}\ 2\ &\ 1\ &\ 1\ \\ \ 6\ &\ 4\ &\ 5\ \\ \ 4\ &\ 1\ &\ 3\ \end{bmatrix}$.
Solution
$$ \begin{align*} \left[\,A\,|\,I_3\,\right] &= \left[\begin{array}{ccc|ccc}\ 2\ &\ 1\ &\ 1\ &\ 1\ &\ 0\ &\ 0\ \\ \ 6\ &\ 4\ &\ 5\ &\ 0\ &\ 1\ &\ 0\ \\ \ 4\ &\ 1\ &\ 3\ &\ 0\ &\ 0\ &\ 1\ \end{array}\right]\ \begin{array}{l} \\ R_2 - 3R_1 \\ R_3 + 2R_1 \end{array} \rightarrow \left[\begin{array}{ccc|ccc} 2 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 2 & -3 & 1 & 0 \\ 0 & -1 & 1 & -2 & 0 & 1 \end{array}\right]\ \begin{array}{l} \\ \\ R_3 + R_2 \end{array} \\ \\ &\rightarrow \left[\begin{array}{ccc|ccc} 2 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 2 & -3 & 1 & 0 \\ 0 & 0 & 3 & -5 & 1 & 1 \end{array}\right]\ \begin{array}{l} R_1-R_2 \\ \\ \\ \end{array} \rightarrow \left[\begin{array}{ccc|ccc} 2 & 0 & -1 & 4 & -1 & 0 \\ 0 & 1 & 2 & -3 & 1 & 0 \\ 0 & 0 & 3 & -5 & 1 & 1 \end{array}\right]\ \begin{array}{l} \\ \\ \frac{1}{3}R_3 \end{array} \\ \\ &\rightarrow \left[\begin{array}{ccc|ccc} 2 & 0 & -1 & 4 & -1 & 0 \\ 0 & 1 & 2 & -3 & 1 & 0 \\ 0 & 0 & 1 & -\frac{5}{3} & \frac{1}{3} & \frac{1}{3} \end{array}\right]\ \begin{array}{l} R_1+R_3 \\ R_2-2R_3 \\ \\ \end{array} \rightarrow \left[\begin{array}{ccc|ccc} 2 & 0 & 0 & \frac{7}{3} & -\frac{2}{3} & \frac{1}{3} \\ 0 & 1 & 0 & \frac{1}{3} & \frac{1}{3} & -\frac{2}{3} \\ 0 & 0 & 1 & -\frac{5}{3} & \frac{1}{3} & \frac{1}{3} \end{array}\right]\ \begin{array}{l} \frac{1}{2}R_1 \\ \\ \\ \end{array} \\ \\ &\rightarrow \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{7}{6} & -\frac{1}{3} & \frac{1}{6} \\ 0 & 1 & 0 & \frac{1}{3} & \frac{1}{3} & -\frac{2}{3} \\ 0 & 0 & 1 & -\frac{5}{3} & \frac{1}{3} & \frac{1}{3} \end{array}\right] \end{align*} $$Thus
$$ A^{-1} = \begin{bmatrix}\ \ \frac{7}{6}\ & -\frac{1}{3}\ &\ \ \frac{1}{6}\ \\ \ \ \frac{1}{3}\ &\ \ \frac{1}{3}\ & -\frac{2}{3}\ \\ -\frac{5}{3}\ &\ \ \frac{1}{3}\ &\ \ \frac{1}{3}\ \end{bmatrix} $$
Exercise 9 - Transpose Inverse¶
Using the matrix from exercise 8, $A = \begin{bmatrix}\ 2\ &\ 1\ &\ 1\ \\ \ 6\ &\ 4\ &\ 5\ \\ \ 4\ &\ 1\ &\ 3\ \end{bmatrix}$, compute the inverse of $A^T$.
Solution
$$ \begin{align*} \left(A^T\right)^{-1} = \left(A^{-1}\right)^T = \begin{bmatrix}\ \ \frac{7}{6}\ & -\frac{1}{3}\ &\ \ \frac{1}{6}\ \\ \ \ \frac{1}{3}\ &\ \ \frac{1}{3}\ & -\frac{2}{3}\ \\ -\frac{5}{3}\ &\ \ \frac{1}{3}\ &\ \ \frac{1}{3}\ \end{bmatrix}^T = \begin{bmatrix}\ \ \frac{7}{6}\ &\ \ \frac{1}{3}\ & -\frac{5}{3}\ \\ -\frac{1}{3}\ &\ \ \frac{1}{3}\ &\ \ \frac{1}{3}\ \\ \ \ \frac{1}{6}\ & -\frac{2}{3}\ &\ \ \frac{1}{3}\ \end{bmatrix} \end{align*} $$Exercise 10 - Product Inverse¶
Using the matrix from exercises 8 and 7,
$$ \begin{align*} A &= \begin{bmatrix}\ 2\ &\ 1\ &\ 1\ \\ \ 6\ &\ 4\ &\ 5\ \\ \ 4\ &\ 1\ &\ 3\ \end{bmatrix} \\ \\ B &= \begin{bmatrix}\ \ 1\ & -2\ &\ \ 2\ \\ \ \ 1\ &\ \ 1\ & -1\ \\ -2\ & -2\ &\ \ 1\ \end{bmatrix} \end{align*} $$
compute the inverse of $C = (AB)^{-1}$.
Solution
$$ \begin{align*} C = (AB)^{-1} &= B^{-1}A^{-1} = \begin{bmatrix}\ \ \frac{1}{3}\ &\ \ \frac{2}{3}\ &\ \ 0\ \\ -\frac{1}{3}\ & -\frac{5}{3}\ & -1\ \\ \ \ 0\ & -2\ & -1\ \end{bmatrix}\begin{bmatrix}\ \ \frac{7}{6} & -\frac{1}{3} &\ \ \frac{1}{6} \\ \ \ \frac{1}{3} &\ \ \frac{1}{3} & -\frac{2}{3} \\ -\frac{5}{3} &\ \ \frac{1}{3} &\ \ \frac{1}{3} \end{bmatrix} \\ \\ \mathbf{c}^1 &= \begin{bmatrix}\ \ \frac{1}{3}\ &\ \ \frac{2}{3}\ &\ \ 0\ \end{bmatrix}\begin{bmatrix}\ \ \frac{7}{6} & -\frac{1}{3} &\ \ \frac{1}{6} \\ \ \ \frac{1}{3} &\ \ \frac{1}{3} & -\frac{2}{3} \\ -\frac{5}{3} &\ \ \frac{1}{3} &\ \ \frac{1}{3} \end{bmatrix} = \begin{array}{lcccc} &\ \frac{1}{3} & [\ \frac{7}{6} & -\frac{1}{3} &\ \ \frac{1}{6}\ ] \\ & \ \frac{2}{3} & [\ \frac{1}{3}\ &\ \ \frac{1}{3}\ & -\frac{2}{3}\ ] \\ \ +\ &\ 0 & [ -\frac{5}{3}\ &\ \ \frac{1}{3}\ &\ \ \frac{1}{3}\ ] \\ \hline \\ & &[ \ \frac{11}{18}\ &\ \ \frac{1}{9}\ & -\frac{7}{18}\ ] \end{array} \\ \\ \mathbf{c}^2 &= \begin{bmatrix} -\frac{1}{3}\ & -\frac{5}{3}\ & -1\ \end{bmatrix}\begin{bmatrix}\ \ \frac{7}{6} & -\frac{1}{3} &\ \ \frac{1}{6} \\ \ \ \frac{1}{3} &\ \ \frac{1}{3} & -\frac{2}{3} \\ -\frac{5}{3} &\ \ \frac{1}{3} &\ \ \frac{1}{3} \end{bmatrix} = \begin{array}{lcccc} & -\frac{1}{3} &[\ \frac{7}{6} & -\frac{1}{3} &\ \ \frac{1}{6}\ ] \\ & -\frac{5}{3} & [\ \frac{1}{3}\ &\ \ \frac{1}{3}\ & -\frac{2}{3}\ ] \\ \ +\ & -1 & [ -\frac{5}{3}\ &\ \ \frac{1}{3}\ &\ \ \frac{1}{3}\ ] \\ \hline \\ & &[ \ \frac{13}{18}\ & -\frac{7}{9}\ &\ \ \frac{13}{18}\ ] \end{array} \\ \\ \mathbf{c}^3 &= \begin{bmatrix}\ \ 0\ & -2\ &\ -1\ \end{bmatrix}\begin{bmatrix}\ \ \frac{7}{6} & -\frac{1}{3} &\ \ \frac{1}{6} \\ \ \ \frac{1}{3} &\ \ \frac{1}{3} & -\frac{2}{3} \\ -\frac{5}{3} &\ \ \frac{1}{3} &\ \ \frac{1}{3} \end{bmatrix} = \begin{array}{lcccc} &\ \ 0 &[\ \frac{7}{6} & -\frac{1}{3} &\ \ \frac{1}{6}\ ] \\ & -2 & [\ \frac{1}{3}\ &\ \ \frac{1}{3}\ & -\frac{2}{3}\ ] \\ \ +\ & -1 & [ -\frac{5}{3}\ &\ \ \frac{1}{3}\ &\ \ \frac{1}{3}\ ] \\ \hline \\ & &[ \ 1\ & -1\ &\ \ 1\ ] \end{array} \\ \\ C = B^{-1}A^{-1} &= \begin{bmatrix}\ \ \frac{11}{18}\ &\ \ \frac{1}{9}\ & -\frac{7}{18}\ \\ \ \ \frac{13}{18}\ & -\frac{7}{9}\ &\ \ \frac{13}{18}\ \\ \ \ 1\ & -1\ &\ \ 1\ \end{bmatrix} \end{align*} $$Exercise 11 - Mathematical Induction¶
If $A$, $B$ and $P$ are $n\times n$ matrices, $P$ is invertible, and $B = PAP^{-1}$, express $B^2$, $B^3$. Show that for any positive integer $k$ $B^k = PA^kP^{-1}$.
Solution
$$ \begin{align*} B^2 &= B\,B = \left( PAP^{-1} \right)\left( PAP^{-1} \right) = PA\left(P^{-1}P\right)AP^{-1} = P(AIA)P^{-1} = PA^2P^{-1} \\ \\ B^3 &= B^2\,B = \left( PA^2P^{-1} \right)\left( PAP^{-1} \right) = PA^2\left(P^{-1}P\right)AP^{-1} = P\left(A^2A\right)P^{-1} = PA^3P^{-1} \\ \end{align*} $$ Suppose that $A$, $B$, and $P$ are $n\times n$ matrices, and $P$ is non-singular such that $B = PAP^{-1}$. This kind of theorem must be proven by mathematical induction. There are three steps.1. Show that the theorem is true for $k=1$.
We assume that $B^1 = B = PAP^{-1} = PA^1P^{-1}$.
2. Show that if the theorem is true for positive integer $k$, then it must also be true for positive integer $k+1$.
Suppose that for positive integer $k$, we have $B^k = PA^kP^{-1}$. Then
$$ B^{k+1} = BB^k = (B)(B^k) = (PAP^{-1})(PA^kP^{-1}) = PA(P^{-1}P)A^kP^{-1} = PAA^kP^{-1} = PA^{k+1}P^{-1} $$
3. Since the theorem is true for $k=1$, it is also true for $k+1=2$. Since the theorem is true for $k=2$, it is also true for $k+1=3$, etc. Using the Principle of Mathematical Induction the theorem is true for all positive integers $k$. $\blacksquare$
Exercise 12 - Singular Matrices¶
Show that if $A$ is a singular $n\times n$ matrix, then there is a nonzero $n\times n$ matrix $B$ so that $AB=O$.
Solution
Suppose that $A$ is a singular $n\times n$ matrix. Then the linear system $A\mathbf{x}=\mathbf{0}$ has a nontrivial (nonzero) solution. There is a nonzero vector $\mathbf{b}\in\mathbb{R}^n$ so that $A\mathbf{b} = \mathbf{0}$. We can construct many $n\times n$ matrices $B$ so that $AB=O$. Consider the $n\times n$ matrix $B = \begin{bmatrix} \mathbf{b} & \cdots & \mathbf{b} \end{bmatrix}$; the columns of matrix $B$ are $n$ copies of vector $\mathbf{b}$. Then $B\neq O$ and$$ AB = A\begin{bmatrix} \mathbf{b} & \cdots & \mathbf{b} \end{bmatrix} = \begin{bmatrix} A\mathbf{b} & \cdots & A\mathbf{b} \end{bmatrix} = \begin{bmatrix} \mathbf{0} & \cdots & \mathbf{0} \end{bmatrix} = O. $$
