📂Matrix Algebra

Inverse Matrix, Reversible Matrix

Definition

Let $A$ be an arbitrary square matrix of size $n\times n$. A matrix $L$ is called the left inverse matrix of $A$ if it satisfies the following equation with $A$ in a matrix multiplication.

$$LA=I_{n}$$

Here, $I_{n}$ is the identity matrix of size $n\times n$. A matrix $R$ that is capable of matrix multiplication with $A$ and satisfies the following equation is called the right inverse matrix of $A$.

$$AR=I_{n}$$

If $A$ has both a left/right inverse, they are equal and denoted as $A^{-1}$, and $A$ is called the inverse matrix.

$$A^{-1}A=I_{n}=AA^{-1}$$

If $A$ has an inverse matrix, it is called an invertible matrix or a nonsingular matrix. If $A$ does not have an inverse matrix, $A$ is called a singular matrix.

Explanation

By the definition, since $LA$ must have the size of $n\times n$, $L$ must be a $n \times n$ matrix, and so must $R$. The reason for limiting $A$ to square matrices is because $A^{-1}$ needs to be capable of being multiplied on both sides of $A$. Similarly, since matrix multiplication is not commutative, both left/right inverses must exist for it to be called an invertible matrix. In fact, if an arbitrary matrix has both a left and a right inverse, they are always the same.

Properties

Let $A$ and $B$ be arbitrary $n \times n$ square matrices. Then the following holds.

(a) If $A$ has a left inverse matrix $L$ and a right inverse matrix $R$, they are the same.

$$L=A^{-1}=R$$

(b) If the inverse matrix of $A$ exists, it is unique.

(c) $AB = I \iff BA = I$

(d) Let $A$ and $B$ be invertible matrices. Then the product of the two matrices $AB$ is also invertible, and its inverse is as follows.

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

(d’) The product of invertible matrices of the same size is also invertible, and its inverse is the same as the product of the inverses in reverse order. That is, if $A_{1},A_{2},\dots,A_{n}$ is an invertible matrix, then the following holds.

$$\left( A_{1}A_{2}\cdots A_{n} \right)^{-1} = A_{n}^{-1}\cdots A_{2}^{-1} A_{1}^{-1}$$

(e) If $AB$ is invertible, then both $A$ and $B$ are also invertible.

(f) If $A$ is invertible, its transpose is also invertible and its inverse is as follows.

$$\left( A^{T} \right)^{-1} = \left( A^{-1} \right)^{T}$$

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Therefore, we can see that (c) $\iff$ (d).

Proof

(a)

Let’s assume a $n\times n$ matrix $A$ is given. Let $L$ be the left inverse matrix of $A$. Then the following equation holds.

$$LA=I_{n}$$

Let’s say $R$ is the right inverse matrix of $A$. Multiplying $R$ to the right side of the equation above results in the following.

$$LAR = I_{n}R =R$$

However, since $R$ is the right inverse matrix of $A$, $LAR=LI_{n}=L$ holds. Therefore, the equation above is as follows.

$$L=R$$

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(b)

Assume an arbitrary square matrix $A$ has two different inverse matrices $B$ and $C$. Then the calculation goes as follows.

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

However, this result contradicts the assumption that $B$ and $C$ are different. Therefore, the assumption is wrong, and if an inverse matrix exists, it is unique.

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(c)

Without loss of generality, let’s only prove $BA = I \implies AB = I$. Assume $BA = I$. Now consider the equation $A \mathbf{x} = \mathbf{0}$.

$$\begin{align*} A\mathbf{x} = \mathbf{0} &\implies B(A\mathbf{x}) = B \mathbf{0} \\ &\implies (BA)\mathbf{x} = \mathbf{0} \end{align*}$$

Since we assumed $BA = I$, $\mathbf{x} = \mathbf{0}$ holds. Therefore, $A \mathbf{x} = \mathbf{0}$ only has trivial solutions.

Equivalent conditions for an invertible matrix

Let $A$ be a square matrix of size $n\times n$. Then the following propositions are equivalent:

• $A$ is an invertible matrix.
• The homogeneous linear system $A\mathbf{x}=\mathbf{0}$ only has trivial solutions.

According to the equivalent conditions for an invertible matrix, $A$ is invertible. Therefore, $A^{-1}$ exists, and

$$BA = I \implies A(BA)A^{-1} = AIA^{-1} \implies AB = I$$

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(d)

Let’s assume $A$ and $B$ are invertible matrices of size $n\times n$. Then $A^{-1}$ and $B^{-1}$ exist. First, multiplying $B^{-1}A^{-1}$ to the right of $AB$ results in the following.

$$\begin{align*} (AB)(B^{-1}A^{-1}) &= ABB^{-1}A^{-1} \\ &= AI_{n}A^{-1} = AA^{-1} \\ &= I_{n}\end{align*}$$

Multiplying to the left results in the following.

$$\begin{align*} (B^{-1}A^{-1})(AB) &= B^{-1}A^{-1}AB \\ &= B^{-1}I_{n}B = B^{-1}B \\ &= I_{n}\end{align*}$$

Therefore, $AB$ is an invertible matrix, and its inverse is $B^{-1}A^{-1}$.

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(d')

This is a corollary to (d).

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(e)

Let’s say the inverse matrix of $AB$ is $C$. Then $ABC=I_{n}$ holds. Therefore, by (c), $A$ is invertible, and $A^{-1}=BC$ holds. Also, since $CAB=I_{n}$, $B$ is also invertible, and $B^{-1}=CA$ holds.

(f)

Check if the product of two matrices results in the identity matrix. According to the properties of transpose matrices, it goes as follows.

$$A^{T} \left( A^{-1} \right)^{T} = \left( A^{-1} A \right) ^{T} = I^{T} = I$$

$$\left( A^{-1} \right)^{T} A^{T} = \left( A A^{-1} \right)^{T} = I^{T} = I$$

Therefore

$$\left( A^{T} \right)^{-1} = \left( A^{-1} \right)^{T}$$

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