Invertible Matrix Theorem
A Guide for Students

Did you know there are two types of square matrices?

Yep. There are invertible matrices and non-invertible matrices called singular matrices.

While singular matrices have their importance and place, invertible matrices draw our attention and focus.

Why?

Because, for lack of a better analogy, invertible matrices get to wield the power of the Invertible Matrix Theorem!

The Invertible Matrix Theorem

The invertible matrix theorem and its vast conditions are possibly one of the most important theorems in all of Linear Algebra.

Okay, so what is the invertible matrix theorem?

The Invertible Matrix Theorem states that if we let \(\mathrm{A}\) be an \(n \times n\) matrix and \(T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) be a matrix transformation \(T(\vec{x})=A \vec{x}\), then the following statements are equivalent:

• \(\mathrm{A}\) is an invertible matrix
• \(\mathrm{A}\) is row equivalent to the \(n \times n\) identity matrix
• A has \(n\) pivot positions
• The equation \(A \vec{x}=\overrightarrow{0}\) has only the trivial solution (zero solution)
• The columns of \(\mathrm{A}\) form a linearly independent set
• The linear transformation \(\vec{x} \mapsto \vec{x}\) is one-to-one
• The equation \(A \vec{x}=\vec{b}\) has a least one solution for each \(\vec{b}\) in \(\mathbb{R}^{n}\)
• The columns of \(\mathrm{A}\) span \(\mathbb{R}^{n}\)
• The linear transformation \(\vec{x} \mapsto A \vec{x}\) maps \(\mathbb{R}^{n}\) onto \(\mathbb{R}^{n}\)
• There is an \(n \times n\) matrix \(\mathrm{C}\) such that \(C A=I\)
• There is an \(n \times n\) matrix \(\mathrm{D}\) such that \(A D=I\)
• \(A^{T}\) is an invertible matrix
• \(T\) is invertible
• \(\mathrm{T}\) is one-to-one
• \(\mathrm{T}\) is onto

Implications of the Invertible Matrix Theorem

So, if a matrix is invertible, then all the statements from the invertible matrix theorem are true. In other words, if one of the statements is true, then so are the others. And if the matrix is not invertible (singular), then all the statements of the invertible matrix theorem are false.

Additionally, it is essential to note that we will add to this list of equivalent statements in future lessons. However, this is definitely a solid list of conditions to get us started understanding and appreciating this important theorem. Alright, so let’s start putting this useful theorem to work.

Applying the Invertible Matrix Theorem

Suppose \(A\) and \(B\) are invertible matrices where \(A B=\left[\begin{array}{ll}1 & 2 \\ 4 & 7\end{array}\right]\) and \(B=\left[\begin{array}{cc}8 & 6 \\ 5 & 4\end{array}\right]\).

What is matrix A?

Well, to isolate \(A\) from the product \(\mathrm{AB}\) we will use the power of inverses to isolate \(\mathrm{A}\)

Remember, when we multiply a number by its inverse it simplifies to 1. For example, \(3\left(\frac{1}{3}\right)=1\).

We are doing the same thing here. B times \(B^{-1}\) becomes the identity matrix, \(I\), which is the same thing as 1 .

And this is how we will isolate matrix A.

\begin{equation}
\begin{aligned}
& A B \\
& A B B^{-1} \\
& A I \\
& A
\end{aligned}
\end{equation}

So, this means that if we multiply AB with the inverse of B, then we have matrix A! Okay, let’s get to it by finding B-inverse.

\begin{aligned}
B^{-1} & =\frac{1}{\operatorname{det}(B)}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \\
& \Rightarrow B^{-1}=\frac{1}{(8 \cdot 4-6 \cdot 5)}\left[\begin{array}{cc}
4 & -6 \\
-5 & 8
\end{array}\right] \\
& \Rightarrow B^{-1}=\frac{1}{2}\left[\begin{array}{cc}
4 & -6 \\
-5 & 8
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 & -3 \\
-5 / 2 & 4
\end{array}\right]
\end{aligned}

Now, let’s find the product of \(A B \cdot B^{-1}\)

\begin{aligned}
\left[\begin{array}{ll}1 & 2 \\ 4 & 7\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ -5 / 2 & 4\end{array}\right] & =\left[\begin{array}{cc}-3 & 5 \\ -19 / 2 & 16\end{array}\right]
\end{aligned}

Thus,

\begin{aligned}
A & =\left[\begin{array}{cc}-3 & 5 \\ -19 / 2 & 16\end{array}\right]
\end{aligned}

all by the power of invertible matrices!

The Invertible Matrix Theorem and Linear Transformations

Now, let’s see how the invertible matrix theorem helps us understand linear transformations.

So, if \(T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) is a linear transformation and \(\mathrm{A}\) is the standard matrix \(T\). Then \(\mathrm{T}\) is invertible if and only if \(\mathrm{A}\) is an invertible matrix.

Consequently, the linear transformation \(T^{-1}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) given by \(T^{-1}(\vec{x})=A^{-1} \vec{x}\) is the unique function satisfying both:

• \(T^{-1}(T(\vec{x}))=\vec{x}\) for all \(\vec{x}\) in \(\mathbb{R}^{n}\)
• \(T\left(T^{-1}(\vec{x})\right)=\vec{x}\) for all \(\vec{x}\) in \(\mathbb{R}^{n}\)

Okay, but what does this mean?

It means that \(A^{-1}\) transforms \(A \vec{x}\) back to \(\vec{x}\).

Understanding Inverse Transformations

Suppose \(T\) is a linear transformation from \(\mathbb{R}^{2}\) to \(\mathbb{R}^{2}\).

Let’s use our knowledge of invertible matrices to show that \(\mathrm{T}\) is invertible and find a formula for \(T^{-1}\) if:

\begin{equation}
T\left(x_1, x_2\right)=\left(6 x_1-8 x_2,-5 x_1+7 x_2\right)
\end{equation}

First, let’s write the standard matrix for \(\mathrm{T}\).

\begin{align*}
A=\left[\begin{array}{cc}
6 & -8 \\
-5 & 7
\end{array}\right]
\end{align*}

Next, we think back to the Invertible Matrix Theorem and recognize that if \(\mathrm{T}\) is an invertible linear transformation, then matrix \(\mathrm{A}\) is also invertible such that \(T^{-1}(\vec{x})=A^{-1} \vec{x}\).

So, all we have to do is find the inverse of \(\mathrm{A}\) and multiply it by vector \(\mathrm{x}\).

\begin{aligned}
A^{-1} & =\frac{1}{\operatorname{det}(A)}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \\
& \Rightarrow A^{-1}=\frac{1}{(6(7)-(-8)(-5))}\left[\begin{array}{ll}
7 & 8 \\
5 & 6
\end{array}\right] \\
& =\frac{1}{2}\left[\begin{array}{ll}
7 & 8 \\
5 & 6
\end{array}\right] \\
& =\left[\begin{array}{cc}
7 / 2 & 4 \\
5 / 2 & 3
\end{array}\right]
\end{aligned}

Therefore, we have shown that \(\mathrm{T}\) is invertible and the formula for \(\mathrm{T}\) inverse is:

\begin{aligned}
T^{-1} & =\left[\begin{array}{cc}
7 / 2 & 4 \\
5 / 2 & 3
\end{array}\right]\left[\begin{array}{l}
x_{1} \\
x_{2}
\end{array}\right] \\
& =\left[\begin{array}{c}
\frac{7}{2} x_{1}+4 x_{2} \\
\frac{5}{2} x_{1}+3 x_{2}
\end{array}\right]
\end{aligned}

or

\begin{aligned}
T^{-1} & =\left(\frac{7}{2} x_{1}+4 x_{2}, \frac{5}{2} x_{1}+3 x_{2}\right)
\end{aligned}

See, not so bad, right?

Next Steps

In this video, you will:

• Examine the Invertible Matrix Theorem in detail
• Answer true-false questions and existence-uniqueness questions to ensure understanding of concepts and theory
• Learn how to determine if a matrix is invertible using as few steps or calculations as possible
• Compare Invertible Linear Transformations with your understanding of composite functions from Algebra 2 and Precalculus

Get ready for a fun learning experience, and let’s get started!

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