Objectives: Topics
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The Invertible Matrix Theorem which is a synthesis of previously introduced concepts.
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Section 2.3 The Invertible Matrix Theorem
Handout 2.3 The Invertible Matrix Theorem
A synonym is a word you use when you can’t spell the other one.―Baltasar Gracián
The theorem introduced in this section provides many equivalent ways of describing the same idea. Depending on the context, some formulations will be more convenient than others.
Objectives: Goals
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Characterize the invertibility of a matrix using the Invertible Matrix Theorem.
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Construct and give examples of matrices that are or are not invertible.
Motivating Question: When is a square matrix invertible? Let me count the ways!
Invertible matrices enjoy a rich set of equivalent descriptions.
Theorem 2.26 The Invertible Matrix Theorem (IMT).
Let \(A\) be an \(n\times n\) matrix. The following statements are equivalent.
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\(A\) is invertible.
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\(A\) is row equivalent to \(I_n\text{.}\)
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\(A\) has \(n\) pivotal columns.
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\(A\vec x=\vec 0\) has only the trivial solution.
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The columns of \(A\) are linearly independent.
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The linear transformation \(\vec x \mapsto A\vec x\) is one-to-one.
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The equation \(A\vec x=\vec b\) has a solution for all \(\vec b\in\mathbb{R}^n\text{.}\)
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The columns of \(A\) span \(\mathbb{R}^n\text{.}\)
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The linear transformation \(\vec x \mapsto A\vec x\) is onto.
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There is an \(n\times n\) matrix \(C\) such that \(CA=I_n\text{.}\) (\(A\) has a left inverse.)
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There is an \(n\times n\) matrix \(D\) such that \(AD=I_n\text{.}\) (\(A\) has a right inverse.)
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\(A^T\) is invertible.
The diagram below provides another perspective on the role of the inverse matrix.
Diagram shows \(\vec x\text{,}\) transformed by multiplication by \(A\text{,}\) and reversed by multiplication by \(A^{-1}\text{.}\)
The matrix inverse \(A^{-1}\) transforms \(A\vec x\) back to \(\vec x\text{,}\) because
\begin{equation*}
A^{-1}(A\vec x)=(A^{-1}A)\vec x=\vec x.
\end{equation*}
Some comments on IMT:
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The Invertible Matrix Theorem divides all square matrices into two classes: invertible and non-invertible.
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As the course progresses, additional equivalent conditions involving determinants, eigenvalues, and other concepts will be added.
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Note that items (j) and (k) of the Invertible Matrix Theorem lead directly to the following result.
Theorem 2.27 Left and Right Inverses.
If \(A\) and \(B\) are \(n\times n\) matrices and \(AB=I\text{,}\) then both matrices are invertible, with \(B=A^{-1}\) and \(A=B^{-1}\text{.}\)
Example 2.28.
Is \(A\) invertible?
\begin{equation*}
A = \begin{bmatrix}
1 \amp 0 \amp -2\\
3 \amp 1 \amp -2\\
-5 \amp -1 \amp 9
\end{bmatrix}
\end{equation*}
Solution.
We can use IMT to determine if \(A\) is invertible by showing that \(A\) either makes at least one statement in the list true or makes at least one statement in the list false. Note that
\begin{equation*}
A = \begin{bmatrix} 1 \amp 0 \amp -2 \\ 3 \amp 1 \amp -2 \\ -5 \amp -1 \amp 9 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp -2 \\ 0 \amp 1 \amp 4 \\ 0 \amp -1 \amp -1 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp -2 \\ 0 \amp 1 \amp 4 \\ 0 \amp 0 \amp 3 \end{bmatrix}\text{.}
\end{equation*}
We see that \(A \) has a pivot in every column. Therefore, by IMT, \(A \) must make every statment on the list true. In particular, we learn that \(A \) is invertible.
Example 2.29.
For each matrix below, determine all values of \(h\) so that the matrix is singular (not invertible). If there are no values of \(h \text{,}\) explain why.
(a)
\(A = \begin{bmatrix}
1 \amp 0 \amp 1\\
1 \amp h \amp 1\\
0 \amp 0 \amp 1
\end{bmatrix}\)
Solution.
Note that
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \amp 1 \\ 1 \amp h \amp 1 \\ 0 \amp 0 \amp 1 \end{bmatrix} \sim \begin{bmatrix} 1\amp 0 \amp 1 \\ 0 \amp h \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{.}
\end{equation*}
Thus, \(A \)is missing a pivot if and only if \(h = 0\text{.}\) Therefore, \(h = 0\) is the only value of \(h \) which makes \(A \) singular.
(b)
\(B = \begin{bmatrix}
1 \amp h \amp 1\\
0 \amp 1 \amp 1\\
0 \amp 0 \amp 1
\end{bmatrix}\)
Solution.
Note that regardless of the value of \(h \text{,}\) \(B \) will have a pivot in each column. Therefore, no value of \(h \) will make \(B \) singular.
(c)
\(C = \begin{bmatrix}
1 \amp 0 \amp 0\\
0 \amp 1 \amp 1\\
0 \amp h \amp 1
\end{bmatrix}\)
Solution.
Note that
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 1 \\ 0 \amp h \amp 1 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 1 \\ 0 \amp h-1 \amp 0\end{bmatrix}\text{.}
\end{equation*}
Thus, \(C \) is missing a pivot in the third row if and only if \(h = 1\text{.}\) Therefore, \(h = 1\) is the only value of \(h\) which makes \(C \) singular.
