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Section 3.11 Equivalent forms of Invertibility and Singularity of Matrices

We have already defined singular and nonsingular matrices in Definition 3.7.5 and the inverse of a matrix in Definition 3.7.1. We now have a new goal: to show that a square matrix is invertible if and only if it is nonsingular. In fact we want to give several other conditions that are equivalent to being invertible. By this we mean that if any one of the conditions are true for a matrix \(A\text{,}\) then all them are true for that matrix \(A\text{,}\) and we then call \(A\) nonsingular or invertible. Equivalently, if any one of the conditions are false for a matrix \(A\text{,}\) then all of them are false for that matrix \(A\text{.}\) In this case the matrix \(A\) is called singular.

The method for showing statements to be equivalent is a little different. Suppose we have (as will be our case) six statements numbered \((1), (2), \ldots,(6)\) that we wish to show equivalent. We start by showing that if \((1)\) is true, then it follows that \((2)\) is true. We denote this by \((1)\Rightarrow(2) \text{.}\) We continue with \((2)\Rightarrow(3), (3)\Rightarrow(4), (4)\Rightarrow(5), (5)\Rightarrow(6), (6)\Rightarrow(1)\text{.}\) Once this has been done, we may visualize our result:

Figure 3.11.1. A cycle of implications

and so once one statement becomes true, all of them are true. This also means that once one statement is false then all of them are false (think about the logic!).

The logic of the proof is to show that

(1) true implies (2) true (which we write as (1) \(\Rightarrow \)(2)),

(2) true implies (3) true (which we write as (2) \(\Rightarrow\) (3)),

(3) true implies (4) true (which we write as (3) \(\Rightarrow\) (4)),

(4) true implies (5) true (which we write as (4) \(\Rightarrow\) (5)),

(5) true implies (6) true (which we write as (5) \(\Rightarrow\) (6)), and

(6) true implies (1) true (which we write as (6) \(\Rightarrow\) (1))

We now proceed to prove the implications one at a time.

(1) \(\Rightarrow\) (2): Assuming (1) means that \(A\) is invertible. If \(x=\vec0\text{,}\) then \(A\vec x=A\vec0=\vec0\text{.}\) Now suppose that \(A\vec x=\vec0.\) Then, since \(A^{-1}\) exists, we may evaluate \(A^{-1}A\vec x\) in two ways:

\begin{equation*} A^{-1}(A\vec x) = A^{-1}\vec0=\vec0\\ (A^{-1}A)\vec x =I\vec x=\vec x \end{equation*}

Equating the two evaluations, we get \(\vec x=\vec0.\)

(2) \(\Rightarrow\) (3): Consider the augmented matrix for the system of linear equations \(A\vec x=\vec0.\) If the reduced row echelon from of \(A\) is not \(I_n,\) then the form for augmented matrix has a free variable; this may be assigned a nonzero value, resulting in a nonzero solution to \(A\vec x=\vec0.\)

(3) \(\Rightarrow\) (4): Suppose \(A\) is reduced to \(I_n\) by a series of elementary row operations whose corresponding elementary matrices are \(E_1,E_2,\dots,E_k\text{.}\) Then \(E_kE_{k-1}\cdots E_2E_1A=I_n\text{,}\) and

\begin{equation*} A=E_1^{-1}E_2^{-1}\cdots E_{k-1}^{-1}E_k^{-1}=F_1F_2\cdots F_{k-1}F_k. \end{equation*}

(4) \(\Rightarrow\) (5): Suppose we want to solve \(A\vec x=\vec b\) where \(\vec b\) is any vector and \(A=F_1F_2\cdots F_{k-1}F_k.\) Then we have

\begin{equation*} F_1F_2\cdots F_{k-1}F_k \vec x=\vec b, \end{equation*}

and

\begin{equation*} \vec x=F_k^{-1}F_{k-1}^{-1}\cdots F_2^{-1}F_1^{-1} \vec b \end{equation*}

which then satisfies

\begin{equation*} \begin{split} A\vec x \amp =(F_1F_2\cdots F_{k-1}F_k)(F_k^{-1}F_{k-1}^{-1}\cdots F_2^{-1}F_1^{-1}\vec b)\\ \amp=(F_1F_2\cdots F_{k-1}F_k)(F_k^{-1}F_{k-1}^{-1}\cdots F_2^{-1}F_1^{-1})\vec b \\ \amp=\vec b \end{split} \end{equation*}

and so gives a solution to \(A\vec x=\vec b.\) Hence \(A\vec x=\vec b\) is consistent.

(5) \(\Rightarrow\) (6): We take the following values for \(\vec b_1,\vec b_2,\dots,\vec b_n\text{:}\)

\begin{equation*} \vec b_1= \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \quad \vec b_2= \begin{bmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \quad \vec b_3= \begin{bmatrix} 0 \\ 0 \\ 1 \\ \vdots \\ 0 \end{bmatrix}, \ldots, \quad \vec b_n= \begin{bmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 1 \end{bmatrix} \text{.} \end{equation*}

Then \(A\vec x=\vec b_1\) has a solution \(\vec c_1\text{,}\) \(A\vec x=\vec b_2\) has a solution \(\vec c_2\text{,}\) \(A\vec x=\vec b_3\) has a solution \(\vec c_3, \dots,\) \(A\vec x=\vec b_n\) has a solution \(\vec c_n\text{.}\)

Now define the matrix \(C\) as the one with \(\vec c_1,\vec c_2,\dots,\vec c_n\) as columns, that is,

\begin{equation*} C= \begin{bmatrix} \vec c_1 \amp\vec c_2 \amp\vec c_3 \amp\cdots \amp\vec c_n \end{bmatrix}. \end{equation*}

Then \(AC=I\text{,}\) and so \(C=A^{-1}\) and \(CA=I.\) Now if \(A\vec x=\vec b\) is consistent, then \(\vec x=I\vec x=CA\vec x=C\vec b\) and so \(\vec x=C\vec b\) is the one and only solution to \(A\vec x=\vec b.\)

(6) \(\Rightarrow\) (1): Use the vectors \(\vec b_1, \vec b_2,\dots,\vec b_n\) given just above. In each case there is exactly one vector \(\vec c_k\) which is a solution to the equation \(A\vec x=\vec b_k\text{.}\) Let \(C\) be the matrix with \(\vec c_1,\dots,\vec c_n\) as columns. This matrix satisfies \(AC=I\) which makes \(C=A^{-1}\) and so \(A\) is invertible.