## Noninvertible square matrices

A square matrix that has an inverse is said to be invertible. Not all square matrices defined over a field are invertible. Such a matrix is said to be noninvertible.

For example, $$A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ is noninvertible because for any $$B = \begin{bmatrix} a& b\\ c & d\end{bmatrix}$$, $$BA = \begin{bmatrix} a & 0 \\ c & 0 \end{bmatrix}$$, which cannot equal $$\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$ no matter what $$a,b,c$$, and $$d$$ are.

There is a testable condition for invertibility without actually trying to find the inverse: A matrix $$A \in \mathbb{F}^{n\times n}$$ where $$\mathbb{F}$$ denotes a field is invertible if and only if there does not exist $$x \in \mathbb{F}^n$$ not equal to $$0_n$$ such that $$Ax = 0_n$$. (Here, $$0_n$$ denotes the $$n$$-tuple of all 0's.)

Notice that the existence of a nonzero $$x$$ such that $$Ax = 0_n$$ does not require the entries of $$A$$ and $$x$$ to be from a field. In particular, the existence of multiplicative inverses is not necessary for the condition to make sense. So, it is said that a matrix $$A$$ is singular if there exists $$x$$ having at least one nonzero entry such that $$Ax = 0$$.

A matrix that is not singular is nonsingular. In the context of square matrices over fields, the notions of singular matrices and noninvertible matrices are interchangeable.

## Testing singularity

Let $$A$$ be an $$m\times n$$ matrix over some field $$\mathbb{F}$$. Recall that $$Ax = 0$$ always has the tuple of 0's as a solution. This solution is called the trivial solution. All other solutions are called nontrivial.

If a nontrivial solution to $$Ax = 0$$ is not readily available, we can determine if there is one by transforming $$A$$ to a matrix $$R$$ in reduced row-echelon form using elementary row operations. As the system always has $$x = 0$$ as a solution, the system has a nontrivial solution if and only if $$R$$ has a nonpivot column.

### Examples

1. The matrix $$A = \begin{bmatrix} 1 & -2 \\ -3 & 6\end{bmatrix}$$ is singular because $$x = \begin{bmatrix} 2 \\ 1\end{bmatrix}$$ as a nontrivial solution to the system $$Ax = 0$$.

2. Let $$A = \begin{bmatrix} 1 & 1 & 1\\0 & 1 & 0\\1 & 0 & 1\end{bmatrix}$$ be defined over $$GF(2)$$. (Recall that $$GF(2)$$ is the field consisting of only the elements 0 and 1 with the rule “1+1 = 0”.) Performing the elementary row operation $$R_1 \leftarrow R_1+R_2$$ and then $$R_3\leftarrow R_3+R_1$$ gives $$\begin{bmatrix} 1 & 0 & 1\\0 & 1 & 0\\0 & 0 & 0\end{bmatrix}.$$ As the third column is not a pivot column, $$A$$ is singular. In particular, $$x = \begin{bmatrix}1\\0\\1\end{bmatrix}$$ satisfies $$Ax = 0$$. So $$A^{-1}$$ does not exist.

### Why aren't singular square matrices invertible?

Let $$A$$ be a square matrix over some field $$\mathbb{F}$$. Suppose that $$x$$ is a nontrivial solution to $$Ax = 0$$. If $$A^{-1}$$ exists, then $$A^{-1}(Ax) = A^{-1}0$$, implying that $$(A^{-1}A)x = 0$$. But $$A^{-1}A = I$$. Therefore $$Ix = 0$$, contradicting that $$x$$ is nontrivial.

Invertible matrices certainly aren't singular because for any $$x$$ such that $$Ax = 0$$, we must have $$A^{-1}(Ax) = A^{-1}0$$, giving $$x = 0$$. This means that the trivial solution is the only solution to $$Ax = 0$$.

With the above observation, we can see why row reducing $$[A \mid I]$$, where $$A$$ is an $$n\times n$$ matrix over some field, will determine if $$A$$ is invertible and gives us $$A^{-1}$$ in the right half of the matrix if it is. If the RREF of $$[A \mid I]$$ does not have the first $$n$$ columns as pivot columns, then we know $$A$$ is singular and therefore is not invertible. Otherwise, we will end up with a matrix of the form $$[I \mid B]$$ and, as we have seen before, $$B$$ will be the inverse of $$A$$.

## Exercises

1. For each of the following matrices, determine if it is singular.

1. $$\begin{bmatrix} 1 & 2 \\ -2 & -4 \end{bmatrix}$$

2. $$\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 5 & 7 & 8\end{bmatrix}$$

2. Let $$A \in \mathbb{R}^{n\times n}$$. Prove that if $$A$$ is invertible, then $$A$$ is nonsingular.