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.
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.
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 \(Ax = 0\).
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.
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 \(A\).
For each of the following matrices, determine if it is singular.
\(\begin{bmatrix} 1 & 2 \\ -2 & -4 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 5 & 7 & 8\end{bmatrix}\)
Let \(A \in \mathbb{R}^{n\times n}\). Prove that if \(A\) is invertible, then \(A\) is nonsingular.