Does a matrix $A$ need to have $\det A \neq 0$ to even have a rank?

Question

Does a matrix $A$ need to have $\det A \neq 0$ to even have a rank?

So I've had this uneasy feeling that the rank could not be calculated for a matrix $4\times 4$ which had two identical columns, and therefore its determinant was zero. Am I wrong?

Answer 1

Any matrix $A\in M_{m\times n}(\Bbb R)$ defines a linear map $T:\Bbb R^n\to\Bbb R^m$ given by $T(\vec x)=A\vec x$. The rank of $A$ is defined to be the dimension of $\DeclareMathOperator{Image}{Image}\Image(T)$.

The point here is that rank is defined for any matrix while the determinant is only defined for square matrices. So the short answer to your question is no: $A$ need not satisfy $\det A\neq 0$ in order for $\DeclareMathOperator{Rank}{Rank}\Rank(A)$ to be defined.

Of course, knowing $\det(A)$ can give information about $\Rank(A)$. If $A$ is $n\times n$ and $\det(A)\neq0$, then $A$ is invertible and $\Rank(A)=n$. Otherwise, if $\det(A)=0$, then $A$ is not invertible so $\Rank(A)< n$.

Unfortunately $\det(A)$ does not give us complete information about $\Rank(A)$. For example each of the following matrices \begin{align*} X &= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} & Y &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} & Z &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \end{align*} has determinant $0$ while \begin{align*} \Rank(X) &= 0 & \Rank(Y) &= 1 & \Rank (Z) &= 2 \end{align*}

Answer 2

The rank is the dimension of the range space, which may be zero (if the matrix is the zero matrix, for example).

If the matrix has two identical columns, then you know that the rank is at most $3$.

I am guessing this was not the point of the question, but every matrix has a rank as defined above, but computing the rank numerically can be a delicate task.