Maximum rank of a matrix based on its eigenvalues

Let $A$ be a square matrix. If $\lambda=0$ is an eigenvalue of $A$, what is the maximum value of $\text{rank}(A)$?

This one was really tricky for me, but I'll show what I did:

So first I tried to find a matrix that actually has an eigenvalue of $0$, and the first one I found was the zero matrix:

$z=\begin{pmatrix}0 & 0\\0 & 0\end{pmatrix}$

To find the eigenvalues of this matrix, I used the formula below:

$\det(z - \lambda I)$

$\det\left(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}\right)$ = $\det\begin{pmatrix} -\lambda & 0 \\ 0 & -\lambda \end{pmatrix}$

I took the determinant and got $\lambda^2$.

To actually find the eigenvalues, I set $\lambda^2=0$, which just yields $\lambda=0$.

I think this is where I did something wrong.

I assumed that the zero matrix was the only matrix that has an eigenvalue of $0$ and because of this I also assumed that the row reduced echelon form of the zero matrix is itself, therefore I concluded that the maximum rank of the zero matrix is $0$.

Can someone shed some light on this problem? Did I even approach this problem correctly?

• It's not the only matrix with a zero eigenvalue. Your general approach is fine; think of some examples and see what you can learn from them, but you made an assumption that led you to the wrong conclusion. Try to think of other matrices with a zero eigenvalue. – dbx Dec 19 '17 at 4:03
• $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ What are the eigenvalues of this? – Mauve Dec 19 '17 at 4:04
• Possibly related: math.stackexchange.com/questions/1349907/… – Mauve Dec 19 '17 at 4:07
• @Soon_to_be_code_mast If you are ok, you can set as solved. Thanks! – gimusi Dec 19 '17 at 15:56

What does it mean to for an $n\times n$ matrix to have an eigenvalue $\lambda=0$?

It means for some nonzero $v \in \mathbb{R}^n$, $Av = 0 * v = 0$.

But this means that at least one nonzero $v$ is in the null space of $A$. A matrix with a null space bigger than just $0$ cannot have "full rank" (i.e. rank $n$). So the most it can be is rank $n-1$.

Notice that this is fundamentally an application of the dimension formula: #rows = rank + dimension of nullspace.

You should also observe that any non-invertible matrix has at least one eigenvalue of zero (corresponding to each of the null space vectors). Of course, no invertible matrix has an eigenvalue equal to zero.

If A has at least one zero eigenvalue then exist at least a subspace with dimension 1 such that Ax=0 thus the maximum rank(A)=n-1.

If $k$ eigen values of a $n\times n$ matrix are $0$(or Eigen value $0$ has multiplicity $k$ in the characteristic equation), then its rank is $n-k$. So maximum value of rank can be $n-1$
An $n\times n$ matrix $A$ has a $0$ as an eigenvalue if and only if $A$ is not invertible. Thus $\operatorname{rank}(A) < n$, saying $\max \operatorname{rank}(A) = n - 1$.