# When a determinant is zero

Is it true that if $C$ is a square matrix of size $n$ and $\det(C) = 0,$ then $C^n = O_n$ or the $0$ matrix? If yes, then why is that?

I know that the reverse is obviously true, so I wondered if there is an equivalence relation between $\det(C) = 0$ and $C^n = \text{ the$0$matrix. }$

• Try it for $n=2$ and $C$ the all-ones matrix. Feb 8, 2015 at 11:50
• Did you mean to say $O(n)$ (this is usually read "big-Oh of $n$")? If so, what did you want to say by it? If not (then I suppose you meant $0$ instead of $O$), then why add "or the $0$ matrix"? Feb 8, 2015 at 17:30
• I did it for clarity. I know now I was supposed to write $O_n$. Feb 9, 2015 at 4:54

Not true ! Take $$A=\begin{pmatrix} 1&1\\1&1 \end{pmatrix}$$ its determinant is obviously $0$ and we have $$A^n=2^{n-1}A$$ never the zero matrix

• I like that this also shows that the norm of the matrix can grow faster than linear (which is how I would have interpreted the question as to whether $A^n=O(n)$). Feb 9, 2015 at 20:45
• You're right! Hadn't thought of it this way Feb 9, 2015 at 22:36

No. If

$$C=\begin{pmatrix} 1&1\\0& 0\end{pmatrix}$$

then $C^n=C\neq 0$ for $n\in \mathbb N$, yet $\det C =0$

If $C^n=0$ whenever $\det C=0$ that means $0$ is the only eigen value of the matrix whenever $0$ is an eigen value of $C$.

But that is not the case anyway ,consider any upper triangular matrix with one entry in the diagonal zero and others non-zero

Over the complex field (or any algebraically closed field) any matrix $C$ can be conjugated to an upper triangular one $D = X^{-1} C X$, with the eigenvalues (of $C$) on the diagonal.

Now $\det(D) = \det(C)$ is the product of the diagonal elements of $D$, i.e. the eigenvalues, so to say that $\det(C) = 0$ means that one of the eigenvalues is zero.

On the other hand $$C^{n} = X D^{n} X^{-1}$$ can only be zero for some $n$ if and only if all the diagonal elements of $D$, i.e. the eigenvalues, are zero.