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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. }$
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
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.
