Let$A∈M_{n×n}(C)$ with characteristic polynomial $p(x) =cx^a\prod _{i=1}^{k}(λ_i−x)$ and $λ_i\not= 0,∀i,a∈Z_{>0}.$ Show that if $\dim(\ker(A)) +k=n$, then A is diagonalizable.

My proof:

Each linear factor in $p(x)$ has algebraic multiplicity 1, and $1 \leq \dim(E_x) \leq$ the algebraic multiplicity of $x$.

So $1 \leq \dim(E_x) \leq 1 \implies \dim(E_x) = 1$.

Therefore the algebraic multiplicity is equal to the geometric multiplicity for each eigenvector and A is diagonizable.

Is this proof valid? It seems too obvious and I didn't even use $\dim(\ker(A)) +k=n$.

  • $\begingroup$ The statement is false, unless you assume that $\lambda_i\ne\lambda_j$, for $i\ne j$. About your question, $0$ is an eigenvalue as soon as $a>0$, and you need to take care of it, too. $\endgroup$ – egreg Apr 6 at 7:46
  • $\begingroup$ I'm finding your first point hard to understand, would you mind elaborating? The question guidelines imply that $λ_i \not= 0$, so would that also solve the second problem? $\endgroup$ – Pivot Apr 6 at 7:56

The statement is false, unless you make some further assumption. For every matrix, the characteristic polynomial has the stated form, but not every matrix is diagonalizable. Simple example: the characteristic polynomial of $$\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ is $p(x)=x\prod_{i=1}^2(\lambda_i-x)$, with $\lambda_1=\lambda_2=1$. This matrix is not diagonalizable.

The assumption under which the statement is true is that $\lambda_i\ne\lambda_j$, for $i\ne j$.

In this case your argument is good: each nonzero eigenvalue has algebraic multiplicity $1$. However, you also need to take care of the $0$ eigenvalue, when $a>0$.

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