# Characteristic polynomial and diagonalizability of a matrix.

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 $$\dim(\ker(A)) +k=n$$.

• 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. – egreg Apr 6 at 7:46
• 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? – 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 $$a>0$$.