Why is it not sufficient that geometric multiplicity is equal to algebraic multiplicity to imply that $A$ diagonalizable

Question

I have been told that for a given matrix $A$: $A \operatorname{diagonalizable} \Rightarrow m_{a}(\lambda)=m_{g}(\lambda)$ for all $\lambda \in \sigma (A)$ where $\sigma (A)$ denotes the spectrum of $A$ and $m_{a}(\lambda)$ is the algebraic multiplicity of $\lambda$ while $m_{g}(\lambda)$ is the geometric multiplicity of $\lambda$.

This then of course would mean that a matrix $B$ who has algebraic multiplicity equal to geometric multiplicity for all eigenvalues is not necessarily diagonalizable.

But I thought that:

$C$ diagonalizable as Linear Endomorphism on finite-dimensional $V$ $\iff$ $\bigoplus_{\lambda \in \sigma (C)}E_{A}(\lambda)=V$

and I thought that $\bigoplus_{\lambda \in \sigma (C)}E_{C}(\lambda)=V$ is simply another way of stating that $m_{a}(\lambda)=m_{g}(\lambda)$ for all $\lambda \in \sigma (\lambda)$. So thus, what is the difference between the two statements?

Answer 1

In your first assertion you wrote $\Rightarrow$, which is correct, but in fact the converse $\Leftarrow$ is also (almost) true. There are several equivalent criterion for diagonalizability. I'll list them out for you.

Let $V$ be a finite-dimensional vector space over a field $F$, and let $T:V \to V$ be a linear map. Then the following statements are all equivalent.

• $T$ is diagonalizable.
• There is an ordered basis $\beta$ of $V$ consisting of eigenvectors of $T$.
• We can express the vector space $V$ as a direct sum of eigenspaces of $T$: \begin{align} V = \bigoplus \limits_{\lambda \in \sigma(T)} \ker(T-\lambda I) \end{align}
• The characteristic polynomial of $T$ splits over $F$, and for every eigenvalue $\lambda$ of $T$, \begin{align} \dim \ker(T - \lambda I) = \text{algebraic multiplicity of $\lambda$} \end{align} (i.e geometric multiplicity = algebraic multiplicity)

There are a few more equivalent statements if you know about Jordan canonical forms and minimal polynomials, however, for now, you should try to prove that the $2^{\text{nd}}$ and $3^{\text{rd}}$ statements are equivalent.

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Edit: After looking at ThorWittich's answer, I modified my answer to include the assumption that the characteristic polynomial splits (I implicitly assumed this fact throughout the discussion as is usually done, but strictly speaking I should have explicitly stated this).

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Edit 2:

The assumption that the characteristic polynomial splits over the given field is important. To show this, consider the matrix \begin{align} A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \in M_{2 \times 2}(\Bbb{R}) \end{align} So we are working over the field $\Bbb{R}$. It is easy to see that the characteristic polynomial is $\chi_A(t) = t^2 + 1$, which doesn't split over $\Bbb{R}$. Hence, $A$ is NOT diagonalizable over $\Bbb{R}$. However, if we consider $A$ to be an element of $M_{2 \times 2}(\Bbb{C})$, then the characteristic polynomial splits over $\Bbb{C}$, and there are two eigenvalues of $A$, namely $i$ and $-i$, and it's easy to verify (either directly or by using the 4th condition above) that $A$ is diagonalizable over $\Bbb{C}$.

Hence, the statement that the characteristic polynomial splits over the given field is actually important.

Answer 2

They are equivalent. Formally, you should add one condition, namely that the characteristic polynomial splits in linear factors (over the field you are considering for your vector space). That is actually rather a plausibility condition to be able to have multiplicities, but I still feel like one should mention that. So the following are equivalent:

1) $A$ is diagonalizable.

2) The characterstic polynomial $\chi_A(T)$ splits into linear factors and the algebraic and geometric multiplicities coincide for each eigenvalue.