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

eigenvalues-eigenvectorsjordan-normal-formlinear algebralinear-transformations

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?

Best Answer

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.

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).

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.

Related Question