# Algebraic or geometric multiplicity?

I am reading a proof of the fact that

every linear transformation $L:V\to V$ can be represented by an upper triangular matrix $M$, with eigenvalues on the diagonal. And if the algebraic multiplicity of an eigenvalue $\lambda$ is $\alpha(\lambda)$, then $\lambda$ appears $\alpha(\lambda)$ times.

The proof is by induction on $m=dim(V)$. It's trivial when $m=1$. Now suppose it's true for all $1,2,3,\cdots,m-1$.

Then for $m$, because there exists at least one eigenvalue, let's call it $\lambda$. Then the author considers $M-\lambda I_V$, which is singular and has a range $W\subsetneq V$, with dimension $k<m$. Then he restricted the operation on $W$, which has dimension smaller than $m$ and can apply the induction assumption. He claimed that $V=W\oplus U$, where $U$ has dimension $l$ and $k+l=m$.

He claimed that $l$ is the algebraic multiplicity $\alpha(\lambda)$, but I think it should be the geometric multiplicity $\gamma(\lambda)$. My understanding is that $\gamma(\lambda)$ is the dimension of the subspace such that when acted on by $M-\lambda I_V$ gives the null vector, and so it should equal $l$.

What's wrong with my understanding? Thank you!

You are correct in that $l$ is the geometric multiplicity of $\lambda$ and doesn't have to be the algebraic multiplicity of $\lambda$. Consider for example the matrix

$$A = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$$

and let $L(x) = Ax$ be the associated linear map on $\mathbb{F}^n$. Then we have $W = \left< e_1 \right>$ and so $k = 1$ and $l = 1$ but the algebraic multiplicity of $\lambda$ is $2$.

• I've changed the example so it works for any $\lambda$. Commented Jun 24, 2016 at 0:31
• right, I made an error computing $\det(A_\lambda-x I) = (\lambda-x)^2$ :D Commented Jun 24, 2016 at 0:33
• Why $W=\langle e_1 \rangle$? Commented Jun 24, 2016 at 0:53
• $W = \left< (L - \lambda I)e_1, (L - \lambda I)e_2 \right> = \left< e_1 \right>$ as $Le_1 = \lambda e_1$ and $(L - \lambda I)e_2 = e_1$. Commented Jun 24, 2016 at 1:04
• Oops sorry, I forgot to subtract $\lambda I$. Commented Jun 24, 2016 at 1:13

I think that the algebraic multiplicity of an eigenvalue $\lambda_0$ of a matrix $A$ is basically the multiplicity that $\lambda_0$ has in the characteristic polynomial of $A$. The geometric multiplicity of $\lambda_0$ should be the dimension of the eigenspace $E_{\lambda_0}(A)$, or the dimension of the nullspace $N(A-\lambda_0I)$. Hence I think that algebraic multiplicity is correct in the definition.

I think your doubt comes from the fact that, when a matrix is diagonalizable we know that the algebraic and the geometric multiplicity of an eigenvalue $\lambda_0$ are the same!

• But isn't $k$ the rank of $M-\lambda I_V$, and $l=N(M-\lambda I_V)$, so that we have rank + nullity = $m$? Commented Jun 24, 2016 at 0:06
• mmm, now you're making me doubting about it ahah Commented Jun 24, 2016 at 0:11