If L is diagonalizable, algebraic & geometric multiplicities are equal

[JD_PM]

TL;DR Summary

I want to prove that if $L$ is diagonalizable then each eigenvalue $\lambda_i$ has algebraic multiplicity equal to its geometric multiplicity (i.e. $a(\lambda_i ) =g(\lambda_i ))$.

I am not used to write down formal mathematical proofs so any advice on that is very much welcomed!

Given a n-dimensional vector space $V$ (where n is a finite number) and a linear operator $L$ (which, by definition, implies $L:V \to V$; reference: Linear Algebra Done Right by Axler, page 86) whose characteristic polynomial (we assume) can be factorized out as first-degree products over $\Bbb R$ and whose collection of eigenvalues is $\{\lambda_1, \lambda_2,..., \lambda_k\}$. If $L$ is diagonalizable then each eigenvalue $\lambda_i$ has algebraic multiplicity equal to its geometric multiplicity (i.e. $a(\lambda_i ) =g(\lambda_i )$

First off, we assume that the matrix $L$ is diagonalizable. That implies that there is a basis of $V$ consisting of eigenvectors of $L$. We made no assumption regarding the degeneracy of the spectrum so let's keep it general. Let us label $a_1$ as the collection of all eigenvectors with eigenvalue $\lambda_i$ and we do the same up to $a_k$. Now we construct the basis $\beta$ of eigenvectors of $L$ i.e.

\begin{equation*}
\beta = \{ a_1, a_2, ..., a_k\}
\end{equation*}

Thus $D=P^{-1} A P$, where $D$ is a diagonal matrix, (with respect to the $\beta$ basis) containing the eigenvalues of $L$ i.e.

We now build up the matrix $P$ as a row matrix containing the columns of $P$

\begin{equation*}
P=(p_1 p_2 ... p_k)
\end{equation*}

We note that

\begin{align*}
&AP=PD \Rightarrow \\
&\Rightarrow (Ap_1 Ap_2 ... Ap_k) = A(p_1 p_2 ... p_k) = (p_1 p_2 ... p_k)D=(\lambda_1 p_1 \lambda_2 p_2 ... \lambda_k p_k)
\end{align*}

So we see that each of the columns of the matrix $P$ satisfies the eigenvector equation i.e. $Ap_i = \lambda_i p_i$.

What I do not see is how to show that $a(\lambda_i) = g(\lambda_i)$ from here.

Thank you!

 

[Infrared]

The sum of the algebraic multiplicities is $n$. Since $L$ is diagonalizable, the sum of the geometric multiplicities is also $n$. Since $a(\lambda)\geq g(\lambda)$ for each eigenvalue $\lambda$, these sums can only be equal if we always have $a(\lambda)=g(\lambda).$

 

[JD_PM]

The sum of the algebraic multiplicities is $n$.

Alright I see this is because we are given that the characteristic polynomial completely factorizes in first degree products over $\Bbb R$

Since $L$ is diagonalizable, the sum of the geometric multiplicities is also $n$.

Naïve question: why?

 

[Infrared]

Since $L$ is diagonalizable, $V$ is the direct sum of the eigenspaces for $T.$ The dimension of each eigenspace is the geometric multiplicity of the corresponding eigenvalue.

 

[jbunniii]

More concretely, suppose that the algebraic multiplicity of eigenvalue $\lambda$ is $m$. Since $A$ is diagonalizable, then (permuting the rows and columns if necessary) we may assume that there is a matrix $D$ such that the first $m$ diagonal entries of $D$ are $\lambda$, and that there is an invertible matrix $P$ such that $PD = AP$.

If $e_j$ denotes the column vector with $1$ in the $j$'th entry and zeros everywhere else, then for $1 \leq j \leq m$ we have $De_j = \lambda e_j$. Also, if $p_j$ denotes the $j$'th column of $P$, then $Pe_j = p_j$. Therefore:
$$PDe_j = P(\lambda e_j) = \lambda Pe_j = \lambda p_j$$
and
$$APe_j = Ap_j$$
Since the LHS's of the above two equations are equal, so are the RHS's:
$$Ap_j = \lambda p_j$$
which shows that $p_j$ is an eigenvector of $A$ associated with $\lambda$.

This holds for $1 \leq j \leq m$, which shows that the geometric multiplicity of $\lambda$ is at least $m$; here we are using the fact that $p_1,p_2,\ldots,p_m$ are linearly independent since $P$ is invertible.

Since the geometric multiplicity cannot exceed the algebraic multiplicity, which is also $m$, the conclusion follows.