Given a diagonal matrix, what is number of times $\lambda$ appears (on the diagonal)

Question

Suppose T ∈ L(V ) has a diagonal matrix A with respect to some basis of V and that λ ∈ F. Prove that λ appears on the diagonal of A precisely dimE(λ, T) times.

Honestly, I have no clue how to prove this statement! Any help would be appreciated!

Answer 1

This is the same problem as Exercise 5.C.7 in "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.

Let $n:=\dim V$.
Let $v_1,\dots,v_n$ be a basis with respect to which $T$ has $A$.
Let $\{\lambda_1,\dots,\lambda_m\}$ be the set of elements of $\mathbb{F}$ which appear on the diagonal of $A$.
Then $\{\lambda_1,\dots,\lambda_m\}$ is the set of the distinct eigenvalues of $T$ by 5.32 on p.152.
Assume that $\lambda_i$ appears on the diagonal of $A$ precisely $n_i$ times for $i\in\{1,\dots,m\}$.
Assume that $n_i\neq \dim E(\lambda_i,T)$ for some $i\in\{1,\dots,m\}$.
Since $n_1+\dots+n_m=n$, there exists $j\in\{1,\dots,m\}$ such that $n_j > \dim E(\lambda_j, T)$.
Since $\lambda_j$ appears on the diagonal of $A$ precisely $n_j$ times, there exists $v_{i_1},\dots,v_{i_{n_j}}\in\{v_1,\dots,v_n\}$ such that $i_k\neq i_l$ if $k\neq l$ and $Tv_{i_k}=\lambda_jv_{i_k}$ for $k\in\{1,\dots,n_j\}$.
So, $v_{i_1},\dots,v_{i_{n_j}}\in E(\lambda_j, T)$.
Since $v_{i_1},\dots,v_{i_{n_j}}$ is linearly independent, $\dim E(\lambda_j, T)\geq n_j>\dim E(\lambda_j, T)$.
This is a contradiction.
So, $n_i=\dim E(\lambda_i,T)$ for all $i\in\{1,\dots,m\}$.
Let $\lambda\in\mathbb{F}$.
If $\lambda$ is not an eigenvalue of $T$, then $\lambda$ appears on the diagonal of $A$ precisely $0$ times by 5.32 on p.152.
And $\dim E(\lambda,T)=\dim \{0\}=0$.
If $\lambda$ is an eigenvalue of $T$, then $\lambda=\lambda_i$ for some $i\in\{1,\dots,m\}$.
From the result above, $n_i=\dim E(\lambda_i,T)=\dim E(\lambda,T)$.
So, $\lambda$ appears on the diagonal of $A$ precisely $\dim E(\lambda,T)$ times.