# Proof on Diagonalizability

What I am given

Definition 1: A endomorphism $F \in \operatorname{End}(V)$ is called diagonalizable iff a basis consisting of eigenvectors ('eigenbasis') exists.

Definition 2: A matrix A is called diagonalizable iff the endomorphism $F(x) = Ax$ is diagonalizable.

Lemma 1: The columns of $S$ (mentioned below) are a eigenbasis.

Question

Prove that if $(v_1, \ldots, v_n)$ is a eigenbasis, $A$ is diagonizable iff the invertible matrix $S = (v_1, \ldots, v_n)$ satisfies $S^{-1} A S = diag(\lambda_1, \ldots, \lambda_n)$, where $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$.

My idea: I will show $A S = S diag(\lambda_1, \ldots, \lambda_n)$. We have $$A S = A (v_1 \ldots v_n) = (\lambda_1 v_1 \ldots \lambda_n v_n) = (v_1 \ldots v_n) (\lambda_1 \ldots \lambda_n) = S \begin{pmatrix} \lambda_1 & & 0 \\ & \ddots & \\ 0 & & \lambda_n \end{pmatrix}.$$ I am pretty sure the third equality doesn't hold because that only works if (v_1 \ldots v_n) only has elements on it's diagonal.

• What issues are you having with the question? Lemma 1 looks like it does all the hard work for you. – Theo Bendit Sep 18 '18 at 15:52
• I'm sorry, I don't really understand the question. If $(v_1, ..., v_n)$ is an eigenbasis, then $A$ is diagonizable by definition. – A. Salguero-Alarcón Sep 18 '18 at 16:44
• The questions asked if the matrix is diagonizable, and the definition of diagonizable matrix is that the corresponding endomorphism is diagonaziable. We want to show that the eigenbasis-matrix $S$ satisfies $A S = S$ diag $(\lambda_1, \ldots, \lambda_n)$. – Viktor Glombik Sep 18 '18 at 17:23

This is correct. Think about the first column of that last product. It gives you $\lambda_1$ times the first column of $S$, which is $\lambda_1v_1$. What about the second column? It gives you $\lambda_2$ times the second column of $S$, which is $\lambda_2v_2$. And so on ...