0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What issues are you having with the question? Lemma 1 looks like it does all the hard work for you. $\endgroup$ – Theo Bendit Sep 18 '18 at 15:52
  • $\begingroup$ I'm sorry, I don't really understand the question. If $(v_1, ..., v_n)$ is an eigenbasis, then $A$ is diagonizable by definition. $\endgroup$ – A. Salguero-Alarcón Sep 18 '18 at 16:44
  • $\begingroup$ 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 $(\lambda_1, \ldots, \lambda_n)$. $\endgroup$ – Viktor Glombik Sep 18 '18 at 17:23
1
$\begingroup$

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 ...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.