1. ## diagonalizable

hi,

what does it mean to say that a matrix A is diagonalizable iff R^n has a basis consisting of eigenvectors for A?

2. If $\displaystyle B=\left\{{e_1,\ldots,e_n}\right\}$ is such a basis then, $\displaystyle Ae_1=\lambda_1,\ldots,Ae_n=\lambda_ne_n$

If we denote

$\displaystyle P=\begin{bmatrix}e_1\hdots e_n\end{bmatrix}\;,\quad D=\begin{bmatrix} \lambda_1 & 0 & \ldots & 0\\ 0 & \lambda_2 & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & \lambda_n\end{bmatrix}$

then

$\displaystyle AP=A\begin{bmatrix}e_1 \hdots e_n\end{bmatrix}=\begin{bmatrix}Ae_1 \hdots Ae_n\end{bmatrix}=\begin{bmatrix}\lambda_1e_1 \hdots \lambda_ne_n\end{bmatrix}=\\D\begin{bmatrix}e_1 \hdots e_n\end{bmatrix}=DP$

This means $\displaystyle P^{-1}AP=D$ i.e. $\displaystyle A$ is similar to a diagonal matrix (definition of diagonalizable matrix).

Try to prove the reciprocal statement, if a is diagonalizable then ...

Regards.

Fernando Revilla

3. so in this case, instead of B consisting of eigenvectors, B consists of elementary matrices?

4. Originally Posted by alexandrabel90
so in this case, instead of B consisting of eigenvectors, B consists of elementary matrices?
If $\displaystyle A$ is similar to a diagonal matrix $\displaystyle D$, (that is there exists $\displaystyle P$ invertible such that $\displaystyle P^{-1}AP=D$) try to prove that the columns of $\displaystyle P$ determine a basis of eigenvectors)

Hint: $\displaystyle P^{-1}AP=D$ is equivalent to $\displaystyle AP=PD$.

Regards.

Fernando Revilla

5. Originally Posted by alexandrabel90
so in this case, instead of B consisting of eigenvectors, B consists of elementary matrices?
Assuming you mean "B" to be the basis, then, no, a basis is a set of vectors, not a set of matrices. I have no idea where you saw anything related to "elementary matrices" in what FenandoRevilla wrote!