hi,

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

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- Nov 24th 2010, 06:47 AMalexandrabel90diagonalizable
hi,

what does it mean to say that a matrix A is diagonalizable iff R^n has a basis consisting of eigenvectors for A? - Nov 24th 2010, 08:14 AMFernandoRevilla
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 - Nov 24th 2010, 08:17 AMalexandrabel90
so in this case, instead of B consisting of eigenvectors, B consists of elementary matrices?

- Nov 24th 2010, 09:15 AMFernandoRevilla
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)

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

Regards.

Fernando Revilla - Nov 25th 2010, 02:32 AMHallsofIvy