diagonalizable

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November 24th 2010, 06:47 AM
alexandrabel90

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?
November 24th 2010, 08:14 AM
FernandoRevilla

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

If we denote

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

then

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

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

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

Regards.

Fernando Revilla
November 24th 2010, 08:17 AM
alexandrabel90

so in this case, instead of B consisting of eigenvectors, B consists of elementary matrices?
November 24th 2010, 09:15 AM
FernandoRevilla

Quote:

Originally Posted by alexandrabel90

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

If $A$ is similar to a diagonal matrix $D$ , (that is there exists $P$ invertible such that $P^{-1}AP=D$ ) try to prove that the columns of $P$ determine a basis of eigenvectors)

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

Regards.

Fernando Revilla
November 25th 2010, 02:32 AM
HallsofIvy

Quote:

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!