[SOLVED] Diagonalizable

**dwsmith** · May 1st 2010, 09:59 AM

If A is nxn diagonalizable matrix, then each vector in $\mathbb{R}^n$ can be written as a linear comination of eigenvectors of A.

This is true but how do I prove this?

**HallsofIvy** · May 2nd 2010, 04:51 AM

Since A is diagonalizable, there exist a diagonal matrix D and an invertible matrix P such that $A= PDP^{-1}$ , $D= P^{-1}AP$

Suppose $\lambda$ is an eigenvalue of A. That means that there exist a non-zero vector v such that $Av= \lambda v$ . Let $u= P^{-1}v$ . Since P is invertible, u is non-zero also and $v= Pu$ . But now, $Du= P^{-1}APu= P^{-1}Av= \lambda P^{-1}v= \lambda u$ .

That is, $\lambda$ is also an eigenvalue of the diagonal matrix D and, since a diagonal matix has its eigenvalues on its diagonal, is one of the numbers on the diagonal of D. Further the eigenvectors of a diagonal matrix are simply the standard basis vectors <1, 0, 0...>, etc.

That means that set of all eigenvectors u, corresponding to the eigenvalues of D are independent and their number is equal to the dimension of the space.. Now, v= Pu and since P is invertible, it maps independent sets of vectors into independent sets of vectors. Since their number is equal to the dimension of the space, it follow that they are a basis for the space.

**dwsmith** · May 2nd 2010, 08:44 AM

Originally Posted by HallsofIvy

Let $u= P^{-1}v$

How is $P^{-1}$ being multiplied by v?

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