Linear map

Beta · **Joined:** Thu, 30 Oct 2008 10:47:09 UTC **Posts:** 182

Let

be a complex vector space and let $A:V\to V$ be a linear map satisfying A^n=I

.
Prove that

has a basis consisting of eigenvectors of the map

.

Any idea ?

outermeasure · **Posted:** Mon, 16 Feb 2009 14:27:50 UTC

Beta wrote:

Let

be a complex vector space and let $A:V\to V$ be a linear map satisfying A^n=I

.
Prove that

has a basis consisting of eigenvectors of the map

.

Any idea ?

Hint: $(A-I)(A-\omega I)(A-\omega^2 I)\cdots(A-\omega^{n-1}I)=0$ , where $\omega=\exp(2\pi i/n)$ .

Beta · **Joined:** Thu, 30 Oct 2008 10:47:09 UTC **Posts:** 182

So the eigenvalues are $\lambda=\omega^k$ , for some $k\in\{0,1,...,n-1\}$ , with multiplicity

.
And the eigenvectors can be any non-zero vectors?

outermeasure · **Posted:** Mon, 16 Feb 2009 15:50:03 UTC

Beta wrote:

So the eigenvalues are $\lambda=\omega^k$ , for some $k\in\{0,1,...,n-1\}$ , with multiplicity

.
And the eigenvectors can be any non-zero vectors?

Err... you can't deduce the algebraic&geometric multiplicities from that. However, depending on what you know, either you play around with that equation and show that $V=\bigoplus_k \mathop{\mathrm{ker}}(A-\omega^k I)$ , or you invoke the minimal polynomial is a product of distinct linears iff diagonalisable. Both will give you what you want.

Shadow · **Posted:** Mon, 16 Feb 2009 22:26:53 UTC

Beta wrote:

Let

be a complex vector space and let $A:V\to V$ be a linear map satisfying A^n=I

.
Prove that

has a basis consisting of eigenvectors of the map

.

Any idea ?

Well, we have that A^n-I=0

, think characteristic polynomial, and since it splits, what does that say about diagonalizability?

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Linear map

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