Suppose T: V----> V has only two distinct eigenvalues k_{1}and k_{2}

Show that T is diagonalizable iff V = E_{k1}+ E_{k2}

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- Mar 12th 2013, 10:57 AMdave52Show that T is diagonalizable
Suppose T: V----> V has only two distinct eigenvalues k

_{1}and k_{2}

Show that T is diagonalizable iff V = E_{k1}+ E_{k2} - Mar 12th 2013, 11:08 AMILikeSerenaRe: Show that T is diagonalizable
- Mar 12th 2013, 11:27 AMjakncokeRe: Show that T is diagonalizable
I think you meant to say that your vector space has Dimension 2.

- Mar 12th 2013, 11:43 AMdave52Re: Show that T is diagonalizable
sorry I just fixed the problem

- Mar 13th 2013, 01:09 PMILikeSerenaRe: Show that T is diagonalizable
Let's first proof T is diagonalizable$\displaystyle \Rightarrow$ V = E1 + E2.

If T is diagonalizable, it can be written as $\displaystyle PDP^{-1}$, where D is a diagonal matrix with its eigenvalues on the main diagonal, and P represents a basis of eigenvectors.

That basis spans V.

Therefore V is the direct sum of the eigen spaces.

Now let's proof $\displaystyle V = E_1 + E_2$ $\displaystyle \Rightarrow$ T is diagonalizable.

This means that that basis of $\displaystyle E_1$ together with the basis of $\displaystyle E_2$ span V.

Put those bases next to each other in a matrix P, and put the eigenvalues diagonally in a corresponding matrix D, and you have that

$\displaystyle TP = PD$

$\displaystyle T = PDP^{-1}$

So T is diagonalizable.

Therefore T is diagonalizable iff $\displaystyle V = E_1+ E_2 \qquad \blacksquare$