# Show that T is diagonalizable

• Mar 12th 2013, 10:57 AM
dave52
Show that T is diagonalizable
Suppose T: V----> V has only two distinct eigenvalues k1 and k2

Show that T is diagonalizable iff V = Ek1+ Ek2
• Mar 12th 2013, 11:08 AM
ILikeSerena
Re: Show that T is diagonalizable
Quote:

Originally Posted by dave52
Suppose T: V----> V has only two distinct eigenvalues 1 and 2

Show that T is diagonalizable iff V = E1 + E2

Hi dave52! :)

It seems to me that it is not true.

$\displaystyle \begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&0&2\end{bmatrix}$

This matrix is diagonal, has distinct eigenvalues 1 and 2, but $\displaystyle \mathbb R^3 \ne E_1 + E_2$.
• Mar 12th 2013, 11:27 AM
jakncoke
Re: Show that T is diagonalizable
I think you meant to say that your vector space has Dimension 2.
• Mar 12th 2013, 11:43 AM
dave52
Re: Show that T is diagonalizable
sorry I just fixed the problem
• Mar 13th 2013, 01:09 PM
ILikeSerena
Re: 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 $\displaystyle V = E_1+ E_2 \qquad \blacksquare$