1. ## Diagonalizable Matrix

$A= \left(\begin{array}{cccccc}1&2&2^2&...&...&2^{n-1}\\2&2^2&2^3&...&...&2^{n}\\2^2&2^3&2^4&...&...&2 ^{n+1}\\.&.&.&&&.\\.&.&.&&&.\\2^{n-1}&2^n&2^{n+1}&...&...&2^{2n-2}\end{array}\right)$

Question: Prove $\lambda=0$ is an eigenvalue of A and find a basis for it's space.

Solution: (My friend gave a different answer so I'd like to know which one is correct)

$\lambda=0$ if matrix A is singular. Matrix A is clearly singular...

To find a basis, I do the elementary operations on matrix A and I'm left with

$A= \left(\begin{array}{cccccc}1&2&2^2&...&...&2^{n-1}\\0&0&0&...&...&0\\0&0&0&...&...&0\\.&.&.&&&.\\. &.&.&&&.\\0&0&0&...&...&0\end{array}\right) = 0$

Matrix A has a rank of 1 ---- dimp=n-p(A) ---- dimp = n-1

$x_1=-2t_1-2^2t_2-...-2^{n-1}t_{n-1}$
$x_2=t_1$
$x_3=t_2$
.
.
.
$x_n = t_{n-1}$

$t_1 \left(\begin{array}{cccccc}-2\\1\\0\\.\\.\\0\end{array}\right)$+ $t_2 \left(\begin{array}{cccccc}-2^2\\0\\1\\.\\.\\0\end{array}\right)$ +...+ $t_{n-1} \left(\begin{array}{cccccc}-2^{n-1}\\0\\0\\.\\.\\1\end{array}\right)$

and those eigenvectors form the basis for that space.

My friend got:

$\left(\begin{array}{cccccc}2\\-1\\0\\.\\.\\0\end{array}\right)$, $\left(\begin{array}{cccccc}0\\2\\-1\\.\\.\\0\end{array}\right)$,..., $\left(\begin{array}{cccccc}0\\0\\.\\.\\2\\-1\end{array}\right)$

So which is the correct answer? (or is it possible they are both correct...or incorrect)

2. ** Did I find the correct basis?
** Also, is there a way to change the Thread topic? I don't need to call it Diagonalizable Matrix...(that's the third part of the question, but it's not hard...)

3. Both bases are correct.

4. Originally Posted by HallsofIvy
Both bases are correct.
If it's not too troublesome, would you mind showing me how the other one is a basis.