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

Question: Prove $\displaystyle \lambda=0$ 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)

$\displaystyle \lambda=0$ 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

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

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

$\displaystyle x_1=-2t_1-2^2t_2-...-2^{n-1}t_{n-1}$

$\displaystyle x_2=t_1$

$\displaystyle x_3=t_2$

.

.

.

$\displaystyle x_n = t_{n-1}$

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

and those eigenvectors form the basis for that space.

My friend got:

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

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