Diagonalization, Solving a System of Linear Equations I have questions regarding the following task, which is to diagonalize the matrix A:
$\begin{pmatrix} 1 & 0 & 0 \\     0 & 9 & -20 \\ 0 & 4 & -9 \end{pmatrix}$
What I have done so far:


*

*Calculating the Characteristic Polynomial I arrive at $(\lambda-1)^2(\lambda+1)$ which means that two eigenvalues are 1 and one is -1.

*Then I start to caculate the eigenvectors. Given we find three independent eigenvectors, the matrix A is diagonalizable and we can find the diagonal matrix with the help of these eigenvectors.


Now I have the following two questions:


*

*For finding the eigenvector to the eigenvalue 1 I solve the following equation:


$(I*\lambda - A)*v$ = 0 i.e.
$(\begin{pmatrix} 1 & 0 & 0 \\     0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} -\begin{pmatrix} 1 & 0 & 0 \\     0 & 9 & -20 \\ 0 & 4 & -9 \end{pmatrix})*(v_1,v_2,v_3)^T = (0,0,0)^T$
which leads to the solution (c, 5/2a,a) with $a,c \in \mathbb R$. My question: How do I know that this eigenvector can be in fact written as the addition of the two independent vectors (1,0,0) and (0,5/2,1) (and hence the eigenspace is 2-dimensional). How for instance do I know that taking  (c, 5/2a,a) into (1,0,0), (0,0,1) and (0,5/2,0) would be wrong?


*

*When it comes to the second eigenvalue,-1, I would calculate the eigenvector as follows by solving:
$(\begin{pmatrix} -1 & 0 & 0 \\     0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} -\begin{pmatrix} 1 & 0 & 0 \\     0 & 9 & -20 \\ 0 & 4 & -9 \end{pmatrix})*(v_1,v_2,v_3)^T = (0,0,0)^T$


Here I get $(\begin{pmatrix} -2 & 0 & 0 \\     0 & -10 & 20 \\ 0 & -4 & 8 \end{pmatrix}*(v_1,v_2,v_3)^T = (0,0,0)$ and hence the eigenvector: (-2,5/2,1) - but this seems to be wrong..where is my error?
Thanks!
 A: The first question, you have to prove to yourself that $W_{\lambda}$, the Eigenspace corresponding to an eigenvalue $\lambda$ is a vector subspace.  To see this, notice that $$A (k v + l w) = k(Av) + l(Aw) = k \lambda v + l \lambda w = \lambda (kv + lw)$$
So any linear combination of eigenvectors corresponding to the same eigenvalue is also an eigenvector.  To see why your choice of choosing a basis for $W_1$ doesn't work, notice that $(0,0,1)$ can never be written in the form $(c, \frac{5}{2} a, a)$.  You can only break the space up into vectors that actually belong in it.  On the other hand, writing $(c, \frac{5}{2} a, a)$ as $c(1,0,0) + a(0, \frac{5}{2}, 1)$ makes it clear that the choices of $(1,0,0)$ and $(0, \frac{5}{2}, 1)$ are suitable for a basis.
For your second question, I think this is just an arithmetic error.  It seems pretty clear that your eigenvectors in $W_{-1}$ will be of the form $k (0, 2, 1)$.  Remember that you should always get infinitely many solutions when solving for eigenvectors corresponding to an eigenvalue you have found by solving $|\lambda I - A| = 0$
A: $\lambda =1:$
When you put $c=1,a=0$ you get eigenvector $(1,0,0)$, when you put $c=0,a=1$ you get eigenvector $(0,\frac{5}{2},1)$. You cannot produce vectors $(0,0,1)$ and $(0,\frac{5}{2},0)$ you are asking for.
$\lambda=-1:$
$v_1$ should be 0 and $v_2 = 2 v_3$ and therefore possible eigenvector could be $(0,2,1).$
@muzzlator did it nicely.
