Find a $3 \times 3$ matrix given its eigenspaces Find a matrix $A_ {3\times3}$ such that $A$ has two eingenvalues  $\lambda_1,\lambda_2$ and their Eigenspaces are $E_{\lambda_1}=\{(x,y,z):x+2y-3z=0 \}$ and $E_{\lambda_2}=\{(x,y,z):2x=-y=z \}$.
In this problem the eigenvalues aren't given and that makes it harder, I have been trying to solve $Av=\lambda_1v$ where $v=(1,1,1)$ and $Au=\lambda_2u$ where $u=(1,2,2)$ but then I get almost trivial equations and get stuck there, since I don't know $\lambda_1,\lambda_2$ I haven't tried by diagonalization. Any help is welcome. Thanks!
 A: A basis of the first eigenspace is given by
$$
v_1=\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}
\quad
v_2=\begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix}
$$
while an eigenvector relative to $\lambda_2$ is
$$
v_3=\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}
$$
Therefore, the matrix $A$ can be written as
$$
A=S
\begin{bmatrix}
\lambda_1 & 0 & 0 \\
0 & \lambda_1 & 0 \\
0 & 0 & \lambda_2
\end{bmatrix}
S^{-1}
$$
where
$$
S=\begin{bmatrix}
-2 & 3 & 2 \\
1 & 0 & -1 \\
0 & 1 & 1
\end{bmatrix}
$$
Doing the computations we get
$$
A=\begin{bmatrix}
(5\lambda_1 - 2\lambda_2)/3 & (4\lambda_1 - 4\lambda_2)/3 & -2\lambda_1 + 2\lambda_2 \\
(-\lambda_1 + \lambda_2)/3 & (\lambda_1 + 2\lambda_2)/3 & \lambda_1 - \lambda_2 \\
(\lambda_1 - \lambda_2)/3 & (2\lambda_1 - 2\lambda_2)/3 & \lambda_2
\end{bmatrix}
$$
A: I think there is a number of matrices that satisfy these conditions.
Try to add more constraints. The values of the eigenvalue. Take for example $\lambda=0$ and $\lambda=1$. Consider also symmetric matrices only. You may reduce the number of unkonows and you will find a solution easily.
A: As @maxmilgram pointed out in the comments section i solved this problem so ill make this solution just for the sake of clarify.
according to the information given by the problem $A=CDC^{-1}$ where $C=\begin{pmatrix}1&\:0\:&\:1\:\\ \:\:\:\:\:\:\:\:0\:&\:1\:&\:-2\:\\ \:\:\:\:\:\:\:\:\frac{1}{3}\:&\:\frac{2}{3}\:&\:2\end{pmatrix}$  and $diag(D)=(x_1,x_1,x_2)$.
Then $A=\begin{pmatrix}\frac{10x_1-x_2}{9}&\frac{2x_1-2x_2}{9}&\frac{x_2-x_1}{3}\\ \frac{2x_2-2x_1}{9}&\frac{5x_1+4x_2}{9}&\frac{2x_1-2x_2}{3}\\ \frac{2x_1-2x_2}{9}&\frac{4x_1-4x_2}{9}&\frac{x_1+2x_2}{3}\end{pmatrix}$. Making sure the answer is correct:

