Calculating Diagonal Matrix, too many zeroes in the eigen vectors, what now?

Given the Matrix $$A = \left(\begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 1 & 0 & 1 \end{matrix}\right)$$ calculate the diagol matrix $diag(A)$

Well, for this I need the eigenvalues and eigenvectors, which I've found out are $\lambda_{1,2}=1$, $\lambda_3=2$ and $E_1=\left(\begin{matrix}0 \\ 0 \\ 1\end{matrix}\right)$ and $E_2=\left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)$.

For the Diagonal matrix $D$ we know

$$D=C^{-1}AC$$

The problem is calculating the inverse of $C$ which is made of the eigen vectors in its columns, because I get zeroes in the main diagonal when I apply the Gauss transformation.

What could I do and how?

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Your matrix is not diagonalizable. That means you cannot express it as $A = CDC^{-1}$ with $D$ diagonal and $C$ non-singular. This is because you have not enough independent eigenvectors.

Anyway, I think you misunderstood the question. In the context of Jacobi iteration (and Gauss-Seidel, SOR, etc.), the notation $diag(A)$ means a diagonal matrix whose entries are taken from the diagonal of $A$. So you just remove off-diagonal entries of $A$ to get the answer. There's no need to find eigenvalues and eigenvectors of this matrix.

You'll usually need to find eigenvalues of a different matrix to guarantee convergence though.

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I was afraid of that for the same reason, however I have to calculate the jacobi iteration starting at $0$, and I see in the lecture they've found out the diagonal matrix $$D = \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{matrix}\right)$$ – Flavius Jan 22 '13 at 11:42
@Flavius I edited the answer. – Tunococ Jan 22 '13 at 11:47
I see, accepted! – Flavius Jan 22 '13 at 11:48

Since $\operatorname{Rank}(A-I)=2$ and the eigenspace of eigenvalue $1$ has dimension $1$, $A$ is not a diagonalizable matrix.

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