Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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?

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Since the algebraic multiplicity of the eigen value $1$ is $2$ and geometric multiplicity is $1$ so the given matrix is not diagonalizable.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.