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I've got a matrix $A = \begin{bmatrix}1&-1&0\\ -1&2&-1\\0&-1&1\end{bmatrix}$ and am asked to determine if it is diagonalisable. I find the eigenvalues to be $3, 1$ and from this the question says we should be able to deduce whether A is diagonalisable or not - without any more work. I know that if

$dim(E_1) + dim(E_3) = 3$

then the matrix is diagonalisable. However, I don't know the dimensions of the eigenspaces without first finding them. I'm not really sure what to do, is the multiplicity of the eigenvalues used? If so what is the multiplicity of them (what does it mean)?


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$0\;$ is also an eigenvalue... – draks ... Oct 4 '12 at 8:58
How do you determine that $0$ is an eigenvalue too? I solve the quadratic and only get two solutions. – user1520427 Oct 4 '12 at 9:26
Since you have a $3\times 3$ matrix, you should have three eigenvalues. Since you already know $3$ and $1$, you also know that the trace of your matrix $\text{tr}(A)=4$, there is one missing, which is $0$. Also your polynomial should be of third degree: $x(...\text{quadratic}...)=0$. Or shorter: $\det(A)=0$. – draks ... Oct 4 '12 at 9:30
Ah cool thanks! – user1520427 Oct 4 '12 at 9:39
up vote 5 down vote accepted

$A$ is symmetric, so you can diagonalize it.

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even more: Bisymmetric – draks ... Oct 4 '12 at 8:55
why you can diagonlize symmetric matrix? – Jim Thio Oct 4 '12 at 10:53
@JimThio due to the Spectral Theorem... – draks ... Oct 4 '12 at 11:13

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