# If a 3x3 matrix is diagonalizable and has eigenvalues 1,2 but has 2 eigenvectors with eigenvalue 2, would we…

If a $3 \times 3$ matrix is diagonalizable and has eigenvalues $1$ and $2$ but has two eigenvectors with eigenvalue $2$, would we have the eigenvalue matrix $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}?$$

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Yes, that is one possibility. Of course you can permute the diagonal elements of the diagonal matrix – Stefan Dec 3 '12 at 23:30
@Stefan ahh, I just wanted to know whether you entered in the value 2 twice for each of the eigenvectors – Becky Dec 3 '12 at 23:35
Are the eigenvectors, with eigenvalue 2, linearly independent? – i. m. soloveichik Dec 3 '12 at 23:36
Also note that both eigenvectors for the eigenvalue 2 have to be linear independent. – Stefan Dec 3 '12 at 23:37
For a matrix to be diagonalizable, $\mathbb R^3$ has to be the direct sum of the eigenspaces. From this it follows, that if you have 2 eigenvalues, one of the two eigenspaces has to have dimension 2, and thus there have to exist two linear independent eigenvectors to that eigenvalue. – Stefan Dec 3 '12 at 23:40

• a one-dimensional eigenspace for eigenvalue $1$, and
it is correct to conclude that the space has a basis composed of eigenvectors, and in this basis the operator has the matrix $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}$$ up to permutation of rows.
If both eigenspaces turned out to be one-dimensional, the Jordan canonical form of this matrix could have been $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$$ or $$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$$