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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
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Are the eigenvectors, with eigenvalue 2, linearly independent? –  i. m. soloveichik Dec 3 '12 at 23:36
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Also note that both eigenvectors for the eigenvalue 2 have to be linear independent. –  Stefan Dec 3 '12 at 23:37
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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

1 Answer 1

Since the linear operator described by this matrix acts on 3-dimensional space, and is known to have

  • a one-dimensional eigenspace for eigenvalue $1$, and
  • two-dimensional eigenspace for eigenvalue 2,

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 $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} $$

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