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 $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}?$$
If a 3x3 matrix is diagonalizable and has eigenvalues 1,2 but has 2 eigenvectors with eigenvalue 2, would we...
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linear-algebra
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1Yes, that is one possibility. Of course you can permute the diagonal elements of the diagonal matrix – 2012-12-03
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0@Stefan ahh, I just wanted to know whether you entered in the value 2 twice for each of the eigenvectors – 2012-12-03
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1Are the eigenvectors, with eigenvalue 2, linearly independent? – 2012-12-03
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1Also note that both eigenvectors for the eigenvalue 2 have to be linear independent. – 2012-12-03
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0They are, but what if that wasn't the case? – 2012-12-03
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1For 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. – 2012-12-03
1 Answers
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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 $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} $$