Find the matrix $T$ that puts $A$ in canonical form.
one eigenvalue I found is $2$ with associated vector $\begin{bmatrix}1 \\ 1 \end{bmatrix} $
How can I found the matrix $T$ with only one vector?
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What does canonical form mean to you? There are a few. – Pedro Jan 30 '14 at 04:29
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It means there are three types of matrices that are canonical form – afsdf dfsaf Jan 30 '14 at 04:31
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yes, since this is repeated eigenvalues – afsdf dfsaf Jan 30 '14 at 04:37
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Hint: To find the second linearly independent (generalized) eigenvector, set up and solve:
$$(A − \lambda I)v_2 = v_1$$
Spoiler
$$v_2 = (-1,0)$$
Amzoti
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when I row reduce the following matrix $\begin{bmatrix}1 & -1 \ 1 & -1 \end{bmatrix} $, I get $\begin{bmatrix}1 & -1 \ 0 & 0 \end{bmatrix} $ which never multiply another vector to get (1,1) which is $v_1$ – afsdf dfsaf Jan 30 '14 at 05:16
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I can solve 2x2 system to find your answer...However, it has been three years for me to do the RREF, could you show me the step – afsdf dfsaf Jan 30 '14 at 05:22
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You end up with $ \begin{bmatrix}1& -1 & -1 \ 0 & 0 & 0\end{bmatrix}$. This give $a = b - 1$. Let $b = 0$, so $a = -1$. Here are more examples if that is not clear – Amzoti Jan 30 '14 at 05:35
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@afsdfdfsaf: Oops, forgot link: http://www.math.tamu.edu/~stecher/Linear-Algebra/Systems/examples.html – Amzoti Jan 30 '14 at 05:43