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Let $A =\left(\begin{array}{ccc}3 & -1 & 0 \\0 & 3 & 0 \\0 & 1 &3\end{array}\right)$

I need to find the Jordan Canonical Basis of A.

I found $p(\lambda) = (3 - \lambda)^3$ and $\ker (A-3I) = [v_1,v_2]$ where $v_1 = (1,0,0)$, $v_2 = (0,0,1)$.

Then the Jordan form looks like this:

$J =\left(\begin{array}{ccc}3 & 0 & 0 \\0 & 3 & 1 \\0 & 0 & 3\end{array}\right)$

I need find a third vector $v_3$ to complete $\{ v_1, v_2 \}$ to a basis.

$v_3$ must satisfy: $A(v_3) = v_2 + 3v_3 \iff (A-3I)v_3 = v_2$

However, $(A-3I)v_3 = \left(\begin{array}{ccc}0 & -1 & 0 \\0 & 0 & 0 \\0 & 1 & 0\end{array}\right)\left(\begin{array}{ccc}x \\y \\z\end{array}\right) = \left(\begin{array}{ccc}0 \\0 \\1\end{array}\right)$ doesn't have a solution.

What am I doing wrong?

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  • $\begingroup$ Have you heard of generalized eigenvectors and chaining due to defective matrices? See: math.upenn.edu/~moose/240S2013/slides7-31.pdf $\endgroup$ – Moo Apr 10 '16 at 1:07
  • $\begingroup$ Thanks for your comment. Your link helped me to solve the problem. $\endgroup$ – Santos Apr 10 '16 at 1:52

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