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$$A =\left(\begin{array}{cc}5 & -4 \\9 & -7\end{array}\right)$$

I found that the eigenvalues are $-1$ (algebriac multiplicity 2)

Therefore, the jordan form looks like this:

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

Also solving $(A-\lambda I$)$=\vec{0}$, when $\lambda = -1$ gives us the nullspace $(3,2)^T$.

However I need one more basis vector, and unsure how to find it?

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from Jordan theorem we know that a matrix M exists such that $$M^{-1}AM=J$$

let $$M=[v,w]$$

then M has to satisfy the following system: $$AM=MJ$$ that is in your case $$Av=-v$$ $$Aw=v-w$$ once you find $v$ from the first equation you can find also $w$ from the second.

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  • $\begingroup$ Thank you I understand $\endgroup$ – K Split X Nov 28 '17 at 23:11
  • $\begingroup$ that's good, I think it's a clear way to see how it works! $\endgroup$ – gimusi Nov 28 '17 at 23:13
  • $\begingroup$ Can you explain why (A-cI)w = v where c is some eigenvalue? What do you do when there is higher multiplicity? eg) mult. 3 in this case $\endgroup$ – skim Dec 5 '17 at 4:12
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    $\begingroup$ To find M you have already to know as J Matrix is done (by Jordan's Theorem). Once you have J you only have to solve the above system AM=MJ. $\endgroup$ – gimusi Dec 5 '17 at 14:08
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You can take any vector which is linearly independent of the eigenvector, except that you need to adapt its length in order to get exactly $1$ (and not some other constant) in the upper right corner of $J$. For example, take something simple like $(k,0)^T$, see what matrix you get in the new basis, and then choose $k$ so that you get exactly $J$.

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