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I have 3 dimensional matrix $$A = \left(\begin{array}{c} 2 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 3 & 1\end{array}\right)$$ and want to find a Jordan Form for it and a basis for the Jordan Form. My procedure: I calculated the characteristic polynomial $\chi_A(\lambda) = -(2-\lambda)^2(1+\lambda)$ and found the roots $\lambda_1 = 2$ with algebraic multiplicity $\mu_1 = 2$ and $\lambda_2 = -1$ with algebraic multiplicity $\mu_2 = -1$, respectively. Then, for $\lambda_1$, I found that a basis for the kernel of $A - 2 I$ is the vector $\left(\begin{array}{c} 1 \\ 0 \\ 1\end{array}\right).$ Clearly, this has subspace has dimension $\gamma_{11} = 1$ which is less than $\mu_1 = 2$, so I have to continue and calculate the kernel of $(A - 2I)^2$. A basis for this space is given by $\left(\begin{array}{c} 1 \\ 0 \\ 1\end{array}\right), \left(\begin{array}{c} 1 \\ 1 \\ 0\end{array}\right).$ Since now the geometric multiplicity equals the algebraic multiplicity, I am finished with calculating kernels. Now I have to pick some vector $w_{12}$ in the kernel of $(A-2I)^2$ which is not in $(A-2 I)$. An obvious choice is $w_{12} = \left(\begin{array}{c} 1 \\ 1 \\ 0\end{array}\right)$. Then: $$w_{11} = (A - 2I) = \left(\begin{array}{c}1 \\ -3 \\ 4\end{array}\right).$$ Now turning to $\lambda_2$, a basis for the kernel is $\left(\begin{array}{c} 1 \\ -3 \\ 4\end{array}\right)$.

But then I get stuck because I have two times the exact same vector in my basis which of course is not enough to span a 3 dimensional space. I cannot see what I did wrong or where my mistake comes from. What do I do in such a situation?

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  • $\begingroup$ You’ve clearly made an error somewhere, since you can’t have both $A(1,0,1)^T = (2,0,2)^T$ and $A(1,0,1)^T = -(1,0,1)^T$, which is what you’re claiming if it’s an eigenvector of both eigenvalues. $\endgroup$ – amd Dec 13 '18 at 0:49
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$$ (A-2I)^2 = \left( \begin{array}{ccc} -1&-2&1 \\ 3&6&-3 \\ -4&-8&4 \\ \end{array} \right) $$ of rank one, with row echelon form $$ (A-2I)^2 \Longrightarrow \left( \begin{array}{ccc} 1&2&-1 \\ 0&0&0 \\ 0&0&0 \\ \end{array} \right) $$

Your vector $w_{12}$ is not in the kernel of $ (A-2I)^2 \; ; \;$ your basis for that kernel is wrong.

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  • $\begingroup$ @Bill, it's alright. I did not particularly expect the OP to pay any attention; something like four hours passed. Meanwhile, I had not noticed your hint, as the display of comments stops at about five unless I make a point of requesting to see all the comments $\endgroup$ – Will Jagy Dec 15 '18 at 18:49
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Where do you get $\pmatrix{1\\1\\0}$?

What do you have for $(A-2I)^2$?

I have $(A-2I)^2 = \pmatrix{-1&-2&1\\3&6&-3\\-4&-8&4}$

And in addition to $\pmatrix{1\\0\\1}$, I see $\pmatrix{0\\1\\2}$ as a candidate for the second eigenvector.

$A\pmatrix{0\\1\\2} = \pmatrix{1\\2\\5} = 2\pmatrix{0\\1\\2}+\pmatrix{1\\0\\1}$ which is exactly what we were hoping for.

$A\pmatrix{1&1&0\\-3&0&1\\4&2&1} = \pmatrix{1&1&0\\-3&0&1\\4&2&1}\pmatrix{-1&0&0\\0&2&1\\0&0&2}$

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