# Jordan Normal Form: Two times the same basis vector?!

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?

• 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. – amd Dec 13 '18 at 0:49

$$(A-2I)^2 = \left( \begin{array}{ccc} -1&-2&1 \\ 3&6&-3 \\ -4&-8&4 \\ \end{array} \right)$$ 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.

• @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 – Will Jagy Dec 15 '18 at 18:49

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.

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