- #1
Mr Davis 97
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Homework Statement
Find the Jordan canonical form of the matrix ##
\left( \begin{array}{ccc}
1 & 1 \\
-1 & 3 \\ \end{array} \right)##.
Homework Equations
The Attempt at a Solution
So my professor gave us the following procedure:
1. Find the eigenvalues for each matrix A. Your characteristic polynomial must definitely SPLIT over ##\mathbb{C}##, but mostly likely you will be able to split it already over ##\mathbb{R}##.
2. For each eigenvalue ##\lambda## of A set the matrix ##B=A-\lambda I##
3. Calculate several consecutive powers of ##B=A-\lambda I##, e.g., ##B^1##,##B^2##,##B^3##, etc., and check the ranks of these matrices. Set ##N## to be the first exponent where the rank stabilizes, i.e., ##\text{rk}(B^N)=\text{rk}(B^{N+1}).##.
4. Find a basis for the kernel of ##B^N##; i.e., a basis for the generalized eigenspace ##K_{\lambda}(T)=\text{Ker}(A-\lambda I)^N##.
5. In each part, put together the bases you found for all ##K_{\lambda}(T)##, verify that you have obtained a basis for the whole space V, and write down the matrix of T in this basis.So, we have the matrix ##
\left( \begin{array}{ccc}
1 & 1 \\
-1 & 3 \\ \end{array} \right)##. It's characteristic polynomial is ##f(\lambda) = (\lambda - 2)^2##. So we have an eigenvalue of 2 with multiplicity 2.
Now, let ##B=A - \lambda I =
\left( \begin{array}{ccc}
-1 & 1 \\
-1 & 1 \\ \end{array} \right)##. This matrix has rank 1. Now, ##B^2 =
\left( \begin{array}{ccc}
0 & 0 \\
0 & 0 \\ \end{array} \right)
##. So it has rank 0. We can see that ##\text{rk}(B^2) = \text{rk}(B^3)##, so we must find a basis for the kernel of ##B^2##.
This is where I can get confused. We got the zero matrix, so the kernel of the zero matrix is all of ##\mathbb{R}^2##. So would we just choose any two linearly independent vectors? Or am I completely off-track and making a big mistake somewhere? I feel like I am following the procedure...