# Jordan canonical bases and form

Find the Jordan bases and the Jordan canonical form for the following matrices:

$\begin{pmatrix} 2 & 3 \\ \\ 0 & 2 \end{pmatrix}$ (sorry about the formatting)

So I found the eigenvalues $\lambda=2,2$ and then I have to find an eigenvector I think:

$\begin{pmatrix} 0 & 3 \\ \\ 0 & 0 \end{pmatrix}$$\begin{pmatrix} x \\ \\ y \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ \\ 0 & 0 \end{pmatrix} So I get the equation 0x+3y=0 so x is a free variable and y=0. So my book tells me the eigenvector v_1\begin{pmatrix} 1 \\ \\ 0 \end{pmatrix} but I ma having a little trouble understanding where the 1 comes from. i need to find another vector v_2 and how we did this in class is we solved for x in terms of y but I can't do that in this case I think because x is multiplied by 0. My book gives v_2=\begin{pmatrix} 0\\ \\ 1/3 \end{pmatrix} but I don't see how they got that. So then J=M^-1AM where M=\begin{pmatrix} 1 & 0 \\ \\ 0 & 1/3 \end{pmatrix} A=\begin{pmatrix} 2 & 3 \\ \\ 0 & 2 \end{pmatrix} and M^-1=\begin{pmatrix} 1 & 0 \\ \\ 0 & 3 \end{pmatrix} so J=\begin{pmatrix} 2 & 1 \\ \\ 0 & 2 \end{pmatrix} Which is of the right form. I just don't understand some of the intermediate steps, thanks! ## 2 Answers You have \ker(A-2I)\varsubsetneq \ker(A-2I)^2=\mathbf R^2 (by Hamilton-Cayley), hence what you have to do is finding a vector v_2 such that (A-2I)v_2=v_1. You'll find v_2 as in your book, an at the same time, you have both Av_1=2v_1 and Av_2=2v_2+v_1, whence the Jordan normal form. the matrix A=\pmatrix{2&3\\0&2} is almost in jordan form. if you want the canonical form J=\pmatrix{2&1\\0&2} change the basis \{e_1=(1,0)^\top, e_2 = (0,1)^\top\} to the basis \{3e_1, e_2\}. \bf edit: the transformation T that is represented by A with respect to the standard basis \{e_1, e_2\}. that is$$Te_1 = 2e_1, Te_2 = 3e_1+2e_2$$suppose we want to choose a basis \{f_1=ae_1+ce_2, f_2=be_1+de_2\} so that$$Tf_1 = 2f_1, Tf_2 = f_1 + 2f_2 $$we have the following equations for the constants:$$2ae_1 + c(3e_1+2e_2)=2ae_1+2ce_2 \to c = 0\\ 2be_1+d(3e_1+2e_2)=ae_1+2(be_1+de_2) $$one choice is$$a = 3, b = 0, c = 0, d = 1$\$

• Hm..I see that it's almost in Jordan canonical form but I don't understand your explanation exact;y – Math Major May 1 '15 at 14:17
• @MathMajor Abel is showing you wrt which basis is your matrix in JCF ... – Timbuc May 1 '15 at 14:19