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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!

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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.

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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 $$a = 3, b = 0, c = 0, d = 1$$

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

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