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$$\begin{bmatrix} a\\\\b\end{bmatrix}\longmapsto \begin{bmatrix}-3&1\\\\0&2\end{bmatrix} \begin{bmatrix} a\\\\b\end{bmatrix}$$

Use the characteristic polynomail to find all eigenvalues for the transformation for each eigenvalues $\lambda$ find all eigenvectors with eigenvalues $\lambda$ and find a basis for $E_\lambda$

Here is what I have so far

$\chi_f(x)=\begin{bmatrix}-3-x&1\\\\0&2-x\end{bmatrix}$ $=(-3-x)(2-x)$ so there are two eigenvalues $\lambda=-3$ and $\lambda=2$

I'm not sure how to find a basis for these eigenvalues though

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  • $\begingroup$ To find $E_\lambda$ you have to find a base for $ker(f-\lambda id)$ can you do that? $\endgroup$ – Stefan Dec 3 '12 at 15:53
  • $\begingroup$ I think they might be of the form $\begin{bmatrix}a\\\\3a\end{bmatrix}$ for $E_0$ and $\begin{bmatrix}a\\\\2a\end{bmatrix}$ for $E_{-1}$ $\endgroup$ – Adam Dec 3 '12 at 15:54
  • $\begingroup$ Your eigenvalues are wrong. $\chi_f(x)$ is $(-3-x)(2-x)$ which has roots $2,-3$. $\endgroup$ – Pedro Tamaroff Dec 3 '12 at 15:57
  • $\begingroup$ @Stefan so a base of everything that maps char(f) to zero? $\endgroup$ – Adam Dec 3 '12 at 16:00
  • $\begingroup$ @PeterTamaroff Yes you are correct, for some reason I though 1 times 0 was 1 :( $\endgroup$ – Adam Dec 3 '12 at 16:01
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To find the subspaces you're looking after you ought to solve the homogeneous system that you obtain by plugging in the eigenvalues. First, note your eigenvalues are wrong. The two roots of $\chi_f(x)$ are $-3$ and $2$, which are precisely those values for which $(-3-x)(2-x)-1\cdot 0=0$. Thus you get that the equations of the spaces are

For $E_{2}$ you get

$$-5x+y=0$$

Thus $5x=y$ and the space is generated by $(5,1)$. In the other case, that is $E_{-3}$, you get

$$0x+y=0$$ $$0x+5y=0$$

Thus, you have $x$ free (it doesn't appear in the equations) and need $y=0$. Whence, the space is generated by $(1,0)$.

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