3
$\begingroup$

Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear transformation that has eigenvalues $\lambda = -1,0,2$.

Find the eigenvalues of $f^2$.

Shouldn't the answer be like this:

  • If $x$ lies in the kernel of the transformation then the eigenvalue is $0$.
  • If $x$ lies in the span of the image of the transformation then the eigenvalue is $1$.

The teacher said that the eigenvalues are $1, 0, 4$.

Any insight is appreciated!

Thanks.

$\endgroup$
  • $\begingroup$ Hint: the eigen values also get squared $\endgroup$ – sashas Jan 15 '15 at 13:17
  • $\begingroup$ Yes but doesn't $f^2$ mean that we apply the transformation once and then once again? $\endgroup$ – Nash Jan 15 '15 at 13:18
  • $\begingroup$ yes it does mean that $\endgroup$ – sashas Jan 15 '15 at 13:19
4
$\begingroup$

Hint: Pick an eigenvector $\bf v$ of $f$ of eigenvalue, say, $2$. Then, applying $f^2$ to $\bf v$ gives $$f^2({\bf v}) = f(f{\bf v}) = f(2 {\bf v}) = 2 f({\bf v}) = 2(2 {\bf v}) = 4 {\bf v}.$$ So, what can we say about $\bf v$ with respect to $f^2$?

More generally, if the eigenvalues of a linear transformation $T: V \to V$ are $\lambda_1, \ldots, \lambda_n$, what can we say about the eigenvalues of $T^2$? What about the eigenvalues of $T^k$?

$\endgroup$
  • $\begingroup$ That the transformation gives back vector $v$ only this time $4$ times longer??? $\endgroup$ – Nash Jan 15 '15 at 13:20
  • $\begingroup$ Travis means to say that if eigen values for $f$ are $a_1,a_2..a_i$ then for $f^k$ the eigen values will be $(a_1)^k,(a_2)^k...$ $\endgroup$ – sashas Jan 15 '15 at 13:26
  • $\begingroup$ @Nash Yes, and what do we call vectors that are scaled by a transformation? $\endgroup$ – Travis Jan 15 '15 at 13:28
  • $\begingroup$ Eigenvectors??? $\endgroup$ – Nash Jan 15 '15 at 13:30
  • 1
    $\begingroup$ Well, it's $4$ simply because $(f^2)({\bf v}) = 4 {\bf v}$. You don't need to know anything about diagonalizability or similarity to conclude this, just the fact that $\bf v$ is an eigenvector of $f$ of eigenvalue $2$. $\endgroup$ – Travis Jan 15 '15 at 13:48
0
$\begingroup$

The fact that you know that the eigenvalues are $-1,0,4$ means that you know that in a suitable basis your linear map is represented by the diagonal matrix $A=$diag$[-1,0,4]$. Therefore, since $f^2 = f\circ f$ you have that $f^2$ is represented by $A^2$, which is diagonal. Hence $f^2$ has eigenvalues $0,1,4$.

$\endgroup$
  • $\begingroup$ Travis has proved in general , your proof only holds true for normal operators i guess $\endgroup$ – sashas Jan 15 '15 at 13:24
  • 1
    $\begingroup$ In this case, all of the eigenvalues are distinct, so the operator is diagonalizable as rafforaffo claims. $\endgroup$ – Travis Jan 15 '15 at 13:29
  • $\begingroup$ oh missed that it was given $R^3$ sorry @rafforaffo $\endgroup$ – sashas Jan 15 '15 at 13:31
  • $\begingroup$ @Travis I still have a doubt having 3 distinct eigen values does not mean that eigen spaces are orthogonal which is necessary condition for diagonalization I guess $\endgroup$ – sashas Jan 15 '15 at 13:34
  • $\begingroup$ That's not true, unless you're restricting to diagonalization by orthogonal matrices. Are you familiar with Jordan Normal Form? In particular, its existence implies that having all distinct eigenvalues implies diagonalizability (over the algebraic closure of the underlying field, anyway). $\endgroup$ – Travis Jan 15 '15 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.