1
$\begingroup$

Let $T:\mathbb{R}^2\to \mathbb{R}^2$ be a linear transformation such that $$(a,b)\longmapsto (a+b, a-b)$$ Find all the eigenvalues and, for each eigenvalue, find the corresponding eigenspace.

My attempt:

I don't know if there is a normal procedure to find the matrix of a linear transformation, but I just "back filled" the entry values to make it work. So I have $$ \begin{pmatrix} 1 & 1 \\ 1 & -1 \\ \end{pmatrix} \begin{pmatrix} a \\ b \\ \end{pmatrix}= \begin{pmatrix} a+b \\ a-b \\ \end{pmatrix} $$ So, denoting the matrix as $A$, I used the characteristic polynomial $$ det(A-\lambda I)= \begin{pmatrix} 1-\lambda & 1 \\ 1 & -1-\lambda \\ \end{pmatrix}=0 $$ $\implies -(1-\lambda)^2-1=0\implies \lambda= 1+i$ or $1-i$.

pluging the former value into the matrix I solve for $$ \begin{pmatrix} -i & 1\\ 1 & -2-i \\ \end{pmatrix} \begin{pmatrix} a \\ b \\ \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix} $$ Which generates the system of equations $$-ia+b=0, \quad a-2b-ib=0 $$ But solving the system gives me $a=b=0$.

There is a previous problem where I got the same thing, so I'm wondering if I am doing something wrong. Any help is much appreciated.

$\endgroup$
  • 1
    $\begingroup$ $-1-\lambda \neq 1-\lambda$ $\endgroup$ – Chinnapparaj R Mar 6 at 8:01
  • 2
    $\begingroup$ Note that the matrix is symmetric; the eigenvalues should be real! $\endgroup$ – Theo Bendit Mar 6 at 8:04
2
$\begingroup$

The eigen values are $\pm \sqrt 2$.

$\endgroup$
  • $\begingroup$ You're right. I can't believe I didn't see my mistake in computing the determinate. Thank you $\endgroup$ – Joe Man Analysis Mar 6 at 8:00
1
$\begingroup$

While computing $det(A- \lambda I)$. You are committing a mistake.

$\endgroup$

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.