2
$\begingroup$

There's this example in my book that goes as follows:

Let $T$ a linear operator on $\mathbb{R}^2$ represented in the standard ordered basis by the matrix $$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ Then the only subspaces of $\mathbb{R}^2$ which are invariant under $T$ are $\mathbb{R}^2$ and $0$. For any other invariant subspace $W$ would necessarily have dimension 1. But if $W$ is the subspace spanned by some non-zero vector $\alpha$ then the statement that $W$ is invariant under $T$ means that $T\alpha=c\alpha$ for some real number $c$.

So far so good, but here comes the step I don't understand.

But this is impossible with $\alpha\neq0$ because one can easily verify that for any $c$ the operator $(T-cI)$ is invertible.

My question is, what does it matter that $T-cI$ is invertible, and why does that make things impossible? Additionally, are there alternative ways to prove such a statement?

$\endgroup$
  • 1
    $\begingroup$ Do you know about eigenvalues? $\endgroup$ – amd Apr 13 '17 at 20:12
  • $\begingroup$ I have an extremely limited knowledge of them. I'm able to recognise what they are when explained but have never worked with them. That being said, the next paragraph goes in to "characteristic values" which seem suspiciously similar. $\endgroup$ – Mitchell Faas Apr 13 '17 at 20:17
  • $\begingroup$ They’re the same thing. “Characteristic value” is a translation of the term. I asked because if you know about them, it’s obvious how invertibility matters, but it sounds like the discussion in your question leads to that, so you can’t really use that. $\endgroup$ – amd Apr 13 '17 at 20:21
  • $\begingroup$ Viewed differently, $T$ is a rotation of the plane by 90 degrees. A subspace is a line through the origin. When you rotate a line by 90 degrees, it cannot be mapped to itself! Do observe that this argument works only in 2D. In 3D a rotation would be about an axis, and the axis is surely mapped to itself no matter how much you rotate. $\endgroup$ – Jyrki Lahtonen Dec 29 '18 at 6:46
5
$\begingroup$

$Tα=cα$ implies $Tα-cα=0$, i.e. $(T-cI)α=0$. Now if $T-cI$ is invertible, we can just multiply the last equation from the left with the inverse of $T-cI$ and we get $α=0$.

$\endgroup$
1
$\begingroup$

There is an elementary alternative proof. Suppose $T(\alpha) = c \alpha$. Write $\alpha = (a, b)$. Then you find that $-b = ca$ and $a = cb$. The only real possibility is $a = b = 0$.

$\endgroup$

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.