How do I find 2-dimensional subspaces that are invariant subspace of T ?

$T\left(\left[ \begin{matrix} x\\ y\\ z\\ \end{matrix} \right]\right) = \left[\begin{matrix} 2y+z \\ -2x+4y+z \\ -2x+2y+3z \end{matrix}\right]$


Here's a Axler technic:

Choose $v=(0,0,1) \in \Bbb R^3$. Then the set $\{v,Tv,T^2v,T^3v\}$ is dependent. Actually the set is $$\{(0,0,1),(1,1,3),(5,5,9),(19,19,27)\}$$ Here $$-(0,0,1)+\frac{5}{6}(1,1,3)-\frac{1}{6}(5,5,9)+0(19,19,27)=0$$ That is $$(-1)v+\frac{5}{6}Tv-\frac{1}{6}T^2v+0 T^3 v=0$$ That is $$\left( -\frac{1}{6} T^2+\frac{5}{6}T-I \right) v=0$$ Now it is easy to chech that $\text{span} \{v,Tv\}$ is a invariant subspace of dimension two!

  • $\begingroup$ Can you explain why you chose the vector $v = (0,0,1)$ and then the line afterward ? aswell as how you got the fractions ? $\endgroup$ – i9-9980XE May 11 at 3:47
  • $\begingroup$ See page number 91 and read theorem 5.24 in this [book]( google.com/url?sa=t&source=web&rct=j&url=http://…) I just imitate this proof with the easy non zero vector (0,0,1) $\endgroup$ – Chinnapparaj R May 11 at 3:58
  • $\begingroup$ Right but if v is the vector (0,0,1) then wouldn’t Tv = (-2,2,3)? $\endgroup$ – i9-9980XE May 11 at 4:24
  • $\begingroup$ How did you get this? For your T , substitute x=y=o and z=1 to get (1,1,3) $\endgroup$ – Chinnapparaj R May 11 at 4:30
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    $\begingroup$ The link to my book Linear Algebra Done Right in one of the comments by Chinnapparaj R is a link to an unauthorized copy of the old second edition of the book. The newer third edition contains many improvements. $\endgroup$ – Sheldon Axler May 12 at 1:36

here's some hint.

I think you can easily find the matrix expression of the given map.

and then try to find eigenvectors. since eigenspaces are T-invariant, you'd find 2-dimensional subspace.

  • $\begingroup$ So I just tried what you said, I got a cubic, haven't solved it yet, but once I find the roots / values, I add both the eigenspaces together so that it'll be $e_{x}+ke_{y}, e_{y}+ke_{z}$ and $e_{x}+ke_{z}$ ? $\endgroup$ – i9-9980XE May 11 at 3:25
  • $\begingroup$ Oh I had some mistakes... you'll get one multiple root (with multiplicity 2) when you solve the characteristic equation. so you can just take the corresponding eigenspace. $\endgroup$ – Juno Seo May 11 at 3:27
  • $\begingroup$ as I mentioned, this case doesn't have to direct sum the eigenspaces. (sorry for confusing you.) but in case that you have to, remind that each eigenvector is independent. (this is why you can take a eigenbasis for a matrix) $\endgroup$ – Juno Seo May 11 at 3:30
  • $\begingroup$ so I got eigenvalues, 2,2 and 3 I then get matrices $e_{2} = \left(\begin{matrix} -2&2&1\\ -2&2&1\\ -2&2&1 \end{matrix}\right)$ & $e_{3}\left(\begin{matrix} -3&2&1\\ -2&1&1\\ -2&2&0 \end{matrix}\right)$ $\endgroup$ – i9-9980XE May 11 at 3:32
  • $\begingroup$ you almost there. the multiple root 2 means there are two eigenvectors corresponding to. all you have to do is find them, and then the span of them will be your answer. $\endgroup$ – Juno Seo May 11 at 3:43

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