Question

Eigenvalues and Eigenvectors...

Find 1-dimensional subspaces of R2 invariant under the linear transformations
given by the following matrix:

[1 2;-3 -6]

Answer 1

so you want the vectors v of the form:
\[A\vec{v} = \lambda \vec{v}\]
for the above matrix (which we call A) and some constant lambda. Do you know how to solve problems like this?

Answer 2

Or you could just do it by looking at it, as 2x2 matrices aren't too hard to see through. Up to you.

Answer 3

only just scratched he surface of eigenvalues/vectors

but from my notes you rearrange \[Av =\lambda v\]
to \[(A-\lambda I)v=0\]
then find \[\det(A-\lambda I)=0\]

and I get \[\lambda = 0, -5\]
so what do I do now

Answer 4

The determinant is
\[(1-\lambda)( -6 - \lambda) +6 = -6 -\lambda + 6\lambda + \lambda^2 + 6 = \lambda^2 + 5\lambda \]
so you're correct. The matrix that came from is
\[\left[\begin{matrix}1-\lambda & 2\\-3 &-6- \lambda\end{matrix}\right]\]
Plug in lambda = 0, and solve
\[\left[\begin{matrix}1 & 2\\-3 &-6\end{matrix}\right]\left(\begin{matrix} a \\b\end{matrix}\right)= 0\]
You get two equations. First,
\[a + 2b = 0\]
second,
\[-3a - 6b = 0\]
these happen to give the same results, namely a = -2b, and so therefore your eigenvector is
\[\left(\begin{matrix}-2 \\ 1\end{matrix}\right)\]
and its multiples.

Answer 5

Ohh ok thanks, I haven't quite covered the eigenvectors yet but this makes complete sense!

there's also another eigenvector [1 -3] which by the look of it you plug in the 5 as lambda in your 2nd step and solve. Thanks again!

Answer 6

Exactly. Therefore, the subspaces defined by those two vectors and their multiples are invariant under the transformation.

Answer 7

Just out of curiosity, what was you other method you described earlier, about reading off the 2x2 matrix?

Answer 8

I just meant looking at the thing and trying to see without doing any computation. It's not so bad sometimes with 2x2 matrices but it's nearly impossible any other time.