Eigenspaces and geometric reasoning

[fogvajarash]

Homework Statement

Let T be the reflection about the line 6x + 1y = 0 in the euclidean plane. Find the standard matrix A of T. Then, write down one of the eigenvalues and its corresponding eigenspace (in the form span {[ ]}). Then, find the other eigenvalue of A and its corresponding eigenspace.

Homework Equations

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The Attempt at a Solution

I found the matrix T of the transformation, which is:

(-35/37 -12/37)
(-12/37 35/35)

I actually found the eigenvalues for the matrix T by finding the characteristic polynomial (they are -1, 1) and then equating to 0 and solve the homogeneous system. However, I was told that i can use "geometric reasoning" to find the answer quickly, and I have no idea where to start using it to find the eigenvalues and eigenspaces (firstly, what do they represent in this case)?

Thank you.

 

[CompuChip]

If T is a linear transformation, then the eigenvalue equation reads
$$T \vec v = \lambda \vec v$$
This means precisely that $\vec v$ is a direction that will be unaffected by the transformation - applying T to it will give back the same vector up to scalar prefactors.

So can you think of a vector that will not change if you reflect it about 6x + y = 0?

 

[HallsofIvy]

You want the matrix
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
satisfying two properties:
\begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}x \\ 6x\end{pmatrix}= \begin{pmatrix}0 \\ 0\end{pmatrix}
for any x, and
\begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}-6y \\ y\end{pmatrix}= \begin{pmatrix}6y \\ -y\end{pmatrix}
for any y.

Do you see why?

 

[fogvajarash]

If T is a linear transformation, then the eigenvalue equation reads
$$T \vec v = \lambda \vec v$$
This means precisely that $\vec v$ is a direction that will be unaffected by the transformation - applying T to it will give back the same vector up to scalar prefactors.

So can you think of a vector that will not change if you reflect it about 6x + y = 0?

I think they would be vectors that go along the line y = -6x. However, how can we find the eigen values in this case?

You want the matrix
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
satisfying two properties:
\begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}x \\ 6x\end{pmatrix}= \begin{pmatrix}0 \\ 0\end{pmatrix}
for any x, and
\begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}-6y \\ y\end{pmatrix}= \begin{pmatrix}6y \\ -y\end{pmatrix}
for any y.

Do you see why?

How does this relate to eigenvalues and eigenspaces though? And how can we establish those two propositions that you stated? (aren't eigenvalues the solution to the homogeneous system Av - cv = 0?

 

[Dick]

I think they would be vectors that go along the line y = -6x. However, how can we find the eigen values in this case?

Yes, and if a vector doesn't change under the reflection, wouldn't that mean it's eigenvalue is 1? And under a reflection some vectors will change into their negatives, yes? Which would those be and what would their eigenvalues be? I think this the 'geometric reasoning' part of the exercise.

 

[fogvajarash]

Yes, and if a vector doesn't change under the reflection, wouldn't that mean it's eigenvalue is 1? And under a reflection some vectors will change into their negatives, yes? Which would those be and what would their eigenvalues be? I think this the 'geometric reasoning' part of the exercise.

Okay i kind of get it (the other vector literally gets rotated by pi degrees so its negative is the same vector. However, can't we as well have have another eigenvalue of 1 with direction (1 -6)? The answers are -1 and eigenspace (1 1/6) and 1 and eigenspace (-1 6). Can't we have more combinations of these?

Thank you.

 

[Dick]

Okay i kind of get it (the other vector literally gets rotated by pi degrees so its negative is the same vector. However, can't we as well have have another eigenvalue of 1 with direction (1 -6)? The answers are -1 and eigenspace (1 1/6) and 1 and eigenspace (-1 6). Can't we have more combinations of these?

Thank you.

Eigenvectors aren't unique. If (-1 6) is an eigenvector with eigenvalue 1 then k*(-1 6) is also an eigenvector with eigenvalue 1 for any constant k. The set of all such vectors is what the eigenspace means. (1 -6) is (-1)*(-1 6). It doesn't define a different eigenspace.

 

[fogvajarash]

Eigenvectors aren't unique. If (-1 6) is an eigenvector with eigenvalue 1 then k*(-1 6) is also an eigenvector with eigenvalue 1 for any constant k. The set of all such vectors is what the eigenspace means. (1 -6) is (-1)*(-1 6). It doesn't define a different eigenspace.

So we can simply state that any scalar of that vector is the eigenspace for that eigenvalue? (As well for the second vector with the other eigenvalue of -1)

 

[Dick]

So we can simply state that any scalar of that vector is the eigenspace for that eigenvalue? (As well for the second vector with the other eigenvalue of -1)

Better to say "any multiple of that vector" rather than "any scalar of that vector", but yes.

 

[HallsofIvy]

or "scalar multiple of that vector".