Question

If you are computing an eigenvector for a eigenvalue and you end up with a 3x3 matrix where the first column equals the third column why is the eigenvector automatically (1,0,-1)^T

Answer 1

Is this for a 0 eigenvalue?

Answer 2

Not necessarily. \[\left[\begin{matrix}3 & 2 & 2\\ 1 & 4 & 1 \\ -2 & -4 & -1\end{matrix}\right]\]
The associated eigenvalues are 1,2,3. How would you (quickly) compute the eigenvectors?

Also, tangentially (but not really), if you could explain how if someone was computing an eigenvector and they ended up with a matrix:
\[\left[\begin{matrix}2 & -1-i \\ 0 & 0 \end{matrix}\right]\]

What is the algebra they deduce to realize the eigenvector is:
\[\left[\begin{matrix} 1+i \\ 2 \end{matrix}\right]\]

I just need a linear algebra refresher :(

Answer 3

About the OP: look at this "formula" for mulitplication between a matrix and a vector:
\[
\begin{pmatrix}
a & b & c\\d& e & f \\ g & h& i
\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix}
= x \begin{pmatrix}a\\d\\g\end{pmatrix}
+ y \begin{pmatrix}b\\e\\h\end{pmatrix}
+ z \begin{pmatrix}c\\f\\i\end{pmatrix}
\]
So, if 1st column = 3rd column, and calling the colums \(v_1^T,v_2^T,v_3^T\), one obtains
\[
xv_1^T + 0 v_2^T + (-x) v_1^T = (0,0,0)^T, \forall x.
\]
Therefore \((1,0,-1)^T\) is an eigenvector with eigenvalue 0. (as said by @Kainui )

Answer 4

Why do you have a -x?

Answer 5

i know pretty much crap all about linear albegra, but there's not much engagement here, and so for the vector you quote, namely:

\(\left( \begin{matrix}
2 & -1 - i \\
0 & 0
\end{matrix} \right)\)

i would just roll out the characteristic equation: \( \lambda^2 - Tr (M) \lambda + \det (M) = 0\)

so here that is \( \lambda^2 - 2 \lambda + 0 = 0 \implies \lambda = 0, 2\)

for \(\lambda = 0 \), you have \(\left( \begin{matrix}
2 & -1 - i \\
0 & 0
\end{matrix} \right)
\left( \begin{matrix}
u \\
v
\end{matrix} \right) = \vec O =
\left( \begin{matrix}
0 \\
0
\end{matrix} \right)\)

so \(2 u - (1 + i) v = 0\) so \(\vec v_1 = [\dfrac{1+i}{2}, 1]^T\) or \(\vec v_1 = [1+i, 2]^T\) or \(\vec v_1 = [1, \dfrac{2}{1+i}]^T\) .... and all the other scalar multiples thereof :-)

for \(\lambda = 2\), it's weird because you have \(\left( \begin{matrix}
0 & -1 - i \\
0 & -2
\end{matrix} \right)
\left( \begin{matrix}
u \\
v
\end{matrix} \right) = \vec O =
\left( \begin{matrix}
0 \\
0
\end{matrix} \right)\)

so to me that means \(\vec v_2 = [1, 0]^T\) and all scalar multiples....

hope that adds to....!

Answer 6

@sh3lsh What I meant is: "any vector of the form \((x,0,-x)^T\) will be an eigenvector of this matrix.

Answer 7

What's an eigenvector? Supposedly "eigen" is German for "same" but I never really looked in to that to see if it was true but whatever, it's good to capture the concept since that's what eigenvectors are about.

An eigenvector is a vector that is basically left the same after matrix multiplication. So imagine in 3D space, your matrix represents rotation around an axis. Then all the vectors along the axis are eigenvectors since rotating didn't change them, so intuitively you should understand what I mean when I say eigenvector, they are basically fixed points of the mapping.

But really eigenvectors are more general than what I just described, they are also the same to within multiplication by a constant, so rotation, then stretching would still stretch the eigenvectors but they'd still be a linear multiple of themselves so it's fine.

\[A v = \lambda v\]

Right, so another way you could say it is, matrix multiplication has the effect of scalar multiplication on eigenvectors.

So let's look at an example, let's say we want to find eigenvectors of this matrix "the easy way"

\[\begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix}\]

So what vectors are the same as if we had just multiplied them by a constant instead of by this matrix? Well the easy way out is if we can find a vector that multiplies this matrix to get the 0 vector, since 0 times any vector is the zero vector, making it an easy eigenvalue to get.

We're already halfway there in a sense with this example, since matrix multiplication by a vector will combine the first column with the second column, we don't even have to worry about the bottom entries, since ALL linear multiples of 0 added together make 0! So all we have to do is find a vector that makes the top two entries become 0, easy peasy really, since \(a*(-b)+b*a=0\)

\[ \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \begin{pmatrix} -b \\ a \end{pmatrix} =\begin{pmatrix} 0\\ 0 \end{pmatrix}\]

This has eigenvalue 0, here's the trick that makes them easy, I'm going to "factor 0 out" of the eigenvector:

\[ \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \begin{pmatrix} -b \\ a \end{pmatrix} =0\begin{pmatrix} -b\\ a \end{pmatrix}\]

Alright so why didn't I pick the vector \(\binom{b}{-a}\) instead? Well, you can see it's actually just the negative of that vector, so you can just multiply the entire equation by -1 to see that it works... In fact you can see that all scalar multiples of any eigenvector is an eigenvector, they fill out an entire eigenspace, it's a subspace.

Here's a quick proof just to show, if you know \(v\) is an eigenvector of \(A\) with eigenvalue \(\lambda\) then all scalar multiples \(u=sv\) are eigenvectors.

\[Av=\lambda v\]\[s(Av)=s(\lambda v)\]\[A(sv)=\lambda (sv)\]\[Au=\lambda u\]

I was generally pretty vague and rambly throughout the whole thing, but I was just trying to give you a quick and dirty refresher/intro. @IrishBoy123 is doing the systematic way, and is much better in general.

One other important fact to know, especially for 2x2 matrices is that it generically has 2 eigenvalues, and the determinant is the product of the eigenvalues and the trace is the sum of eigenvalues (even for nxn matrices, but 2 equations with 2 unknowns is where it's best to use):

\[\mathrm{tr}(A) = \lambda_1 +\lambda_2\]\[\det(A) = \lambda_1*\lambda_2\]

Answer 8

someone should bump this. it's quite interesting.

Answer 9

Hi, sorry for my atrocious writing skills, but here's the solution: