MHB Help to understand this eigenvector case

[ognik]

Did a practice problem finding eigenvalues &-vectors, ended with this row-reduced matrix: $ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$; Solving to get the eigenvectors, I could get $x_1 = x_2 = x_3$, in which case the eigenspace would be the zero vector.

Instead the answer is $ \begin{bmatrix}0\\0\\1\end{bmatrix}$ I think its because I need to choose 1 independent variable(right word?) and a method is to set the last variable to some real t - $x_3=t$, then $x_1 = x_2 = 0t$ ...but above I found $x_3 = 0$? So I know to set $x_3$ to t - but it doesn't make sense to me in this particular case where there are no dependent variables?

(I can see that it is an identity matrix, which is also a Markov matrix, but I don't thank those are relevant?)

 

[Deveno]

It's not clear "what" matrix you row-reduced, nor why, so it's impossible to verify you did it correctly.

That said, finding eigenvectors usually is a two-step process:

Step 1-given the matrix $A$, calculate the determinant of $xI - A$ (this will be a polynomial in $x$).

Step 2 - for any root of that polynomial, say $\lambda$, solve the system of equations:

$(\lambda I - A)v = 0$.

This system will typically be "under-determined", since we know that $\lambda I - A$ is singular (why?).

Typically, one will get a solution of the form (in $3$ dimensions, for example): $(at,bt,ct) = t(a,b,c)$. Any non-zero value of $t$ gives an eigenvector. Often, one of $a,b,c$ is chosen to be $1$, and it is often possible to have all $3$ coordinates be chosen to be integers.

 

[ognik]

Hi - I'm sure it's correct - and it's not the first I've encountered, where solving $ (λI−A)v=0 $ could have all the variables = 0. (it is singular 'cos we found eigenvalues for which the characteristic poly = 0, in turn from det(λI−A)=0)

I know that the method is to choose one variable (in this case $v_3$) to be 1, so in this case we get t(0,0,1). But I'd like to understand why t(0,0,0) is not valid (because solving $ (λI−A)v=0 $ gives $v_1 = v_2 = v_3 = 0)?

 

[Deveno]

Eigenvectors are always non-zero.

 

[ognik]

Eigenvectors are always non-zero.

Hi, just came across a lemma that says 'an eigenspace is a subs-space - which I knew. But it then goes on to say

"Proof. An eigenspace must be nonempty — for one thing it contains the zero vector — and so we need only check closure"

Doesn't this suggest that the zero vector CAN be an eigenvector?

 

[Deveno]

An eigenspace is composed of all eigenvectors belonging to a certain eigenvalue, AND the zero vector (which is NOT an eigenvector).