Finding eigenvectors when one column is zero

I have the matrix: $\begin{pmatrix} 3 & a & 0 \\ 0 & 4 & 1 \\ 0 & 2 & 5 \\ \end{pmatrix}$

and I should find every eigenvector for every eigenvalue.

After calculations I get that my eigenvalues are: $\lambda_1 = 6$ and $\lambda_2 = 3 = \lambda_3$.

The eigenvector for $\lambda_1$ is: $t\begin{pmatrix} a\\3\\6 \end{pmatrix}$.

However, for $\lambda = 3$ I get (after row reduction): $\begin{pmatrix} 0 & a & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix}$ which gives me:

$ax_2 = 0$ and $x_2 = - x_3$

The problem is that I don't really know where to go from here. I mainly have 2 questions:

1. How do I handle the variable a?

2. What do I do with $x_1 = 0$ ? (Especially unsure about this in general when it comes to eigenvectors)

Assuming your prior calculations are correct, the situation hinges on whether $a$ is zero or not. If $a \neq 0$, then $x_2$ is forced to be zero, so $x_3$ is as well, and $x_1$ is free. If $a=0$, then one of $x_2$ and $x_3$ can be made free, while $x_1$ is again free.

• Yeah, I kind of got that it depended on whether a was 0 or not. However, I don't really get why $x_1$ is free. Don't we know what $x_1 = 0$? – gbgult Mar 27 '18 at 19:05
• @gbgult No, there's no equation involving $x_1$ at all: all the coefficients that would wind up hitting $x_1$ when you multiply everything out are zero. – Ian Mar 27 '18 at 19:08
• Yeah, I see the logical fallacy now. So basically, if there is no equation concerning a variable - I should consider it to be a free variable? Is that a "general rule" for similar situations? – gbgult Mar 27 '18 at 19:10
• @gbgult If you have a reduced echelon form of a linear system, each (nontrivial) equation binds its left hand side as a basic variable. Every variable that isn't a basic variable is a free variable. This includes those that don't appear on the left or right sides of any of the equations in the reduced echelon form. – Ian Mar 27 '18 at 19:11

If $a \ne 0$ then $\pmatrix{a \\ 1 \\ 0}$ and $\pmatrix{0 \\ 1 \\ 0}$ are linearly independent so the eigenspace is given only by $\operatorname{span}\left\{\pmatrix{1 \\ 0 \\ 0}\right\}$, since the rank of your matrix is $2$.

If $a = 0$ then the second and third columns are equal so the eigenspace is $\operatorname{span}\left\{\pmatrix{1 \\ 0 \\ 0}, \pmatrix{0 \\ 1 \\ -1}\right\}$, since rank of your matrix is now $1$.

• Why is the span of the eigenspace two vectors if the rank of the matrix is 1? I believe I've missed that theorem. – gbgult Mar 27 '18 at 19:05
• @gbgult In the case $a=0$ after subtracting off $3I$ you have a rank 1 matrix. This means it has a two-dimensional kernel, by the rank-nullity theorem. – Ian Mar 27 '18 at 19:09
• Alright, thanks, I believe I need to do some reading on the connection between the eigenspace and kernel. Any suggestions? – gbgult Mar 27 '18 at 19:15
• @gbgult That connection is quite simple: the eigenspace with eigenvalue $\lambda$ is exactly the kernel of $A-\lambda I$. This is pretty much the entire reason why eigenvalues/eigenvectors have anything to do with solving linear systems. – Ian Mar 27 '18 at 19:16

Solving

$$\begin{pmatrix} 0 & a & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix}v=0$$

for $a=0$ we obtain $(0,-1,1)$ and $(1,0,0)$ while for $a \neq 0$ we obtain $(1,0,0)$.

• Okay, I believe the problem is the second question in the end of my post. Why do $x_1$ become a free variable? Don't get that intuitively. – gbgult Mar 27 '18 at 19:07
• @gbgult simply solve the system for $a\neq 0$ from the first equation we find $x_2=0$ from the second $x_3=0$ and we haven't no condition on $x_1$ that is free. In this case we can't diagonalize the matrix but we can find the jordan form. – user Mar 27 '18 at 19:10
• So basically, if a variable doesn't have a condition - it should be considered free? – gbgult Mar 27 '18 at 19:11
• @gbgult yes of course, for $a\neq 0$ indeed rank =2 then the null space has dimension 1. – user Mar 27 '18 at 19:14