Calculating Null Space: Xn = 0

Question

So, I am calculating the null space of bases and matrices in order to get eigenvectors, and I occasionally come across matrices which will have a row like the following: (Put in system format) $x_2 = 0$

According to the textbook the eigenvector for something like this would be something like: \begin{bmatrix}1\\0\\1\end{bmatrix} But I haven't been able to figure out why this is. Could anyone explain why this is? Or if I'm misunderstanding something, what one does in a situation where such rows appear when calculating eigenvectors?

Answer 1

You're getting $x_2=0$ and hence $x_1$ and $x_3$ are $free$ to be assigned any value, in particular $1$.

Answer 2

Ok. Let's say that you have set up $(A-\lambda I){\bf x} = {\bf 0}$ in an augmented matrix and row reduced and got... $$ \begin{bmatrix} 1 & 0 & -1 & : & 0 \\ 0 & 1 & 0 & : & 0 \\ 0 & 0 & 0 & : & 0 \end{bmatrix}$$

Interpreting this in terms of a "system" we have: $x_1-x_3=0$ and $x_2=0$. Notice that we can treat $x_3$ like a free variable (since the 3rd column contains no leading $1$). Let's call $x_3=t$. Then we get $x_1=x_3=t$ and $x_2=0$. Written as a vector this is... $$ \begin{bmatrix} t \\ 0 \\ t \end{bmatrix}$$

Any $t \not=0$ will yield a non-zero solution of the system $A{\bf x} = \lambda{\bf x}$. So for any $t \not=0$ you get an eigenvector. Setting $t=1$ gives your particular choice.

Why $t=1$? Because any (non-zero) choice will do. We may as well pick a nice one.