misunderstanding of vectors in the eigenspace

Question

I'm not too sure if I understand why two different scalars were chosen for the vectors in this eigenspace. I know that a zero row means a free variable(parameter), thus you can choose any scalar. But doesn't it have to be distinct for each row meaning row 2 and 3 should have different scalars? Then, the additional scalar s in the first row throws me off completely. I don't get it at all.. Can anybody help me? Thanks!

edit: i understand how to expand the characteristic polynomial and everything, just the vectors in eigenspace is confusing for me

Link of Example: http://puu.sh/A9IJe/21a1eb38e8.jpg

You might find this answer helpful. – amd Apr 24 '18 at 19:01 — amd, Apr 24 '18 at 19:01
I'll check it out, thanks. – Michael Johnson Apr 24 '18 at 19:11 — Michael Johnson, Apr 24 '18 at 19:11

score 2 · Accepted Answer · answered Apr 24 '18 at 19:41

The nullspace of $A-I$ consists of all vectors $(x_1,x_2,x_3)$ such that $$ \begin{bmatrix}0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix} \iff \begin{bmatrix}x_2 - x_3 \\ 0 \\ 0\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix} \iff x_2 = x_3. $$ So whatever the first two coordinates are (say, $r$ and $s$) the third coordinate has to be equal to the second. Hence, $(r,s,s)$.

Another way of thinking about it is that in this row-reduced version of $A-I$, the pivot variable is $x_2$, and $x_1, x_3$ are the free variables. (Because $x_2$'s column is the one that has the pivot entry in it.) So the first and third components are the parameters we can vary, and the second component is given in terms of the first and third. (In this case, it happens to be equal to the third.)

misunderstanding of vectors in the eigenspace

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