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The last part is something I don't understand.

The equation should be $x_1-x_3=0$ right? Therefore, $x_1=x_3$

I know how to write the eigenspace and find the basis if there is only one free variable in a 3x3 matrix. I don't understand what happened here though. Some help required :)

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Everything is worked out pretty nicely in the solution you posted. Do you have a specific question? – Jeff Dec 3 '11 at 19:34
Yea, where do they get the linear combination s and t from? – Andrew Dec 3 '11 at 19:41
up vote 2 down vote accepted

Finding a basis for the eigenspace is the same as finding a basis for a null space of $\left[ \matrix{ -1&0&1\cr 3&0&-3\cr 1&0&1 }\right] $.

So, we find the solution set of the equation $$\tag{1}\left[ \matrix{ 1&0&-1\cr 0&0&0\cr 0&0&0 }\right] {\bf x}={\bf 0} $$

There are two free variables (corresponding to the columns that don't have a "leading row entry"):

$x_3$ is free, so set $x_3$ equal to an arbitrary value, say $s$. $x_2$ is free, so set $x_2$ equal to an arbitrary value, say $t$. The first row of the matrix in (1) will tell you $x_1=x_3=s$.

So the general solution to (1) is ${\bf x}=\left[ \matrix {s\cr t\cr s }\right]$, where $s$ and $t$ are arbitrary scalars.

Two free variables tells you the dimension of the eigenspace is 2. To find a basis for the eigenspace, we need to find two independent eigenvectors. We may do so by "spliting up the vector $\bf x$ into parts" (as the sum of the "$s$ part" and the "t part"): $$ {\bf x}= \left[ \matrix {s\cr t\cr s }\right]=\left[ \matrix {s\cr 0\cr s }\right] +\left[ \matrix {0\cr t\cr 0}\right]= s\left[ \matrix {1\cr 0\cr 1 }\right] +t\left[ \matrix {0\cr 1\cr 0}\right]. $$ The vectors $\left[ \matrix {1\cr 0\cr 1 }\right]$ and $\left[ \matrix {0\cr 1\cr 0}\right]$ form a basis for the eigenspace.

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I liked the first "version" better. But I got it :) I didn't understand the part of splitting the vector x into parts. Now I got it. – Andrew Dec 3 '11 at 19:52

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