# Thread: help finding the basis of an eigen space

1. ## help finding the basis of an eigen space

How did they get that x_2 = t?

I get that x_1 = -S and x_3 = s but why does x_2=t? Shouldn't x_2 just = zero?

2. ## Re: help finding the basis of an eigen space

also how did they know to create two separate vectors for s and t? Is it that a single variable goes for a single combination, meaning if x1 and x3 can are versions of one another then you can describe them both in terms of S, but x2 is not so it gets it's own variable t, and therefore it's own vector? But if t=0 then the second vector should be all zeros not 0,1,0 no?

3. ## Re: help finding the basis of an eigen space

Originally Posted by kingsolomonsgrave

How did they get that x_2 = t?

I get that x_1 = -S and x_3 = s but why does x_2=t? Shouldn't x_2 just = zero?
hey there..

x2=0 is just one of the values that x2 can have.

4. ## Re: help finding the basis of an eigen space

Originally Posted by kingsolomonsgrave
also how did they know to create two separate vectors for s and t? Is it that a single variable goes for a single combination, meaning if x1 and x3 can are versions of one another then you can describe them both in terms of S, but x2 is not so it gets it's own variable t, and therefore it's own vector? But if t=0 then the second vector should be all zeros not 0,1,0 no?
Here the eigen space consists of vectors (x1 x2 x3) such that 1.x1 + 0.x2 + 1.x3 = 0. So, we find the values that x1, x2, x3 can have to satisfy the equation. We find that if x1= s, which is any scalar, then x3 must be -s and since the co-eff. of x2 is 0 it can have any value and still satisfy the equation and so we assign x2 the value t.

Now (x1 x2 x3) looks like (s t -s).

We need to express (s t -s) as a linear combination of linearly independent vectors.

(s t -s) = (s 0 -s) + (0 t 0) = s(1 0 -1) + t(0 1 0).

Here (1 0 -1) and (0 1 0) are linearly independent and from the above expression it it clear that any vector of the eigen space( which is of the form (s t -s)) can be expressed as a linear combination of these two vectors, in other words these two linearly independent vectors span the eigen space. So, (1 0 -1) and (0 1 0) form a basis for the eigen space.

Note: c(1 0 -1), where c is a scalar, is a scalar multiplication of (1 0 -1) which means any vector of the form c(1 0 -1) lies on the same straight line passing through (0 0 0) as (1 0 -1) does and they form a subspace of R3.

5. ## Re: help finding the basis of an eigen space

Originally Posted by kingsolomonsgrave

How did they get that x_2 = t?

I get that x_1 = -S and x_3 = s but why does x_2=t? Shouldn't x_2 just = zero?
$\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1\\ x_2 \\ x_3\end{bmatrix}= \begin{bmatrix}x_1+ x_3 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$
reduces to the single equation $\displaystyle x_1+ x_3= 0$. That simplifies to $\displaystyle x_3= -x_1$ but there is NO condition on $\displaystyle x_2$. That means that $\displaystyle \begin{bmatrix}x_1 \\ x_2\\ -x_1\end{bmatrix}$ will satisfy the equation for any numbers, $\displaystyle x_1$ and $\displaystyle x_2$:
$\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1\\ x_2 \\ -x_1\end{bmatrix}= \begin{bmatrix}x_1- x_1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$
no matter what $\displaystyle x_2$ is.

6. ## Re: help finding the basis of an eigen space

Originally Posted by kingsolomonsgrave
also how did they know to create two separate vectors for s and t? Is it that a single variable goes for a single combination, meaning if x1 and x3 can are versions of one another then you can describe them both in terms of S, but x2 is not so it gets it's own variable t, and therefore it's own vector? But if t=0 then the second vector should be all zeros not 0,1,0 no?
As before an eigen vector is of the form $\displaystyle \begin{bmatrix}x_1 \\ x_2 \\ -x_1\end{bmatrix}= \begin{bmatrix}x_1 \\ 0 \\ -x_1\end{bmatrix}+ \begin{bmatrix}0 \\ x_2\\ 0\end{bmatrix}= x_1\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}+ x_2\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}$

No, the second vector should not be "all zeros" because you are wrong about $\displaystyle x_2$ being 0. Since it is not required to satisfy any equation, it can be any number, not just 0.

thanks!

8. ## Re: help finding the basis of an eigen space

in the attached image, for one matrix we get a t parameter for a row of zeros, but in the next we don't. Why is that? Does it have to do with the algebraic and geometric multiplication?

EDIT: nm I see now: zero x_3 = 0 places no restrictions on the value of x_3

but -4x_3 = 0 means that x_3 = 0