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    help finding the basis of an eigen space

    help finding the basis of an eigen space-eigen-question4.jpg
    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?
    Last edited by kingsolomonsgrave; April 12th 2013 at 07:01 AM.
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    Re: help finding the basis of an eigen space

    help finding the basis of an eigen space-eigen-question5b.jpgalso 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?
    Last edited by kingsolomonsgrave; April 12th 2013 at 07:17 AM.
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    Re: help finding the basis of an eigen space

    Quote Originally Posted by kingsolomonsgrave View Post
    Click image for larger version. 

Name:	eigen question4.jpg 
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Size:	325.6 KB 
ID:	27917
    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..

    help finding the basis of an eigen space-linear-algebra-questions1.jpeg

    x2=0 is just one of the values that x2 can have.
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    Re: help finding the basis of an eigen space

    Quote Originally Posted by kingsolomonsgrave View Post
    Click image for larger version. 

Name:	eigen question5b.jpg 
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ID:	27918also 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.
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    Re: help finding the basis of an eigen space

    Quote Originally Posted by kingsolomonsgrave View Post
    Click image for larger version. 

Name:	eigen question4.jpg 
Views:	11 
Size:	325.6 KB 
ID:	27917
    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?
    \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 x_1+ x_3= 0. That simplifies to x_3= -x_1 but there is NO condition on x_2. That means that \begin{bmatrix}x_1 \\ x_2\\ -x_1\end{bmatrix} will satisfy the equation for any numbers, x_1 and x_2:
    \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 x_2 is.
    Last edited by HallsofIvy; April 13th 2013 at 07:31 AM.
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    Re: help finding the basis of an eigen space

    Quote Originally Posted by kingsolomonsgrave View Post
    Click image for larger version. 

Name:	eigen question5b.jpg 
Views:	5 
Size:	738.6 KB 
ID:	27918also 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 \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 x_2 being 0. Since it is not required to satisfy any equation, it can be any number, not just 0.
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    Re: help finding the basis of an eigen space

    thanks!
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    Re: help finding the basis of an eigen space

    help finding the basis of an eigen space-screen-shot-2013-04-15-4.15.14-pm.png

    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
    Last edited by kingsolomonsgrave; April 15th 2013 at 12:28 PM.
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