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Math Help - homogenous

  1. #1
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    homogenous

    im trying to prove that 0 is an eigenvalue of A iff A is non invertible. may i know why for a homogeneous solution Ax = 0 to have a non trivial solution, it means that A in non-invertible?

    cant it be that x is a zero vector?

    if det of A is non-zere, where A is invertible, then there doesnt exist a non trivial solution in the homogenous solution Ax=0?

    i think im confusing myself:/
    thanks.
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    im trying to prove that 0 is an eigenvalue of A iff A is non invertible. may i know why for a homogeneous solution Ax = 0 to have a non trivial solution, it means that A in non-invertible?

    cant it be that x is a zero vector?

    if det of A is non-zere, where A is invertible, then there doesnt exist a non trivial solution in the homogenous solution Ax=0?

    i think im confusing myself:/
    thanks.
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  3. #3
    Senior Member jakncoke's Avatar
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    "may i know why for a homogeneous solution Ax = 0 to have a non trivial solution, it means that A in non-invertible?"

    Certainly, let us suppose A is an nxn matrix. Having a non trivial solution for the homogeneous equation means that there is a redundancy in the one of the column vectors, namely that one of the column vectors can be written as a linear combination of the others. To find an invertible matrix for an nxn matrix we can solve

    [ A I ] and try to get [ I A^-1 ] , for example if i had

    A = \begin{bmatrix} 2 && 1 \\ 1 && 2 \end{bmatrix} to find its inverse i can do

     \begin{bmatrix} 2 && 1 && 1 && 0 \\ 1 && 2 && 0 && 1 \end{bmatrix} using our row operations i can make this matrix into
     \begin{bmatrix} 1 && 0 && \frac{2}{3} && \frac{-1}{3} \\ 0 && 1 && \frac{-1}{3} && \frac{2}{3} \end{bmatrix}

    so the inverse of  \begin{bmatrix} 2 && 1 \\ 1 && 2 \end{bmatrix} is  \begin{bmatrix} \frac{2}{3} && \frac{-1}{3}\\ \frac{-1}{3} && \frac{2}{3} \end{bmatrix}

    this can also be done by solving : Ax=b where b are the column vectors of the identity for nxn matricies.

     \begin{bmatrix} 2 && 1 \\ 1 && 2 \end{bmatrix} x = \begin{bmatrix} 1 \\ 0 \end{bmatrix}  for which the solution is  x = \begin{bmatrix} \frac{2}{3}\\ \frac{-1}{3} \end{bmatrix} so this x is the first column of our inverse matrix, so u can do with all the column vectors of the identity.

    Since the matrix u asked has a non trivial soln to Ax = 0. it means that even though A is nxn that it can really be modified into a matrix with 1 less column vector and still span as much space as the original nxn matrix. clearly a nx(n-1) matrix cannot span R^n which is the space spanned by the identity nxn matrix. so u will come upon a column vector from the nxn identity for which Ax=b dosent have a solution, so there can't exist an inverse matrix for ur nxn matrix. i hope this view point clear it up. if you have any question plz feel free to ask.
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  4. #4
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    Simpler, I think:

    If A is invertible, then if Ax= v, we have A^{-1}(Ax)= A^{-1}v or x= A^{-1}v so that Ax= v has a unique solution.

    If A is invertible, Ax= 0 has the unique solution x= A^{-1}0= 0. Therefore, if Ax= 0 has a non-trivial (non-zero) solution then A is not invertible.

    Yes, x can be 0- in fact, x= 0 is always a solution to Ax= 0. The question is whether that (trivial) solution is unique or not. If it is not unique, if Ax= 0 has other, non-zero, solutions, then A must not have an inverse.
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