homogenous

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May 5th 2010, 12:23 PM
alexandrabel90

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.
May 5th 2010, 02:03 PM
dwsmith

Quote:

Originally Posted by alexandrabel90

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.

http://www.mathhelpforum.com/math-he...rexamples.html
May 5th 2010, 03:39 PM
jakncoke

"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.
May 6th 2010, 02:40 AM
HallsofIvy

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.