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I totally get how I can get the middle and bottom line in the right matrix, but the one on top just seems wrong (the ratio).
I just would like to know if the right matrix is right or if there must be a mistake.
Yes, it looks like a typo. You could get rid of the fractions If you multiply the 2nd row (after it is 0 1 2 | 0 ) by 8/5 and add to the top row you can get 1 2 3 | 0 for the top row but that does not match their row.
I don´t know if this is important I am supposed to find the eigen-vector but still I think I can not do some trick to get that row, right?
1 2/5 -1/5 0 18/5 36/5 ; 5/18 0 36/5 72/5 ; 5/36 1 2/5 -1/5 0 1 2 0 1 2 ; this row is a multiple of the other 1 2/5 -1/5 ; *5 0 1 2 0 0 0 5 2 -1 0 1 2 0 0 0 i think it cant be turned
now if direction is what is important and not size, <5,0,0> = <1,0,0> but i cant verify that to be a good move
I'm sure it is a typo (somewheres). But exactly what are you trying to do here?
Who are you talking to, phi? Me or amistre64? If you meant me. I was just trying to figure out if there is some way to rewrite the top line as it shown in the picture of if it is a typo. I thought it also might be a special case of operation that one can do because I am supposed to find the eigen-vectors with this matrix so perhaps there is someway to get that top line because you can do stuff with this eigen-stuff your normally can not do with a matrix.
I meant Tom. The reason I asked is that row operations do not preserve the eigenvalues (it would be nice if they did, because the eigenvalues sit on the diagonal of a triangular matrix) however, once you get an eigenvalue, you subtract it from the diagonal of the matrix and do gaussian elimination to find its corresponding eigenvector.
That is what is happening here. The eigenvalue was 0 and I wanted to solve this matrix. The solution is the one above but because of the typo I was not sure I couldn´t follow it or if it simply was wrong. So, you are saying we can not multiply by five?
I tend to reduce the matrix all the way to reduced row echelon form starting with the original matrix on the left I get 1 0 -1 0 1 2 0 0 0 now I read off the answer negate (-1 2) and append a 1 to get (1 -2 1) as the eigenvector
I got this from Strang's course http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-7-solving-ax-0-pivot-variables-special-solutions/ and maybe the following lecture
Ok. Thanks a lot for your help phi. I saved my day. :)