Simple Linear Algebra problem that's confusing me.

I'm stuck for the longest time on this problem: Find a row vector l such that lA = l, with A = [.2 .8; .7 .3]. This is somehow related to eigenvalues, is there such thing as a row eigenvector? thanks - Yingquan

[Richard Brown]

a "row" eigenvector is usually called a left eigenvector

Answer 1

[Wayne King]

I don't want to just give you the answer, but think of it this way

1.) Think of the typical eigenvalue problem, I'll use B as the matrix

Bx = \lambda x

2.) if 1 were an eigenvalue of B with some corresponding eigenvector x, then

Bx = x

3.) Now what if you transposed both sides of the above equation and let B'=A (where ' is the transpose)

Now do you see?

[Antony Chung]

Are you saying to do B'*x'=x'?

3.) confused me a little bit.

[Richard Brown]

don't forget the rule about transposing products ...