I think I got it now.
After multiplying by x, I have ASx = SBx.
Bx has already been shown equal to \lambdax, so I substitute that in, giving
ASx = S\lambdax
\lambda can be moved to the other side of S since it's a scalar, giving ASx = \lambdaSx.
What do I use to show that?
The only new information I've got that might be helpful is that A = S * B * S^-1
Multiplying on the left by S gives A*S = S*B
After doing that, I'm stuck again. I feel like this is the right track, but I don't know how to relate this back to what I'm trying...
Homework Statement
Let B = S^-1 * A * S and x be an eigenvector of B belonging to an eigenvalue \lambda. Show S*x is an eigenvector of A belonging to \lambda.
Homework Equations
The Attempt at a Solution
The only place I can think of to start, is that B*x = \lambda*x.
However...
Homework Statement
Suppose A is a vector \in R^{2x2}.
Find whether the following set is a subspace of R^{2x2}.
S_{1} = {B \in R^{2x2} | AB = BA}
The Attempt at a Solution
I know that S isn't empty, because the 2 x 2 Identity matrix is contained in S.
The problem I'm having...
If I'm understanding you correctly, I should take the inverses of all the elementary matrices and multiply those, and it should give me A?
Essentially, (E1E2...En)-1 = A
Homework Statement
Given
A = \left( \begin{array}{cc}
2 & 1 \\
6 & 4 \end{array} \right)
a) Express A as a product of elementary matrices.
b) Express the inverse of A as a product of elementary matrices.
Homework Equations
The Attempt at a Solution
Using the following EROs
Row2 --> Row2 -...