Let A′ A ′ denote the standard (coordinate) basis in ℝn R n and suppose that T:ℝn→ℝn T : R n → R n is a linear transformation with matrix A A so that T(x)=Ax . Further, suppose that A is invertible. Let B be another (non-standard) basis for ℝn R n , and denote by A(B) A ( B ) the matrix for T with respect to B .
a) Prove that A ( B ) is also an invertible matrix.
b) If {x1,…,xk} is a linearly independent set in ℝn , prove that { A ( B ) [ x 1 ] ( B ) … A ( B ) [ x k ] ( B ) } is also linearly independent, where x denotes the B -coordinate vector of x.
Ok I think I understand now that A_B = P(A_E)P^-1, where P is the change of coordinates matrix. And since A_E is invertible (it says so in the problem) and obviously P is invertible, then the product PAP^-1 gives an invertible matrix? Am I close here? And how is the second set also linearly independent? I'm lost here.
a) Yes you are almost there. Use the fact that $A_E$ and $P$ are invertible to explicitly write down an inverse for $PA_E P^{-1}$.
b) Suppose otherwise, that the set of $A x_i$ is not linearly independent. Then write down what that means as an equation, and multiply by invertible matrices to show that the $x_i$ are not linearly independent, which is a contradiction.