Proving Nonsingularity of Similar Matrices: A vs. B | Matrix Similarity Q1 & Q2

[Bertrandkis]

Question 1
If A and B are similar matrices, then prove that A is nonsingular if and only if B is nonsingular.

MY SOLUTION:
B is nonsingular if it's columns are linearly independent
P^{-1}AP=B where the main diagonal of B is made of eigenvectors of A.
How does this affect the nonsingularity of A ? this is where I am stuck.

Question 2
If A and B are similar matrices, Show that if B=PAP^{-1} then det(B)=det(A)

MY SOLUTION
As B=PAP^{-1} this implies that BP=PA
Taking the determinant of both sides.
det(BP)=det(PA)
when expanded, we have det(B).det(P)=det(P)det(A) therefore det(B)=det(A)

 

[Defennder]

1. Here's an easier way to think about it: Let A, B be square matrices of same size. AB = C. Note that C^{-1} = B^{-1}A^{-1}. Which effectively means that if C is invertible (non-singular), then it must be the case that A,B whose matrix product make up C must also be invertible. You can generalise this to mean that if a square matrix A is invertible, then its "product components" (ie. all the possible square matrices which maybe multiplied together to give A) must also be invertible.

2. I think your answer is ok.