Matrix similarity

  • #1
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[tex]^{-1}[/tex]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[tex]^{-1}[/tex] then det(B)=det(A)

MY SOLUTION
As B=PAP[tex]^{-1}[/tex] 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)

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Answers and Replies

  • #2
Defennder
Homework Helper
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1. Here's an easier way to think about it: Let A, B be square matrices of same size. AB = C. Note that [tex]C^{-1} = B^{-1}A^{-1}[/tex]. 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.
 

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