- #1

- 613

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invertible symmetric matrices and AB = BA.

Show that A*B^-1 is symmetric

(A*B^-1)^T

=A^T * (B^-1)^T

=A^T * (B^T)^-1

Since A and B are symmetric

=A*B^-1

Is this right? Is (B^-1)^T = (B^T)^-1?

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- Thread starter pyroknife
- Start date

- #1

- 613

- 3

invertible symmetric matrices and AB = BA.

Show that A*B^-1 is symmetric

(A*B^-1)^T

=A^T * (B^-1)^T

=A^T * (B^T)^-1

Since A and B are symmetric

=A*B^-1

Is this right? Is (B^-1)^T = (B^T)^-1?

- #2

HallsofIvy

Science Advisor

Homework Helper

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Yes, that is true, but "(A*B^-1)^T= A^T*(B^-1)^T"

invertible symmetric matrices and AB = BA.

Show that A*B^-1 is symmetric

(A*B^-1)^T

=A^T * (B^-1)^T

=A^T * (B^T)^-1

Since A and B are symmetric

=A*B^-1

Is this right? Is (B^-1)^T = (B^T)^-1?

Rather, (A*B^{-1})^T= (B^{-1})^T*B^T.

- #3

- 613

- 3

Yes, that is true, but "(A*B^-1)^T= A^T*(B^-1)^T"isn't.

Rather, (A*B^{-1})^T= (B^{-1})^T*B^T.

Oh I memorized the identity wrong. (AB^T)=B^t*A^T

- #4

- 613

- 3

IF AB=BA, then B^-1 * A *B=A, so B^-1*A=AB^-1

For the last type, the "B^-1*A=AB^-1", part how did they go from B^-1 * A *B=A to

B^-1*A=AB^-1.

Did they divide both sides by B?

- #5

Dick

Science Advisor

Homework Helper

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IF AB=BA, then B^-1 * A *B=A, so B^-1*A=AB^-1

For the last type, the "B^-1*A=AB^-1", part how did they go from B^-1 * A *B=A to

B^-1*A=AB^-1.

Did they divide both sides by B?

They multiplied both sides on the right by B^(-1). Talking about 'dividing' matrices by B is ambiguous. You can 'divide' on the left or the right.

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