- #1

tysonk

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In other words, A +I = B , it B still invertible? Why?

Thanks.

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- Thread starter tysonk
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- #1

tysonk

- 33

- 0

In other words, A +I = B , it B still invertible? Why?

Thanks.

- #2

Dick

Science Advisor

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-I is invertible. Adding I to it gives you something pretty noninvertible.

- #3

tysonk

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- #4

tysonk

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Solve for X. If you need to invert a matrix explain why that matrix is invertible.

A,B,X, are nbyn matrices. A,X, A-AX are invertible.

(A-AX)^-1 = X^-1 B

So I got X = BA(I + BA) ^-1

I'm just not sure why (I + BA) is invertible.

Thanks for the help

- #5

Dick

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Solve for X. If you need to invert a matrix explain why that matrix is invertible.

A,B,X, are nbyn matrices. A,X, A-AX are invertible.

(A-AX)^-1 = X^-1 B

So I got X = BA(I + BA) ^-1

I'm just not sure why (I + BA) is invertible.

Thanks for the help

You got there by showing BA=(1+BA)X, right? If X-AX is invertible and X is invertible then B is invertible from your original equation, yes? Now if B is invertible, A is invertible and X is invertible, then (1+BA) is invertible. If you have an equation like J*K=L and two of the three matrices are invertible, then the third is invertible. And haven't you got the BA and (1+BA)^(-1) factors in reverse order?

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- #6

tysonk

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Makes sense, Thanks!

Edit: yes, the ordering should be the other way.

Edit: yes, the ordering should be the other way.

- #7

Rad021

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- #8

Dick

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Starting from (A-AX)^(-1)=X^(-1)B, begin by inverting both sides. Then expand and try to get all the X's on one side.

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