Linear Algebra-Inverses

  • Thread starter craigy
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  • #1
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1. Homework Statement

Determine which of the formulas hold for all invertible nxn matrices A and B

1. (In-A)(In+A) = In - A^2
2. (A+B)(A-B) = A^2 - B^2
3. A^8*B^5 is invertible
4. (A+A^-1)^9 = A^9 + A^-9
5. AB = BA
6. A+A^-1 is invertible

I know 5 is right, and number 2 wrong, but what else?
 

Answers and Replies

  • #2
radou
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A starting hint: a product of regular matrices is a regular matrix.
 
  • #3
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I know that but that doesn't solve the problem
 
  • #4
HallsofIvy
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For 1, 4, and 6, just go ahead and do the multiplication! What is (I-A)(I+A)?
If you don't want to actually multiply (A+A-1) nine times, try one or two times and see if you can't find a pattern. That's how you solve math problems- you try things, you don't just sit and try to "remember" how to solve it! As for (A+ A-1) try to guess an inverse and then do the calculation to see what happens. Try it with a simple diagonal matrix first. (Another general math method- to do complex problems, try a few simple examples first.)
 
  • #5
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I sub n is an identity matrix. 2, 3 and 5 seems to work when using a diagonal matrix.
 
  • #6
radou
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I sub n is an identity matrix. 2, 3 and 5 seems to work when using a diagonal matrix.
You're not interested in diagonal matrices, you're interested in all regular (i.e. invertible) matrices. Take, for example [tex]A=\left(\begin{array}{cc}1 & 2\\0 & -1\end{array}\right)[/tex], and [tex]B=\left(\begin{array}{cc}-1 & 0\\2 & 3\end{array}\right)[/tex]. Obviously [tex]AB\neq BA[/tex], although both det(A) and det(B) are non-zero.

Edit: after reading this, you should easily see if (2) holds or not for two regular nxn matrices A and B.
 
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  • #7
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Thanks radou, after some thought, I figured out that 1 and 3 are the only formulas that hold for all invertible n x n matrices A and B.
 
  • #8
radou
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Thanks radou, after some thought, I figured out that 1 and 3 are the only formulas that hold for all invertible n x n matrices A and B.
Assuming that by " (In-A)(In+A) = In - A^2 " you meant " (In-A)(In+A) = In^2 - A^2 ", where In = A^-1.
 
  • #9
HallsofIvy
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Assuming that by " (In-A)(In+A) = In - A^2 " you meant " (In-A)(In+A) = In^2 - A^2 ", where In = A^-1.
In= A-1? Why would you assume that? Craigy specifically said In is the n by n identity matrix.
(In- A)(In+ A)= In2+ InA- AIn+ A2. Since InA= AIn= A, and In2= In, yes, (In- A)(In+ A)= In- A2.
 
  • #10
radou
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In= A-1? Why would you assume that? Craigy specifically said In is the n by n identity matrix.
(In- A)(In+ A)= In2+ InA- AIn+ A2. Since InA= AIn= A, and In2= In, yes, (In- A)(In+ A)= In- A2.
The notation confused me, it's okay now.
 

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