Which Inverses of Matrices are Invertible?

[craigy]

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?

 

[radou]

A starting hint: a product of regular matrices is a regular matrix.

 

[craigy]

I know that but that doesn't solve the problem

 

[HallsofIvy]

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.)

 

[craigy]

I sub n is an identity matrix. 2, 3 and 5 seems to work when using a diagonal matrix.

 

[radou]

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 A=\left(\begin{array}{cc}1 & 2\\0 & -1\end{array}\right), and B=\left(\begin{array}{cc}-1 & 0\\2 & 3\end{array}\right). Obviously AB\neq BA, 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.

 

[craigy]

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.

 

[radou]

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.

 

[HallsofIvy]

Assuming that by " (In-A)(In+A) = In - A^2 " you meant " (In-A)(In+A) = In^2 - A^2 ", where In = A^-1.

I_n= A^-1? Why would you assume that? Craigy specifically said I_n is the n by n identity matrix.
(I_n- A)(I_n+ A)= I_n²+ I_nA- AI_n+ A². Since I_nA= AI_n= A, and I_n²= I_n, yes, (I_n- A)(I_n+ A)= I_n- A².

 

[radou]

I_n= A^-1? Why would you assume that? Craigy specifically said I_n is the n by n identity matrix.
(I_n- A)(I_n+ A)= I_n²+ I_nA- AI_n+ A². Since I_nA= AI_n= A, and I_n²= I_n, yes, (I_n- A)(I_n+ A)= I_n- A².

The notation confused me, it's okay now.