Can You Calculate Matrix Inverses with Nilpotent Matrices?

[songoku]

Homework Statement

Let B = \left(\begin{array}{cc}2&4\\1&2\end{array}\right) and A be a matrix satisfying A² = 0, AB = 0

1. Calculate (I+A) (I-A), I = identity

2. Calculate (I+A)^-1 (I+2A)^-1B

Homework Equations

The Attempt at a Solution

1. Ans = I

2.
In the manual, it's written : (I+2A)^-1 = 1/4 (I-2A)
I don't understand how 1/4 pops up...

Thanks

 

[Hurkyl]

What is the definition of the inverse of a matrix? What happens if you tried plugging (I-2A) into the definition?


Incidentally, it's worth making an analogy with the power series for 1/(1+2a).

 

[songoku]

What is the definition of the inverse of a matrix?

The inverse of a matrix A is a matrix A^-1 such that AA^-1=I

What happens if you tried plugging (I-2A) into the definition?

(I-2A) (I-2A)^-1 = I...I still don't get it

Incidentally, it's worth making an analogy with the power series for 1/(1+2a).

power series of 1/(1+2a) = (1+2a)^-1 = 1 - 2a - 8a² - ...

What is the relation to the inverse?


Thanks

 

[Mark44]

The inverse of a matrix A is a matrix A^-1 such that AA^-1=I


(I-2A) (I-2A)^-1 = I...I still don't get it

power series of 1/(1+2a) = (1+2a)^-1 = 1 - 2a - 8a² - ...

Should be an alternating series, and the coefficient on a² is wrong.

What is the relation to the inverse?

1/(1 + 2a) can be written as (1 + 2a)^-1.

(I + 2A)^-1 has a similar expansion. That's the connection.

Thanks

I also don't get what your solution manual shows for (I + 2A)^-1; namely, the 1/4 factor. I think it's a typo.

 

[Hurkyl]

The inverse of a matrix B is a matrix B^-1 such that BB^-1=I


(I-2A) (I-2A)^-1 = I...I still don't get it

(I've changed the variable in the above to reduce confusion)

Sorry, I meant plugging in (I-2A) in for the "inverse of I+2A" -- i.e. B=I+2A and B^-1=I-2A.

My mistake, though -- when you make the substitution, things do work out. The equation

(I+2A)^-1 = (1/4) (I-2A)​

is, indeed, in error.

 

[songoku]

Should be an alternating series, and the coefficient on a² is wrong.

yeah, it should be 4

1/(1 + 2a) can be written as (1 + 2a)^-1.

(I + 2A)^-1 has a similar expansion. That's the connection.

a is variable and A is matrix. Can matrix also be expanded?

I also don't get what your solution manual shows for (I + 2A)^-1; namely, the 1/4 factor. I think it's a typo.

(I've changed the variable in the above to reduce confusion)

Sorry, I meant plugging in (I-2A) in for the "inverse of I+2A" -- i.e. B=I+2A and B^-1=I-2A.

My mistake, though -- when you make the substitution, things do work out. The equation

(I+2A)^-1 = (1/4) (I-2A)​

is, indeed, in error.

So, (I + 2A)^-1 should be equal to (I-2A). Am I right?

Thanks

 

[Mark44]

Yes, since (I + 2A)(I - 2A) = I. This means that I + 2A is the inverse of I - 2A.

 

[songoku]

Thanks a lot Mark and Hurkyl

 

[Hurkyl]

yeah, it should be 4



a is variable and A is matrix. Can matrix also be expanded?

Yes; you can do calculus with matrices! It's a little trickier, but useful. However, you don't have to know matrix analysis to use it as inspiration -- e.g. armed with the knowledge of geometric series, you can guess at a formula for the inverse of (I - A) is for any nilpotent matrix A... and once you have made the guess, it's easy to show directly that the formula does give the inverse.

 

[songoku]

Yes; you can do calculus with matrices! It's a little trickier, but useful. However, you don't have to know matrix analysis to use it as inspiration -- e.g. armed with the knowledge of geometric series, you can guess at a formula for the inverse of (I - A) is for any nilpotent matrix A... and once you have made the guess, it's easy to show directly that the formula does give the inverse.

hm...haven't studied about it yet but I read about it a little on wiki. Thanks a lot hurkyl