Prove Invertibility of I+A When A^k=0

[kidsmoker]

Homework Statement

Let A be an nxn matrix such that A^k=0 for some natural integer k (0 is the nxn zero matrix). Show that I + A is invertible, where I is the nxn identity matrix.

Homework Equations

Invertible implies det(I+A) not equal zero.

The Attempt at a Solution

I really don't know where to start with this one. I can see that A itself must be non-invertible, but I can't seem to get any more conditions on A based on that fact that A^k=0. Could anyone give me a hint please?

Thanks.

 

[boaz]

A hint :
I-A^k=(I+A)(I-A+A^2-A^3+...+(-A)^{k-1})

 

[tiny-tim]

Let A be an nxn matrix such that A^k=0 for some natural integer k (0 is the nxn zero matrix). Show that I + A is invertible, where I is the nxn identity matrix.
…
Invertible implies det(I+A) not equal zero.

Hi kidsmoker!

Forget determinants … use algebra, and construct an inverse!

Hint: I = I - A^k

Edit: ooh, boaz … that's too near a complete solution!

 

[statdad]

To get you thinking, suppose A^3 = 0. What happens when you fully multiply and collect terms for the product

<br /> (I + A) (I - A + A^2) <br />

Use the distributive rule for multiplication, remember that A^3 = 0. If you make this work, you will have the idea for general case. (The pattern should remind you of the geometric series for numbers:

<br /> \frac 1 {1 + x} = 1 -x + x^2 - x^3 + \dots<br />

with the big difference that the matrix problem doesn't involve an infinite series.)

 

[statdad]

Man, some of these other people are FAST on the keyboard.

 

[kidsmoker]

Okay thanks you guys, got it :p