Proving invertibility of a matrix

  • Thread starter chipotleaway
  • Start date
  • Tags
    Matrix
In summary, we want to show that A+I_n is invertible, given that A is an nxn matrix such that A^{2013}=0. The hint provided is to look at the pattern in the entries of A^{2013}=0 and consider using a similar approach as the geometric series 1/(1+r)=1-r+r^2-r^3+r^4-... to find the inverse of A+I_n. This involves looking at a smaller situation first, where a matrix B^4=0, and using the expansion (I-B)^(-1) = I - B + B^2 - B^3 to find the inverse of I-B. The same approach can be applied to the
  • #1
chipotleaway
174
0

Homework Statement



If A is an nxn matrix such that [itex]A^{2013}=0[/itex], show that [itex]A+I_n[/itex] is invertible and find an expression for [itex](A+I_n)^-1[/itex]

The Attempt at a Solution



...some hints would be nice :biggrin:
 
Physics news on Phys.org
  • #2
chipotleaway said:

Homework Statement



If A is an nxn matrix such that [itex]A^{2013}=0[/itex], show that [itex]A+I_n[/itex] is invertible and find an expression for [itex](A+I_n)^-1[/itex]

The Attempt at a Solution



...some hints would be nice :biggrin:

If r is a real number and |r|<1 then 1/(1+r)=1-r+r^2-r^3+r^4-... It's a geometric series. How is that proved? See if you can apply the same pattern to your matrix problem. That should be a good hint.
 
  • #3
Thanks, we haven't covered that yet but it seems as though there's lots of stuff on the problem sets that we have to find for ourselves.
 
  • #4
I think I need a bigger hint :p

I'm looking at [itex]A^{2013}=0[/itex] and wondering what kind of pattern the entries in the matrix would need to be so that the result of taking it to that power is 0 (other than all zero entries)

EDIT: Ah, found something - 'nilpotent matrices'...but the characterisations listed on WIki are mostly alien to me
 
Last edited:
  • #5
If I have a number x, and I say what is (1+x)-1 (as a 1x1 matrix), you could do the following:
[tex] (1+x)^{-1} = 1-x+x^2-x^3+x^4... [/tex]

It only works if x has certain properties (otherwise the series won't converge). You can do a similar thing with matrices
 
  • #6
Look at a "smaller" situation first.
Suppose you have a matrix B such that B^4 = 0

With real numbers, if x is small enough, it can be shown that

(1-x)^(-1) = 1 -x + x^2 - x^3 + x^4 -

that is,

(1-x)(1 -x + x^2 - x^3 + x^4 -...) = 1

and this expansion goes on "forever" (it is an infinite series, if you've had calculus and know that term)

At least formally, let's try the same thing with our matrix (B)

Start with

(I - B)^(-1) = I - B + B^2 - B^3 + B^4 - B^5 +...

This may look bad ("how do I work with an infinite series when the terms are matrices?") but remember, for our matrix B, B^4 = 0: that means all higher powers of B are zero also, so our candidate for (I-B)^(-1) is

(I - B)^(-1) = I - B + B^2 - B^3

just a finite sum.

Now (this is for you) work through this product:

(I-B)(I-B+B^2-B^3)You should end up with the product equal to I - that means the inverse of (I-B) is given by (I-B)^(-1).

If this work makes sense, ask yourself: what is different about this small example and the question I asked?
 

Related to Proving invertibility of a matrix

1. What does it mean for a matrix to be invertible?

For a matrix to be invertible, it means that it has an inverse matrix that when multiplied together, give the identity matrix (a square matrix with 1s along the main diagonal and 0s everywhere else).

2. How do you know if a matrix is invertible?

A matrix is invertible if its determinant is non-zero. The determinant is a numerical value that can be calculated for a square matrix and reflects certain properties of the matrix. If the determinant is zero, the matrix is not invertible.

3. What is the process for proving invertibility of a matrix?

The process for proving invertibility of a matrix involves calculating its determinant and showing that it is non-zero. Then, using this determinant, the inverse matrix can be found using a specific formula. The inverse matrix can be verified by multiplying it with the original matrix to get the identity matrix.

4. Can every matrix be inverted?

No, not every matrix can be inverted. Only square matrices (with the same number of rows and columns) can be inverted, and even then, only if their determinant is non-zero.

5. What is the significance of an invertible matrix?

An invertible matrix is significant because it allows us to solve systems of linear equations, which have a wide range of applications in fields such as engineering, economics, and physics. It also allows for efficient computation and manipulation of matrices.

Similar threads

  • Calculus and Beyond Homework Help
2
Replies
40
Views
4K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
951
  • Calculus and Beyond Homework Help
Replies
1
Views
801
  • Calculus and Beyond Homework Help
Replies
4
Views
990
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
451
  • Calculus and Beyond Homework Help
2
Replies
36
Views
3K
  • Calculus and Beyond Homework Help
Replies
25
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top