## Linear algebra & fields problems

Forum for the GRE subject test in mathematics.
Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

### Linear algebra & fields problems

(1)A is a N dimension matrix, if (A-I)^2 =0
Then:
(1) A=I (X)
(2) det A = I
(3) tr(A) = N

I don't have any clue about such type of questions except for trying to find some examples to disprove. Could you please also give me some advices. Thanks

(2) A 2x2 invertible matrix whose element is from a field of n elements, how many of such matrixes exist?
(a b)
(c d) ---> ad!=bc what's next?

mathfun
Posts: 9
Joined: Fri Nov 19, 2010 3:52 am

### Re: Linear algebra & fields problems

Hi,

I suppose your question is which of 1-3 must be true.

--> (A-I)^2=0, so the minimal polynomial of A, let's say p(x), is either p(x)=x-1 or p(x)=(x-1)^2. In the first case A=I and in the second one, A is un upper triangular matrix with 1s at the diagonal (see Jordan form of a matrix). Therefore, in both cases it is true that detA=1 and trA=N.

For the second question I have no idea!

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

### Re: Linear algebra & fields problems

^ The Jordan Normal Form is extremely useful for proving stuff and something you should definitely learn for the GRE! Here's an example of a matrix A with the properties you want:

$$A = \left( \begin{array}{c c} 1 & 1 \\ 0 & 1 \end{array} \right)$$

For the second problem, first count that there are n^4 2x2 matrices with entries from a field. How many have determinant zero? Let's do this by case analysis:

- Suppose all entries are non-zero. Given any non-zero entries in the first three coordinates, there exists exactly one entry for the last coordinate that makes the matrix singular. There are (n-1)^3 singular matrices of this type.

- Suppose exactly one entry is zero. Then the matrix is always invertible.

- Suppose exactly two entries are zero. Then the matrix is singular if and only if these two entries lie in the same row or column. There are 4 * (n-1)^2 of these singular matrices.

- If exactly three entries are zero, then the matrix is always singular. There are 4 * (n-1) of these matrices.

- And of course there's the zero matrix.

Adding up, there should be n^4 - (n-1)^3 - 4 * (n-1)^2 - 4 * (n-1) - 1 = n^4 - n^3 - n^2 + n = (n-1)^2 * n * (n+1) invertible matrices.

Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

### Re: Linear algebra & fields problems

mathfun wrote:Hi,

I suppose your question is which of 1-3 must be true.

--> (A-I)^2=0, so the minimal polynomial of A, let's say p(x), is either p(x)=x-1 or p(x)=(x-1)^2. In the first case A=I and in the second one, A is un upper triangular matrix with 1s at the diagonal (see Jordan form of a matrix). Therefore, in both cases it is true that detA=1 and trA=N.

For the second question I have no idea!
owlpride wrote:^ The Jordan Normal Form is extremely useful for proving stuff and something you should definitely learn for the GRE! Here's an example of a matrix A with the properties you want:

$$A = \left( \begin{array}{c c} 1 & 1 \\ 0 & 1 \end{array} \right)$$

For the second problem, first count that there are n^4 2x2 matrices with entries from a field. How many have determinant zero? Let's do this by case analysis:

- Suppose all entries are non-zero. Given any non-zero entries in the first three coordinates, there exists exactly one entry for the last coordinate that makes the matrix singular. There are (n-1)^3 singular matrices of this type.

- Suppose exactly one entry is zero. Then the matrix is always invertible.

- Suppose exactly two entries are zero. Then the matrix is singular if and only if these two entries lie in the same row or column. There are 4 * (n-1)^2 of these singular matrices.

- If exactly three entries are zero, then the matrix is always singular. There are 4 * (n-1) of these matrices.

- And of course there's the zero matrix.

Adding up, there should be n^4 - (n-1)^3 - 4 * (n-1)^2 - 4 * (n-1) - 1 = n^4 - n^3 - n^2 + n = (n-1)^2 * n * (n+1) invertible matrices.
Thank you both very much. That's really helpful.

I found myself having very little knowledge about the Jordan Normal Form. Will definitely spend some time looking at this. Thanks again.

rahulshr
Posts: 25
Joined: Mon Sep 19, 2011 12:16 pm

### Re: Linear algebra & fields problems

For the first question, what is wrong with this reasoning?

(A-I)^2=0 => A^2-2*A + I=0.
Which means characteristic equation = x^2-2x+1=0 which means det(A) = 1..but trace is inconclusive for dim(A)>2.

e.g. take a 3x3 matrix...
$$A = \left( \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0.5 \end{array} \right)$$

the determinant is 1, it does satisfy (A-I)^2 = 0, and trace !=N.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

### Re: Linear algebra & fields problems

Hi,
I thought (A-I)^2 = 0 is not satisfied?

the result of (A-I)^2 I got is
(0 | 0 | 0
0 | 1. | 0
0 | 0 | 0.25)

rahulshr
Posts: 25
Joined: Mon Sep 19, 2011 12:16 pm

### Re: Linear algebra & fields problems

yoyobarn wrote:Hi,
I thought (A-I)^2 = 0 is not satisfied?

the result of (A-I)^2 I got is
(0 | 0 | 0
0 | 1. | 0
0 | 0 | 0.25)
my bad - I miscalculated (A-I)^2 as det((A-I)^2)

but, still - what is wrong with the approach - forming the characteristic equation and 'guessing' trace and determinant from there.

rahulshr
Posts: 25
Joined: Mon Sep 19, 2011 12:16 pm

### Re: Linear algebra & fields problems

I think got my mistake..
because characteristic polynomial satisfies det(A-kI) = 0...and not (A-I)^2 is something different.

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

### Re: Linear algebra & fields problems

Be careful to distinguish between the characteristic and the minimal polynomial. If (A - I)^2 = 0, then the minimal polynomial of A is either (x-1) or (x-1)^2, but its characteristic polynomial could be (x-1)^100. The Jordan Normal Form captures precisely this difference between the minimal and the characteristic polynomial.

jstanczyk
Posts: 4
Joined: Thu Nov 10, 2011 12:31 pm

### Re: Linear algebra & fields problems

Although the jordan form is useful, it is not necessary for this problem. It does give a quick answer, but I would like to note that the problem has a quick solution even without it.

If (A-I)^2=0 then we have A^2-2A+I=0. Since the right side is the zero matrix, it will map every x to zero. Every eigenvalue h must have a non-zero eigenvector x. Apply x to both sides of the last equation and you have:

h^2x -2hx+x=(h^2-2h+1)x=0. Since x is not zero h must satisfy (h^2-2h+1)=0 meaning h must be 1. Therefore all eigenvalues are equal to 1. That, of course, tells you that 2 and 3 must be true.