I suppose your question is which of 1-3 must be true.
Answer: (2) and (3)
--> (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:
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.
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)
Users browsing this forum: Bing [Bot], omhk and 2 guests