Question

Hello,
I had a test yesterday and I need your help to understand some questions.

1. is it possible that a matrix A^3 = I , but A != I (over M2x2(R)).
The answer is yes but I can't seem to find such a matrix. Notice your matrix value must be real values and cannot be complex numbers or something else.

2. If A and B both matrix with the same order nxn, and X is also nxn matrix and invertible, and AX=B then rank(A)=rank(B).

I guess that because X is invertible its rank doesn't change the rank of A. Am I right?

3. If A is a nxn matrix then det(A - A^t) = 0.

Answer 1

Regarding the 3rd question :
I'm not sure if its correct. any example of matrix I'm trying to use does satisfies this equation.

Could someone tell me if it's ture and if it is true tell me how to prove it please.

Thank you!!!!

Answer 2

A != I

what does this mean

Answer 3

i guess A is not the identity matrix

Answer 4

It means that your A matrix isn't the Identity matrix.
I = Identity matrix

Answer 5

so

!=

means is not equal to

Answer 6

yes @Coolsector is right.

Answer 7

yep. A isn't the identity matrix. what is it then...

Answer 8

i found this solution \[\mathbf A=\left(\begin{array}{cc} (-1)^{2/3}&0\\ 0&(-1)^{2/3}\end{array}\right)\] but \((-1)^{2/3}\) is complex

Answer 9

yeah. you can't use complex. that's why it's so wierd.

Answer 10

3. If A is a nxn matrix then det(A - A^t) = 0.
this is not correct

Answer 11

as for the rank.. there is a property Rank(AB) <= Min(Rank(A),Rank(B))
since x is invertible its rank is so Rank(Ax)<=Rank(A)
and it means RanK(B) <= Rank(A)

but we can multiply both sides of Ax=B by x-1 and get
A = Bx-1
so
Rank(Bx-1) <= Rank(B)
since rank(x-1) = n
and it means
Rank(A) <= Rank(B)

so we have

Ran(B) <= Rank(A)
Rank(A) <= Rank(B)

-> Rank(A) = Rank(B)

Answer 12

WHY

Answer 13

First of all thank you very much #Coolsector

Answer 14

But Why If A is a nxn matrix then det(A - A^t) = 0.
is not correct?

Answer 15

I forgot to mention the n is odd

Answer 16

I don't know if it changes something.

Answer 17

oh so let me check since i checked one even n and get it wrong ..
as for the rank you understand my answer ?

Answer 18

I'm trying to understand it, just a second.
I'm sorry I forogt to mention about the n odd....

Answer 19

Yeah I think I understand the rank proof. thank you!

Answer 20

for the odd n it looks correct .. i see the pattern but not sure how to come with a proof

Answer 21

Okay, thank you very much for you help! I'll think about it more. There must be a way to prove it beccause it always work

Answer 22

yes im sure.. i just dont know how :(

Answer 23

you see you always left with a matrix with zero in the diagonal
and it is an anti symmetric matrix

Answer 24

so if we have anti symmetric matrix
A-A^T = - (A-A^T)^T

Answer 25

now lets call A-A^T = B
so from what i wrote we get B = -B^T

so
Det(B) = Det(-B^T) = (-1)^n * Det(B^T)
but for every matrix
Det(B) = Det(B^T)

now for an odd n we have
Det(B) = -Det(B) -> Det(B) = 0

Answer 26

now the only thing you have to do is to show that A-A^T is always an anti symmetric matrix

Answer 27

but im happy with this proof

Answer 28

@mtaOS

Answer 29

You are good man!! thank you!