Question

Linear Algebra problem.

Let A be an 2x2 matrix over the real numbers.
1. The subset {A^2, A^5, A^11} is always linearly dependant.
2. The subset {I, A, A^2} is always linearly dependant.
3. It is possible that the subset {A^2, A^5, A^11} is linearly independant.
4. It is possible that the subset {I, A^2, A^5, A^11} is linearly independant.
5. It is possible that the subset {I, A, A^2} is linearly independant.


Appreciate your help!

notice I means the identity matrix.

Answer 1

@ash2326

Answer 2

?

Answer 3

@Coolsector - It is pretty hard question, huh?

Answer 4

no you are asked the opposite

Answer 5

linearly dependant.

Answer 6

yes lol sorry

Answer 7

It is really hard to find such an example. I'm trying....

Answer 8

Theorticlly It should be possible somehow because you don't exceed the 4 element limit (the dimension of the 2x2 matrices is 4). If you provide one answer it will provide an answer almsot to all the others.

Answer 9

are you here? still trying?

Answer 10

@imron07 do you have an idea?

Answer 11

It seems like this post is not going to be resolved.

Answer 12

try asking it here
http://math.stackexchange.com/

Answer 13

read FAQ first
http://math.stackexchange.com/faq

Answer 14

if you get response ... also let me know.

Answer 15

thank you @experimentX

Answer 16

honestly i would also like to know it.

Answer 17

@experimentX if you would like to know the absolute answer , see :
http://www.wolframalpha.com/input/?i=linear+independence%28x%2Cy%2Cz%2Ct%29%2C+%28x%5E2%2Byz%2C+ty%2Bxy%2C+tz%2Bxz%2C+t%5E2%2Byz%29%2C+%281%2C0%2C0%2C1%29

Answer 18

It means that (2) is right.
But I still want to understand why... I can't row-reduce x^2....and such terms.

Answer 19

try putting matrices as
{{1,2},{3,4}}

Answer 20

I did..... I tried anything... like 20 times

Answer 21

Nothing works. all are depended

Answer 22

aI + bA + cA^2 = 0 ...
aI + A(bI + cA) = 0
-------------------
try playing around!!

Answer 23

Okay @experimentX, I have the answers and I promised to include them here as I find them so here it is.

By the "Cayley-Hamilton" theorem we can actually say that every square matrix over a commutative ring satisfies its own characteristic equation.

That is, \[A^n = \lambda_1(I) + \lambda_2(A)\]
Which means the the dimension of the subspace spanned by the A powers has at most dimension 2. It follows that every set with 3 or more members is linearly dependent.

Lets see some concrete example: see (!) and (2) above.

These must be linearly depenedent subsets because every one of them is a subset of the A powers.

As I mentioned earlier the basis of a subset spanned by the A powers has at most dimension 2.

I'll show you why. Lets look at the second one.
We have the subset \[{I, A, A^2}\]
Now the identity matrix is simply A^ 0. so we have a subset of A powers.
I will show you now that A^2 can be expressed with only I and A. If I show that It's true, we are done because this is exactly the defintion of linearly dependent subset.

A^2 is a sqaure matrix so we can use "Cayley-Hamilton" theorem.

characteristic polynomial is given by \[p_A(λ)=λ^2−(a+d)λ+(ad−bc)\]so the Cayley–Hamilton theorem states that \[p(A)=A^2−(a+d)A+(ad−bc)I\] and therefore :
\[A^2=(a+d)A-(ad−bc)I\] and we are done!

The same thing can be done with any subset of A powers.

I hope everything is clear.
Let me know if you have questions.

Answer 24

Also @Coolsector , you might find this one interesting.

Answer 25

ty