What Are the Possible Values of Det A if A^4 Equals 2A for a 6x6 Matrix?

[bmxicle]

Homework Statement

Suppose that the 6x6 matrix A obeys A^4 = 2A. Find all possible values of det A.

Homework Equations

Det(AB)=(detA)(DetB)

The Attempt at a Solution

Well my first guess is to show that A is either invertible or not invertible thus making Det A either non zero or zero but I'm really not sure.

 

[lanedance]

how about trying your relevant equation first?

 

[bmxicle]

Here are some of the equations I've found from that relevant equation.

Det(A^4) = Det((AA)(AA)) = (Det(AA)Det(AA)) = (Det(A))^4

Also since the matrix is 6x6 i know that Det 2A = 2^6(Det(A))

Is it valid to let Det(A)^4 = x therefore Det(A) = x^1/4 which means Det(A) must be positive.

 

[lanedance]

I would start from
(Det(A))^4 = 2^6(Det(A))

 

[bmxicle]

So if i just solve for the zeros of that equation those should be the only possible values of Det (A).

 

[lanedance]

well whenever the equation is satisfied you have a valid determinant

however you should also consider that
A^4 = 2A = I.2A
then
((1/2)A^3)(2A) = I.(2A)
and then
(((1/2)A^2).A)(2A) = I.(2A)

so whilst zero satisfies the equation, do you still think it is a valid solution?

...consider what you know about matrcies with zero determinant

 

[bmxicle]

hmm ok, so

let x= Det(A)
this gives
x^4 = 2^6x
x = 2^1.5 * x^0.25
0 = x(x^(-0.75)-1)
x = 0, 1

However looking at the above equation
A^4 = I2A
((1/2)A^3)(2A) = I(2A)

seems to imply that A must be the identity matrix, and that Det (A) = 1 is the only possible solution, but I'm still a little shaky on the reasoning you used in that equation right above. I have to go to bed now, but this has been very helpful. Thanks.

 

[lanedance]

hmm ok, so

let x= Det(A)
this gives
x^4 = 2^6x
x = 2^1.5 * x^0.25
0 = x(x^(-0.75)-1)
x = 0, 1

However looking at the above equation
A^4 = I2A
((1/2)A^3)(2A) = I(2A)

seems to imply that A must be the identity matrix, and that Det (A) = 1 is the only possible solution, but I'm still a little shaky on the reasoning you used in that equation right above. I have to go to bed now, but this has been very helpful. Thanks.

note quite, i think there's some errors in your algebra, and as mentioned you nee dto be careful with the x=0 solution

let x= Det(A)
this gives
x^4 - 2^6x = x(x^3 - 2^6) = 0

now clearly one solution is x=0,
then other

you should be able to get a number

consider det(A)=0, this implies that A is not invertible... but from before you have
(((1/2)A^2).A)(2A) = I.(2A)
which means
((1/2)A^2).A = I

 

[HallsofIvy]

You have A^4= 2A so that det(A^4)= det(2A). Now, as you correctly say, det(A^4)= (det(A))^4 and det(2A)= 2^6det(A).


Perhaps it would be clearer if you let x= det(A). Now the problem is to solve x^4= 64x. That equation has two real solutions.

 

[lanedance]

but only one relevant one...

 

[bmxicle]

Ah yes ops i did make a mistake.

x^4 = 64x
0 = x^0.25(2^1.5-x^0.75)

which makes detA 0 or 4, but because there exists a matrix ((1/2)A^2)A = I. So A must be invertible therefore detA cannot be zero.

 

[spamiam]

consider det(A)=0, this implies that A is not invertible... but from before you have
(((1/2)A^2).A)(2A) = I.(2A)
which means
((1/2)A^2).A = I

Aren't you assuming that A is invertible in this implication by canceling the 2A on both sides? If I let A be the zero matrix, then it certainly satisfies A⁴ = 2A and det(A) = 0.

 

[lanedance]

good point & missed that & i think you're probably right, will have a bit more of a think... the zero matrix is definitely a solution and obvioulsy det(0)=0.

so in effect its like saying 1=2, because 0x1 = 0x2

looks like det(A) = 0 is a valid solution

 

[lanedance]

Ah yes ops i did make a mistake.

x^4 = 64x
0 = x^0.25(2^1.5-x^0.75)

which makes detA 0 or 4, but because there exists a matrix ((1/2)A^2)A = I. So A must be invertible therefore detA cannot be zero.

out of interest why do you keep going to fractional powers? I'd just do the following
x^4 = (2^6)x
then
x(x^3-2^6) = 0
so either x=0 or
x^3=2^6
x= 2^2