Can A2+I=0 for an Odd n x n Matrix A?

[Dustinsfl]

Homework Statement

Let A be an n x n matrix. Is it possible for A²+I=0 in the case when n is odd?

So A is a 2p+1 x 2p+1; however, I don't see this making a difference to the proof if n is odd or even.

The only way I view A²+I=0 is if A has zero has every elements except when i=j where all a₁₁ to a_(2p+1)(2p+1) elements are equal to i=$$\sqrt{-1}$$.

Other then this observation I have made I am lost on this problem.

 

[rasmhop]

Are we talking about real matrices or complex matrices? If complex matrices are allowed then your construction is perfectly valid. If not note that det(-I) = -1 when n is odd (this is where the oddness is required), and therefore we have,
$$-1=det(-I) = det(A^2) = det(A)^2$$
Is this possible for a real matrix A?

 

[Dustinsfl]

The book doesn't state whether it is speaking of real or complex. But when -I², we will obtain I+I not 0.

 

[rasmhop]

The book doesn't state whether it is speaking of real or complex. But when -I², we will obtain I+I not 0.

Due to the fact that the condition of being odd is necessary for the real case, but not the complex I would assume the real case is meant. I'm sorry but I don't understand your last sentence. If it's about my solution, then maybe I should expand a bit. I simply subtracted I to get $A^2 = -I$ and then took the determinant of both sides.

 

[Dustinsfl]

I saw it a different way but given your method and how I thought of it ((-I²)+I=0). When is a real number squared to obtain -1? So I am guessing A²+I=0 when n is odd is never unless we consider complex solutions; however, isn't this the same when n is even?

 

[rasmhop]

I saw it a different way but given your method and how I thought of it ((-I²)+I=0). When is a real number squared to obtain -1? So I am guessing A²+I=0 when n is odd is never unless we consider complex solutions; however, isn't this the same when n is even?

No because if I is the n x n identity matrix, then -I is the nxn diagonal matrix with -1 as its only diagonal element. Thus the determinant is,
$$det(-I) = (-1)^n$$
In the odd case this gives us -1 which as you rightly observed is impossible for real matrices. However in the even case we get 1 and then my equation would simply say
$$1 = det(A)^2$$
which is easy to find matrices satisfying. Consider for instance the matrix,
$$A = \left[\begin{array}{cc} 1 & 1 \\ -2 & -1 \end{array} \right]$$
which actually satisfy A^2 = -I (if my quick calculations are correct).

 

[Dustinsfl]

This only works then when n is even?

 

[rasmhop]

This only works then when n is even?

What does "this" refer to? I only made two claims regarding the case where n is even:
1) If n =2, then your statement definitely isn't true (as I provided a counterexample).
2) The proof I gave does not work when n is even because det(I)=1 IF n is even.
As I have mentioned previously det(-I) is the product of the diagonal elements and therefore it's (-1)^n which is -1 if n is odd. Thus if n is odd and A^2 = -I we get:
$$-1 = det(A)^2$$
As a real matrix has a real determinant and no real number squared is -1 we get a contradiction.

 

[Dustinsfl]

You example matrix A when squared is -I and is a 2 x 2

 

[rasmhop]

You example matrix A when squared is -I and is a 2 x 2

That just shows that it's possible for A^2 + I =0 when n is even (namely when n is 2) so a proof could never possibly prove it in the even case too.

 

[Dustinsfl]

That doesn't make sense to me. "That just shows that it's possible...so a proof could never possibly prove it..."

How can you show it and then say it isn't possible?

 

[rasmhop]

That doesn't make sense to me. "That just shows that it's possible...so a proof could never possibly prove it..."

How can you show it and then say it isn't possible?

I haven't proved it. I presented a proof that says the following:
- (1) If n is odd, then there does not exist a real nxn matrix A such that A^2 + I = 0.
I then gave an example that shows,
- (2) If n is 2, then there does exist a real nxn matrix A such that A^2 + I = 0.
What I simply tried to emphasize is that my proof doesn't work when n is even, and you can't possibly construct a proof that would work when n is even because then it would contradict (2).

 

[Dustinsfl]

Based on det(A²)=(-1)ⁿ couldn't there exist matrices when n is odd?

 

[rasmhop]

Based on det(A²)=(-1)ⁿ couldn't there exist matrices when n is odd?

No. Because imagine that such a matrix A existed. When n is odd $(-1)^n=-1$ so,
$$det(A^2) = (-1)^n = -1$$
Also since det is multiplicative we have $det(A^2) = det(AA) = det(A)det(A) = det(A)^2$ so we would have,
$$det(A)^2 = -1$$
However det(A) is real so in that case we would have a real number whose square is negative, but this is a contradiction so such a matrix can't exist.

 

[Dustinsfl]

So it isn't true in either case but there are some in the even case? However, if the elements can be complex, it is true for both cases then?