# Linear Algebra - Determinant Properties

1. Nov 11, 2008

### lubricarret

1. The problem statement, all variables and given/known data

1. Give an example of a 2x2 real matrix A such that A^2 = -I
2. Prove that there is no real 3x3 matrix A with A^2 = -I

2. Relevant equations

I think these equations would apply here?
det(A^x) = (detA)^x
det(kA) = (k^n)detA (A being an nxn matrix)
det(I) = 1

3. The attempt at a solution

1.
Would I use above equations with this question? This is what I did so far; I don't know if I'm off in answering this question...
I wrote:
It is a 2x2 matrix, so n = 2
det(A^2) = det(-I)
(detA)^2 = (-1^2)detI
(detA)^2 = detI (and detI = 1)

Therefore, detA * detA must = 1; so could I use the identity matrix itself as a matrix example for A:
A =
[1 0
0 1]
Then, detA * detA = 1 = detI
Does this make sense? Or am I not allowed to use the identity matrix here?

2.
I wrote:
It is a 3x3 matrix, so n = 3
det(A^2) = det(-I)
(detA)^2 = (-1^3)detI
(detA)^2 = -(detI )
(detA)^2 = -1

Then, can I just say that since (detA)^2 is always positive since it is squared... therefore, (detA)^2 can never equal -1, and there is no real 3x3 matrix A with A^2 = -I

Thanks a lot for the help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 11, 2008

### gabbagabbahey

Your solutions look great to me!

3. Nov 11, 2008

### HallsofIvy

Staff Emeritus
I hate to disagree with gabbagabbahey, but you haven't answered the questions at all!

Have you forgotten what the question asked? You were asked to find A such that A2= -I. The example you give has A2= I, not -I.

Yes, this part is correct.

4. Nov 11, 2008

### gabbagabbahey

Sorry, I had a small brain fart there ...As Halls said, you are looking for an example of a matrix such that $A^2=-I$.

5. Nov 11, 2008

### lubricarret

Yeah, thanks! Oops!

Well, since I am looking for an example of a matrix A where A^2 = -I, I could use the example A =
[0 -1
1 0]
Since this squared = -I

But, I found this using trial and error. Is there any other way to answer this question...? It seems the question is too easy if it's just asking for an answer... is there some sort of formula or something I can use here?

Thanks again.

6. Nov 11, 2008

### Dick

You could think geometrically. The linear operator -I represents rotation by 180 degrees in the plane. The square root of that just might be rotation by 90 degrees, right? What's the matrix for that?

7. Nov 12, 2008

### lubricarret

Okay, so I could use properties from linear transformations and say:
[0 -1
-1 0]
= -I, which equals:
[cos180 -sin180
sin180 cos180]

Then, since A^2 =
[cos180 -sin180
sin180 cos180]

I need the square root of this, which is:
[cos90 -sin90
sin90 cos90]

So then A =
[0 -1
1 0]

This may sound like a stupid question, but how do I take the square root of
[cos180 -sin180
sin180 cos180]
to obtain:
[cos90 -sin90
sin90 cos90]

How do you take the square root of sin180, cos180 etc... Sorry I haven't done math before linear algebra in 4 years... I can't remember this stuff.

Thanks!!

8. Nov 12, 2008

### Dick

A^2(x)=A(A(x)). It just means, 'do A twice'. If you rotate by 90 twice, you get a rotation by 180. That's why I said think geometrically. Of course, you could also rotate by -90. There are two 'square roots'.

9. Nov 12, 2008

### lubricarret

Ok got it. Thanks for the help!

10. Nov 12, 2008

### HallsofIvy

Staff Emeritus
Or write $$A= \left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$$ so that $$A^2= \left[\begin{array}{cc}a^2+ bc & ab+ bd \\ ac+ cd & bc+ d^2\end{array}\right]= \left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$$

So we have $a^2+ bc= -1$, $ab+ bd= 0$, $ac+ cd= 0$, $bc+ d^2= -1$, four equations to solve for a, b, c, and d.

11. Dec 5, 2008