Linear Algebra - Determinant Properties

[lubricarret]

Homework Statement

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

Homework 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

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!

 

[gabbagabbahey]

Your solutions look great to me!

 

[HallsofIvy]

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

Homework Statement

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

Homework 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

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?

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.

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

Yes, this part is correct.

Thanks a lot for the help!

 

[gabbagabbahey]

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

 

[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.

 

[Dick]

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.

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?

 

[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!

 

[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'.

 

[lubricarret]

Ok got it. Thanks for the help!

 

[HallsofIvy]

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.

 

[universedrill]

Please help me with these:
1) Prove that: nxn real matrix A is a root of f(X)= a[n].X^n+...+a[0].I, where a[n],...,a[0] are coefficients of the polynomial P(t)= det [A-t.I]
2) Let 5x5 real matrix A be satisfied: A^2008 = 0. Prove that: A^5=0.
Thanks.

 

[gabbagabbahey]

Please help me with these:
1) Prove that: nxn real matrix A is a root of f(X)= a[n].X^n+...+a[0].I, where a[n],...,a[0] are coefficients of the polynomial P(t)= det [A-t.I]
2) Let 5x5 real matrix A be satisfied: A^2008 = 0. Prove that: A^5=0.
Thanks.

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