# Thread: Prove: A^2+I=0 for A=[3x3] is imossible

1. ## Prove: A^2+I=0 for A=[3x3] is imossible

Hello everyone!

I'm stuck on this linear quetion that is supposed to be easy: Show that no 3x3 matrix $A$exists such that $A^2+I=0$.

Since the question is asking for a 3x3 matrix I figured it needed the long and boring way, so I considered a matrix and initialized its entries to $[A_1 A_2 A_3]$ where $A_1=[a \, x \, u]^t$, $A_2=[b \, y \, v]^t$, and $A_3=[c \, z \, w]^t$, carried out matrix product... LOST!

2nd way: $A^2+I=0,A^2+AA^{-1}=0 \,\text{ so }A+A^{-1}=0$. But what then??

2. What is your field? Remember, a matrix is a linear map between two vector spaces over a specific field...Basically, are you working over $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{C}$?

I am asking you this because the field does matter; let $I$ be the identity matrix, and assume we are working over $\mathbb{C}$. Then $(iI)^2+I=0$, as (not) required.

Thus, your field does matter so you'll need to be careful and you will need to use your field. However, your second approach looks promising. Essentially, you have that $A^{-1} = -A$. So, hit your matrix with a stick - stick in variables for your 9 entries and see what happens when you take its inverse or negate it (there will be a more subtle way, but I can't think of it at the moment!).

EDIT: Note that $A^{-1}$ exists because $A^2=-I$ so $A^4=I$. So splitting $I$ into $AA^{-1}$ makes sense...(I'm sure you know this, but I thought I should probably point it out anyway).

3. Can't you use determinants? If you've got $A^{2}=-I$, then it must be that $\det^{2}(A)=-1$. (We know that $\det(-I)=-1$, since for a diagonal matrix, the determinant is simply the product of the elements along the diagonal, and you have an odd number of $-1$'s there.)

EDIT: Swlabr's point about the field over which you have the vector space is, I agree, important.

4. I am working in the field of real numbers (never worked otherwise)...

The thing is I'm plugging these variables, but then when carrying out the multiplication, I'm getting sums of products of two terms, e.g. $a^2,ab,cv,xy,y^2,etc.$
and by equating my result to [tex]-IMATH] I am set to get like 9 equations with 24 unkowns!!! This is because I preferred working with linear-algebra than non-linear algebra (which I know nothing about)...

5. Ackbeet: The question I was asking has a 2nd part can be rephrased as: find a 2x2 matrix where the above relation holds.
(W. Keith Nicholson, Linear Algebra with Applications, 3.2 #16)

so if $(det\,A)^2=-1$ do you mean that there is something wrong? Suppose it is absurd in general, but HEY, I got a 2x2 matrix that satisfies the relation!

6. If the determinant squared is equal to a negative number, then the determinant must be complex. Can that ever happen for a real matrix?

For the 2 x 2 case, the determinant of $-I$ is actually positive. Generalizing, for n x n matrix $-I$, the determinant is $(-1)^{n}$. So the even-n case and the odd-n case are fundamentally different.

7. Aha

to cut the proof short, can we say: $(\det\,A)^2=-1\Rightarrow \ \det A \ \text{doesn't belong to} \ R$ so we cannot find a real valued 3x3 matrix that has this property. i.e. does what I've said complete the proof?

8. You'd need to reference a theorem somewhere that says if you have an n x n matrix over the reals, its determinant must be real. The determinant of an n x n matrix just involves "basic four" arithmetic, so the result is, at least, intuitive. Does Nicholson have a theorem like that?

9. Well so far he haven't mentioned complex numbers, although I see a few later on in the book.
Well my professor says that he mentions things like span, rank and nullity very early in the book but then goes into details later on. So I guess your suggested theorem stands right now...

10. Originally Posted by rebghb
Hello everyone!

I'm stuck on this linear quetion that is supposed to be easy: Show that no 3x3 matrix $A$exists such that $A^2+I=0$.

Since the question is asking for a 3x3 matrix I figured it needed the long and boring way, so I considered a matrix and initialized its entries to $[A_1 A_2 A_3]$ where $A_1=[a \, x \, u]^t$, $A_2=[b \, y \, v]^t$, and $A_3=[c \, z \, w]^t$, carried out matrix product... LOST!

2nd way: $A^2+I=0,A^2+AA^{-1}=0 \,\text{ so }A+A^{-1}=0$. But what then??

I think you also can say that the only way that A^2+I=0 is if to A^2 has -1 on diagonal and 0 on all other places, you deduce that A is not in M_3(R)

11. If you use the Laplace expansion or the Leibniz formula for the determinant, I think you'll see this result pop right out.

12. Nevermind, I was wrong