# Thread: Squares of nxn matrices

1. ## Squares of nxn matrices

The problem:

Let A be an nxn matrix. Is it possible for A^2 + I = 0 in the case where n is odd? What about when n is even? (I is the identity matrix)
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Squaring any nxn matrix to achieve (-1) down the main diagonal is impossible, right? Certainly this can't be the answer that they're looking for though. I have a hunch that there's a more eloquent way to say this using determinants, but I don't know for sure... any suggestions?

2. Originally Posted by n3mo
The problem:

Let A be an nxn matrix. Is it possible for A^2 + I = 0 in the case where n is odd? What about when n is even? (I is the identity matrix)
------
Squaring any nxn matrix to achieve (-1) down the main diagonal is impossible, right? Certainly this can't be the answer that they're looking for though. I have a hunch that there's a more eloquent way to say this using determinants, but I don't know for sure... any suggestions?

If $A^2 + I = 0 \implies A^2 = - I \implies [\det (A) ]^2 = \det (-I) = (-1)^n$
And if $n$ is odd then RHS is negative and LHS are is non-negative.

3. Originally Posted by ThePerfectHacker
If $A^2 + I = 0 \implies A^2 = - I \implies [\det (A) ]^2 = \det (-I) = (-1)^n$
And if $n$ is odd then RHS is negative and LHS are is non-negative.
I follow you until you get to $=(-1)^n$. With either a 3x3 or 4x4 matrix you get a determinant of 1 after squaring them. In both cases the determinants alone become $1 + 1 = 0$ for the original equation. Thus it fails in either case-- odd or even. Can you explain what I'm missing in regards to the $=(-1)^n$ part?

Thanks for the help by the way!

4. Originally Posted by n3mo
Let A be an nxn matrix. Is it possible for A^2 + I = 0 in the case where n is odd? What about when n is even?
If $A = \begin{bmatrix}\phantom{-}0&1\\-1&0\end{bmatrix}$ then $A^2=-I$. A similar construction works for any even value of n.

When n is odd it can't be done (with real scalars), as explained by ThePerfectHacker.