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this is not a question per se, just a simple cool solution to a potentially difficult question, that I want to share. I liked it.

The question is:

Let $A$ be a $3\times3$ matrix with real values. Show that $A^2 \neq -I_3$

There are probably many solutions for this problem, but i really thought this one was simple.

Solution is posted down below.

yo'
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Oria Gruber
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    Best practice for this site, for those who want to ask and answer their own question, is to do exactly that: please edit your question to only contain the question, then post your answer as an answer. You can even accept your own answer afterward! That way it doesn't sit around looking like an open question. – Nick Peterson Nov 22 '13 at 16:47

4 Answers4

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Hint: What can $\operatorname{det}(A)$ be?

2

Solution:

Let's suppose that $A^2 = -I_3$, then $A^2+I_3=0$

That means that $A$ is a root of the polynomial $t^2+1$.

However, since we are dealing with real values only, this polynomial has no roots, and so it is irreducible in $\mathbb R$. But it is still a polynomial that $A$ is a root of, and so, from the definition of the minimal polynomial (let's call it $m_A$):

$m_A|t^2+1$ but $t^2+1$ is irreducible and so $m_A=t^2+1$

Invoking the Cayley-Hamilton theorem, we know that the characteristic polynomial of $A$ , $p_A$ can be divided by the minimal polynomial, and it has the same roots, so we can say that:

$p_A =(m_A)^r = (t^2+1)^r$, $r \in \mathbb N$

But since $A$ is a 3-by-3 matrix, the characteristic polynomial should be in a degree of 3:

$deg(p_A)=deg((t^2+1)^r) = 2r = 3$ so 3 is an even number. Contradiction!

Oria Gruber
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Well, you can go much simpler. You have that the eigenvalues of $-A^2$ are $-\lambda^2$, where $\lambda$ runs through the eigenvalues of $A$. Since $A$ is real odd-dimensional, it has a real eigenvalue $\lambda$ and then $-\lambda^2<0$, which is not possible for the identity matrix.

yo'
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Suppose $A^2=-1$. Then $$(\det A)^2=\det (?)={?^?}\det(?)={?},$$ a contradiction

dfeuer
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