# Determine the eigenvalue of a real matrix

I think to this question for two days :

Let $A$ be a $3\times3$ real matrix such that $\det(A) = 1$ and $A^{-1}= A^T$. Prove that one of the eigenvalues is equal to $1$.

I used the fact that determinant of $A$ is the product of its eigenvalues and then I wrote many equations that I couldn't solve the question.

• Pedantic matter of taste: I always prefer the transpose to be denoted by a lower case $t$, as OP originally had it. Jul 25, 2016 at 21:32
• @User8128: My apologies for changing it. I changed it without much thought because in browser $A^t$ looks almost identical to $A'$. Jul 25, 2016 at 21:40
• @MårtenW Totally fine! As I said, I was merely being pedantic. Jul 25, 2016 at 21:49

Let $\lambda \in \mathbb C$ be an eigenvalue with eigenvector $0 \neq x \in \mathbb C^n$. Then $$(x,x) = (A^tAx,x) = (Ax,Ax) = \lvert\lambda \rvert^2 (x,x).$$ Thus $\lvert \lambda \rvert = 1$ for all eigenvalues. Since $A$ is real, eigenvalues come in conjugate pairs, so either all three are real, or two are complex and one is real, but the two which are complex form a conjugate pair. Together with $\det A = 1$, do you see why this implies that one of the eigenvalues is $1$?

• det$A=1$ does not imply necessarily that $1$ is eigenvalue. Jul 25, 2016 at 21:44
• Of course not, but orthogonality + $\det A=1$ does. This is exactly what the question is asking us to prove. Jul 25, 2016 at 21:48
• I mean orthogonality and det$A=1$ does not imply it. I give a counterexample below. Jul 25, 2016 at 21:49
• The eigenvalues in the second case are $-1$, $-1$ and $1$ if I'm not mistaken. Isn't the characteristic poynomial $p(\lambda) = -(\lambda -1)(\lambda+1)^2$. Jul 25, 2016 at 21:56
• In the second case you have $1$ as eigenvalue iff $\theta=k\pi$. In other cases the only eigenvalue is $-1$. Jul 25, 2016 at 21:58