$AA^T = A^TA=I$ polynomial of $A$? If for a matrix $A \in M_3(R)$ we know that $AA^T=A^TA=I$ and $\det(A)=1$ is it true that $p_A(1)=0$? I am first supposed to say if this is true, than give an explanation, but I am really confused by this one. Any help is appreciated..
 A: This is true. \begin{align} p_A(1) & = \det(A - I) \\ & = \det(A^T-I)\\ & =\det(A)\det(A^T-I)\\ & =\det(AA^T-A)\\ & =\det(I-A)\\ & =\det((-1)(I-A))\\ & =(-1)^3 \det(A-I) \\ & = -\det(A-I)\end{align}
So, $p_A(1) = 0$.
A: The answer given is a nice direct approach.  A more typical idea is as follows:
The values $\lambda$ such that $p_A(\lambda) = 0$ are the eigenvalues of $A$.  We wish to show that $1$ is an eigenvalue of $A$.  
Note that $\det(A)$ is the product of the $3$ eigenvalues of $A$.  Because $A^TA = 1$, we know that all eigenvalues satisfy $|\lambda| = 1$.  Because $A$ is a matrix with real entries, all strictly complex eigenvalues (i.e. eigenvalues with non-zero imaginary part) must come in conjugate pairs.  
Suppose that $A$ has real eigenvalues.  Then every eigenvalue $\lambda_i$ is from $\{\pm 1\}$, and we can confirm that $\lambda_1\lambda_2\lambda_3 = 1$ implies that one of eigenvalues is $1$.
Suppose that $A$ has a strictly complex eigenvalue $\lambda_1$.  The conjugate $\lambda_2 = \overline{\lambda_1}$ must also be an eigenvalue.  The determinant is
$$
\det(A) = \lambda_1 \overline{\lambda_1}\lambda_3 = |\lambda_1|^2 \lambda_3 = \lambda_3
$$
since $\det(A) = 1$, the remaining eigenvalue must be $1$.

Here's another approach: write 
$$
p_A(\lambda) = \det(A - \lambda I) = -\lambda^3 + a_2\lambda^2 + a_1 \lambda + a_0
$$
This polynomial is a cubic, so it either has $3$ roots or one root.
 If it has $3$ roots, then it's easy to see that any real eigenvalue (i.e. any real root) of $A$ is $\pm 1$: suppose that $Ax = \lambda x$, where $\lambda$ is real.  Then 
$$
\|Ax\|^2 = \lambda^2 \|x\|^2
$$
but we also have $\|Ax\| = \|x\|$, which means that $\lambda^2 =1$, so $\lambda = \pm1$.
With that, we can use the above proof for the three real eigenvalue case as written. 
Otherwise, consider the graph $y = p_A(x)$.    Suppose that $p_A(0) = 1$ and $p_A(-1) = 0$.  We know that $\lim_{x \to \infty}p_A(x) = - \infty$.  So, by the intermediate value theorem, $p_A$ has another root, which contradicts our assumption.
So, if $p_A$ has only one real root, and if $\det(A) = p_A(0) = 1$, then we must have $p_A(1) = 0$.
