$(AA^t)^3$, $(AA^tA)^5$, real eigenvalues and determinant based on a variable $\lambda$ $A=\begin{bmatrix}9&3&4\\1&1&2\\3&0&\lambda\end{bmatrix}$
So I have this matrix and I am supposed to:
1) Find for which values of $\lambda$ the matrix $(AA^t)^3$ has real eigenvalues.
2) Find the determinant of the matrix $(AA^tA)^5$ for different values of $\lambda$.
So for the first one my answer would be to find the matrix (without that $^3$ ), which shows that all lines are non zero (don't know if it plays a role) and then also say that '$(AA^t)$ is symmetric and that does not depend on the λ so it has real eigenvalues for every $\lambda$.' completely ignoring that $^3$. Is that an acceptable solution?
As for the second one I do not know what to do, a lot of stuff going on, determinant + that $AA^tA$ + that $()^5$ and I do not know where to start and what to do.
 A: $A=\begin{bmatrix}9&3&4\\1&1&2\\3&0&λ\end{bmatrix}$
Exercise 1 If $A$ is a square matrix then $AA^T$ is a symmetric matrix. 
Exercise 2 If $A$ is real symmetric matrix, then $A$ is diagonalizable.
So for problem (1), if you take $\lambda$ to be real, then $B=AA^t$ will be a real, symmetric matrix hence diagonalizable. That is there exist a invertible matrix $P$ such that $B=P^{-1}DP$.
We know if$\lambda$ is an eigenvalue of $B$, then $\lambda^3$ is an eigenvalue of $B^3$   $$B^3x = B^2(Bx) = B^2\lambda x = \lambda(B^2x) = \lambda^2Bx=\lambda^3 x$$
So, In our case $B=AA^t$ is diagonalizable, and has say 3 real eigenvalues $\lambda_1, \lambda_2, \lambda_3$, then $B^3$ has eigenvalues(all real) $\lambda_1^3, \lambda_2^3, \lambda_3^3$
2) Find the determinant of the matrix $(AA^tA)^5$ for different values of λ.
Let $C=AA^tA$ and you want to find the det($C^5$). 
Now determinant of $A$ is $6\lambda +6$, therefore determinant of $C$ is $(6\lambda+6)^3$(using the fact that $det(AB)=det(A)det(B)$ and $det(A^t)=det(A)$)
Determinant of $C^5$ is $(6\lambda+6)^{15}$
A: *

*$AA^t$ is a real symmetric matrix, and so is $(AA^t)^3$. Real symmetric matrices always have real eigenvalues.

*We have $\det A = 6\lambda + 6$ and so
$$\det (AA^tA)^5 = \left(\det(A) \det(A^t) \det(A)\right)^5 = (\det A)^{15} = (6\lambda + 6)^{15}$$
because the determinant is multiplicative and $\det A = \det A^t$.

