$(AA^t)^3$, $(AA^tA)^5$, real eigenvalues and determinant based on a variable $\lambda$

Question

$A=\begin{bmatrix}9&3&4\\1&1&2\\3&0&\lambda\end{bmatrix}$

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.

Answer 1

1. $AA^t$ is a real symmetric matrix, and so is $(AA^t)^3$. Real symmetric matrices always have real eigenvalues.
2. 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 $\det A = \det A^t$.

Answer 2

$A=\begin{bmatrix}9&3&4\\1&1&2\\3&0&λ\end{bmatrix}$

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 $(6\lambda+6)^{15}$