Relationship between matrix equations and determinants? What is the relationship between a matrix equation and the determinants of the matrix?
Say if you have the matrix equation $A^3 - 4A = 0$, what can I learn about the matrix $A$'s determinants?
Or even if you have $A - 3B = 0$, does that imply $\det(A) = 3*\det(B)?$
 A: One of the properties of the determinant (more here: http://en.wikipedia.org/wiki/Determinant) is 
$$
\det(AB) = \det(A)\det(B)
$$ 
and from here you can deduce 
$$
\det(A^n) = (\det(A))^n
$$
Another property of determinant is
$$
\det(cA) = c^n\det(A)  
$$
if $A$ is a $n \times n$ matrix.
If you apply the determinant on your first equation you get
$$
\det(A^3) = \det(4A)
$$
or 
$$
(\det(A))^3 = 4^n \det(A)
$$
implying (if $\det(A)$ is not zero)
$$
(\det(A))^2 = 4^n
$$
and finally
$$
\det(A) = (\pm2)^n
$$
From the equation
$$
A-3B=0
$$
you can deduce
$$
\det(A) = 3^n \det(B)
$$
where $A$ and $B$ are square $n \times n$ matrices.
A: In general, an annihilating polynomial doesn't give you information about the determinant but rather the eigenvalues of the matrix. There is a well known theorem that says if $m$ is the minimal polynomial of $A$ and $p$ is any polynomial for which $p(A)=0$ then $m\mid p$. What this says is that the eigenvalues of the matrix necessarily forms a subset of the roots of any annihilating polynomial $p$.
For simpler cases, we can extract more information. For your first polynomial, we know that $A$ satisfies the polynomial $p(x)=x^3 - 4x$. This implies that the eigenvalues of $A$ are amongst the roots of $p$, being $\{0,\ 2,\ -2\}$. If the matrix has $0$ as an eigenvalue, that necessarily implies that the determinant is $0$. Otherwise, the determinant is of the form $\pm 2^n$ where $n$ is the size of the matrix. This is the most information that we can extract since all of these cases are possible.
For your second equation, we get $A = 3B$. Taking determinants gives $\det A = 3^n \det B$ where $n$ is the size of the matrix.
A: More generally, the eigenvalues of $p(A)$ are the images of the eigenvalues of $A$ under the polynomial $p$.  So if $A$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ (counted by algebraic multipicity), $p(A)$ has eigenvalues $p(\lambda_1), \ldots,
p(\lambda_n)$ (again counted by algebraic multiplicity), and 
in particular $\det p(A) = p(\lambda_1) \ldots p(\lambda_n)$.
