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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 $\det(A) = 3*\det(B)?$

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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.

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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.

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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 $\det p(A) = p(\lambda_1) \ldots p(\lambda_n)$.

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