Relation between minimal polynomials of $AA^T$ and $A^TA$ Let $A$ be a real $m\times n$ matrix with $m<n$, of full row rank. Let $f$ be the minimal polynomial of $AA^T$, and $g$ be the minimal polynomial of $A^TA$. Prove that $g=xf$.
What I know is that $A, A^T, AA^T, A^TA$ all have the same rank, so $AA^T$ must be invertible and $A^TA$ is not invertible, thus having eigenvalue 0, which means that $g$ must have a factor $x$. However, I don't know how to proceed further. Does anyone have idea?
 A: $A^TA$ and $AA^T$ are both diagonalizable and have the same eigenvalues, except for 0.
This means that the minimal polynomials are the same, except for a factor $x$.
A: The minimal polynomial is the polynomial of smallest degree that has the matrix as a root. Now to look at polynomials, let's take for example $f = x^3$:
$$f(AA^T) = AA^TAA^TAA^T = 0.$$
Now consider $g(A^TA)$ and insert parentheses in a clever way:
$$g(A^TA) = A^TA f(A^TA) = A^TAA^TAA^TAA^TA = A^T(AA^TAA^TAA^T)A = A^Tf(AA^T)A = 0.$$
Extending this argument to all powers of $x$ appearing in $f$ and then through linearity to an arbitrary polynomial should give you $g(A^TA) = 0$. 
For the other direction assume $h$ is a polynomial having $A^TA$ as root. Then you already showed that $h$ is divisible by $x$. Using the same trick as above, you can show that $xh$ is a polynomial that has $AA^T$ as a root, thus it is divisible by $f$. But $f$ does not contain $x$ as a factor (here, the full rank comes into play!), so $\deg(h) \geq \deg(f) + 1 = \deg(g)$ and thus $g$ has indeed minimal degree.
A: We claim that $g=x^{\alpha}f$; $\alpha\in \mathbb Z^+$.
As $A'A$ is symmetric, it must be diagonalizable and hence must contain only linear factors in its minimal polynomial $g$. So, $\alpha=1$.
