1) Relation b/w eigen values of $AB$ and $BA$ ?
let, $AB \cdot x = \lambda \cdot x $
multiplying B both sides
$B \cdot (AB \cdot x) = B \cdot (\lambda \cdot x) $
$(BA) \cdot (B \cdot x) = \lambda \cdot (B \cdot x) $
it's like, $C \cdot y = \lambda \cdot y$
where, $C = BA$ and $y = B \cdot x$
on observation we can say that $BA$ has $B \cdot x$ as eigen vector and $\lambda$ as eigen value.
so, we can say that $AB$ and $BA$ share $\lambda$ as their eigen value.
2) Why $AB$ and $BA$ share only non-zero eigen values ?
we have proved that $B \cdot x$ is eigen vector of $BA$ corresponding to eigen value $\lambda$ (which is common b/w $AB$ and $BA$).
but we are forgetting one important thing here, by defintion eigen vectors should be non-zero vectors.
so, we have to prove that $B \cdot x$ is a non-zero vector.
we know that $x$ is eigen vector of $AB$, that means it is a non-zero vector. hence if we strict $\lambda$ to be non-zero $\lambda \cdot x$ will also be non-zero.
now, if we see the equation $ AB \cdot x = \lambda \cdot x $
non-zero $\lambda$ will keep the right hand side of this equation also non-zero.
so, left hand side has to be non-zero to be equal to right hand side.
so, $ AB \cdot x = A \cdot (B \cdot x)$ is also not zero. that means $(B \cdot x)$ is also non-zero. because if any of $A$ or $(B \cdot x)$ is zero than their product will also become zero.
by doing all this exercise we can understand that a zero $\lambda$ will make $B \cdot x$ zero and that is against the definition of eigen vectors.
so, we can say that $AB$ and $BA$ only share non-zero eigen values.
3) If $A_{m \times n}$ and $B_{n \times m}$ where $m \geq n$ then if $BA_{n \times n}$ has $a, b, c$ as non-zero eigen values then $AB_{m \times m}$ have eigen values like $a, b, c, 0, 0, 0, ...$ why ? means why the eigen values which are not common b/w $AB$ and $BA$ are zero ?
given, $A_{m \times n}$ and $B_{n \times m}$ where $m \geq n$
$AB_{m \times m}$ and $BA_{n \times n}$
$A$ can have atmost $n$ linearly independent columns in $\mathbb{R}^{m}$ and similarly $B$ can have atmost $n$ linearly independent columns in $\mathbb{R}^{n}$. so, we can have atmost $n$ linearly independent columns in $A$ or $B$ and that means we can only have $n$ linearly independent columns in $AB$ or $BA$.
for example suppose, $m = 7, n = 3$ and eigen values of $ BA_{3 \times 3} = a, b, c $ (where, $a, b, c$ are non-zero)
now, as we know that in $AB$ and $BA$ we can have atmost $n = 3$ (when $m \geq n$) linearly independent columns.
so, that means in $AB$ we have $m - n = 7 - 3 = 4$ free columns. that means $det(A) = 0$ and again that means we have atleat one $\lambda = 0$.
now, when we solve for eigen vectors corresponding to $\lambda = 0$ using,
$(AB - \lambda \cdot I) \cdot x = 0$
$(AB - 0 \cdot I) \cdot x = 0$
$(AB - 0) \cdot x = 0$
$AB \cdot x = 0$
as we have 4 free columns in $AB$ we will be getting 4 L.I. solutions of $x$ which are actually 4 L.I. eigen vectors corresponding to $\lambda = 0$.
and we know, arithmetic muliplicity(AM) $\geq$ geometric multiplicity(GM) $\Rightarrow$ AM $\geq$ 4 that means we have atleast 4 in the power of $\lambda$ in characteristic polynomial $|AB - \lambda I| = 0$ of matrix $AB$.
so, characteristic polynomial will be like this:
$\lambda^{4} \cdot (\lambda - \alpha)^{p} \cdot (\lambda - \beta)^{q}...... = 0$
here, from the characteristic polynomial we can say that we have $\lambda = 0$ four times and rest are non-zero eigen values.
previously we proved that if $BA$ has $a, b, c$ as non-zero eigen values then they will be common b/w AB also.
so, eigen values of $AB = a, b, c, 0, 0, 0, 0$
Therefore, we can say that eigen values which are not common b/w $AB$ and $BA$ are zero.
