Elementary proof for $\text{rank} \left(A \otimes B \right) = \text{rank}A\cdot \text{rank}B$ (Where $\otimes$ denotes Kronecker product)
Hello,
All the proofs I found for this equality are using SVD decomposition and singular values.
Let $A,B\in \mathbb{F}^{n\times n}$ I want to prove the above.
I'm supposed to do it in 2 steps:


*

*First, for the case that $A$ and $B$ are invertible. I didn't do it formally yet but it doesn't look difficult.

*Using the mixed-product property, namely
$$\left(A\otimes B \right)\left(C\otimes D \right)=AC\otimes BD$$ deduce for all matrices.


I can't figure out how the mixed-product property helps finishing the proof.
Thanks. 
 A: I will prove the general case for arbitrary sized matrices.
Lemma: Suppose that $A$ is an $n\times n$ matrix and that $B$ is an $m\times m$ matrix, both invertible. Then $A\otimes B$ is invertible.
Proof: Suppose that $A$ and $B$ are invertible. By the mixed product property we then have
$$(A^{-1}\otimes B^{-1})(A\otimes B) = I_n \otimes I_m = I_{nm}$$
Therefore $A\otimes B$ is invertible. $\square$
We now need two elementary and well known facts. First, let $I_{m\times n}^r$ denote the $m\times n$ matrix which is zero everywhere except the first $r$ diagonal entries. Then given any $m\times n$ matrix $A$ with rank $r$, there exists invertible matrices $P$ and $Q$ such that
$$PAQ = I_{m\times n}^r$$
This is just an application of successive elementary column and row operations. 
Second, note that right or left multiplying by an invertible matrix does not change rank, i.e. if $A$ is an invertible matrix, then
$$\mathrm{rank}(AB) = \mathrm{rank}(B)\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \mathrm{rank}(CA) = \mathrm{rank}(C)$$
for all appropriately sized matrices $B$ and $C$.
Theorem: Ranks for Kronecker products are multiplicative, i.e.
$$\mathrm{rank}(A\otimes B) = \mathrm{rank}(A)\cdot\mathrm{rank}(B)$$
Proof: Let $A$ and $B$ be $m\times n$ and $p\times q$ matrices respectively. Suppose that $A$ has rank $r$ and $B$ has rank $s$. There exists invertible matrices $P_A,\ Q_A$ and $P_B,\ Q_B$ such that
$$P_AAQ_A = I^r_{m\times n}\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ P_BBQ_B = I^s_{p\times q}$$
Applying the mixed product property, we therefore have
$$(P_A\otimes P_B)(A\otimes B)(Q_A\otimes Q_B) = I_{m\times n}^r \otimes I_{p\times q}^s$$
By our first lemma, both $P_A\otimes P_B$ and $Q_A\otimes Q_B$ are invertible matrices, therefore
$$\mathrm{rank}(A\otimes B) = \mathrm{rank}(I_{m\times n}^r \otimes I_{p\times q}^s)$$
But the latter matrix has precisely $rs$ non-zero entries, each on a different row and column by construction. Therefore
$$\mathrm{rank}(A\otimes B) = rs = \mathrm{rank}(A)\cdot\mathrm{rank}(B)$$
as required. $\square$
A: I have a suggestion. Suppose that $S,T:\mathbb{F}^n\rightarrow\mathbb{F}^n$ are linear, and consider their tensor product $$S\otimes T:\mathbb{F}^n\otimes\mathbb{F}^n\rightarrow\mathbb{F}^n\otimes\mathbb{F}^n.$$ To prove your claim, it suffices to prove that the rank of $S\otimes T$ is a product of the ranks $S$ and $T$. 
Let $\{v_1,\ldots,v_k\}$ and $\{w_1,...,w_l\}$ be bases of the images of $S$ and $T$, respectively. Note that $\{v_i\otimes w_j\}$ is then a basis of the image of $S\otimes T$. There are $kl$ elements in this basis. Since $k$ and $l$ are the ranks of $S$ and $T$, respectively, this means that the rank (ie. dimension of the image) of $S\otimes T$ is the product of the ranks of $S$ and $T$. 
