If $A^2B^2 = ABAB$, for square matrices A,B having same size, then prove that $AB = BA$. I actually know it's false if A and B are not invertible.
But how can i prove that not by using counterexamples
You can prove that by considering the null-space and value space of matrices. Recall that we know that if a matrix is non-invertible then the dimension of the null-space is greater than $0$ and the dimension of the value space is less than the row count. Also we know that $\dim N(M) + \dim V(M)$ is the actual row count of $M$.

What we use that given two subspaces $U$ and $W$ with $\dim U + \dim W$ being the row count then there exist a matrix $M$ that has $U = N(M)$ and $W = V(M)$. This can be seen as the ortogonal complement to $U$ has the same dimension as $W$ and therefore it exists a linear bijection between $U^\perp\to W$ and then we can define a mapping of the whole space by projecting the argument on $U^\perp$.

Now we know that if there's at least two dimensions there therefore exists $A$ and $B$ such that $N(A) = V(B)$ and $N(B) \perp V(A)$, and $\dim V(A), \dim V(B)&gt; 0$. This means that $B$ would map anything to something that $A$ maps to $0$ which means that $AB$ always maps to zero which means that $AB=0$. On the other hand $V(A)$ contains vectors not in $N(B)$ which means that $A$ will map some vectors to vectors not in $N(B)$ which means that $B$ will not map those vectors to $0$, that is there exists vectors $x$ such that $BAx\ne 0$ which means $BA\ne 0$. This clearly means that $AB-BA\ne 0$. But sice $AB=0$ then so is $A^2B^2 = A (AB) B$ and also $ABAB=(AB)(AB)$.

We have concluded that there exists $A$ and $B$ such that $AB\ne BA$ yet $A^2B^2 = ABAB$. 



Note that although this may seem to be a proof without using counter-example it's probably not that free of counter-example. For example showing that there exists linear bijection between spaces of the same dimension is probably shown by actually constructing such a mapping. Also see that we construct the mapping with prescribed zero and value space from this through direct construction from the linear bijection to the value space. Also that there are subspaces $N(A), V(A), N(B), V(B)$ with the desired properties is also probably done by direct construction.

Actually normally when we want to show the existence of something one often relies on concrete examples, otherwise one constructs an example abstractly starting from something that is known to exist. Only occationally one proves the existence non-constructively. And disproving a general assertion means that we're to prove that there exists an object that falsifies the assertion.
It is indeed false. For a counterexample, consider:
$$
A = \begin{bmatrix}
0 &amp; 1 \\
0 &amp; 0
\end{bmatrix}
\qquad\text{and}\qquad
B = \begin{bmatrix}
1 &amp; 0 \\
0 &amp; 0
\end{bmatrix}
$$
Note that $BA = A$, and:
$$
A^2 = AB = \textbf{0} = \begin{bmatrix}
0 &amp; 0 \\
0 &amp; 0
\end{bmatrix}
$$
so we have that $A^2B^2 = \textbf{0}(B^2) = \textbf{0} = \textbf{0}(AB) = ABAB$, yet $AB = \textbf{0} \neq A = BA$.