Let $(e_1,\dots,e_n)$ be a basis of $U$. Then $\text{Im}(A)=\text{span}(Ae_1,\dots,Ae_n)$ (why?), so $\text{rank}(A)=\dim(\text{Im}(A))\le n$. I'll leave it to you to show that $\text{rank}(A)\le\dim(V)$.
I assume $A$ is applied first in $BA$. Obviously, $\text{Im}(BA)\subseteq\text{Im}(B)$, so
\[
\text{rank}(BA)=\dim(\text{Im}(BA))\le \dim(\text{Im}(B))=\text{rank}(B).
\]
In addition, $\text{Im}(BA)=B(\text{Im}(A))$, so using statement a),
\[
\text{rank}(BA)=\dim(\text{Im}(BA))\le \dim(\text{Im}(A))=\text{rank}(A).
\]
Thanks for reply. My work is Problem 1.pdf
at problem1, if you have time, please, check the next one and guide me more if I am not right. Thanks in advance. This is my practice review for final. I desperate need help.
I think problem 1 (a) and (b) are largely correct. If I could make a general advice, it would be to not hesitate to use more English text in complete sentences. It would make the role of the grader, or any reader, much easier.
I assume Col(A) means the column space of A. But in the beginning of problem 1, A and B are defined as linear operators, not matrices. You need to say something about a basis.
I believe you mean, "If $x\in\text{Col}(BA)$, then $x=BAy$ for some $y$, so $x=B(Ay)$, which means $y\in\text{Col}(B)$".Let x ∈ R^p $\to$ x ∈ Col BA, then BA(x) = B(Ax) ∈ Col B
Note the the solution in post #4 does not use the rank-nullity theorem.
Problem 6.
Is $a$ a fixed vector? This should be said explicitly. The right-hand side of the last equality is a number while it should be a vector, i.e., three numbers. The matrix of A is not Av, and it should not depend on v1, v2, v3 (the first matrix). I am not sure how you determine the matrix of $A$. In my book, its columns are images of the basis vectors. For example, (1, 0, 0) is mapped to $(0, a_3,-a_2)$, so that would be the first column. Overall, the matrix of $A$ isLet a =(a1,a2,a3) and v = (v1,v2,v3) then, a x v = a2v3 – a3v2 –a1v3+a3v1 +a1v2-a2v1
\[
\begin{pmatrix}
0&-a_3&a_2\\
a_3&0&-a_1\\
-a_2&a_1&0
\end{pmatrix}
\]