I solved the following question, and I think I am correct, but I am not sure.
Given are three matrices, A,B and C. And $AB = C$.
Now the question is to fill in a couple of things. I will put these in bold.
1) The rows of C are linear combinations of the rows of B.
I answered this because when you write it down, you can see this.
2)So, the rank of C is not bigger than the rank of B
Explanation. If A = mn, B=np C will be m*p. So the dimension of the column space can only be maximum p.
3) Because $B^TA^T=C^T$ the rank of C isn't bigger than the rank of A. This is due to the fact that the columns of C are linear combinations of the columns of A
I think this is because, when you have a matrix C of m*p, m can be smaller then p or m can be bigger than p, When p is smaller than m, the amount of columns of C limits the rank. When p is bigger than m, the amount of rows limits the dimension of the column space and the rank.
Are all these explanations true, or did I make som major mistakes? What can I say more ?
Thank you for your help,
Kind regards
Thomas
Write $A$ as a matrix of rows: $$ A=\begin{bmatrix}a_1\\a_2\\\vdots\\a_m\end{bmatrix} $$ Then, as matrix of rows, $$ AB=\begin{bmatrix}a_1B\\a_2B\\\vdots\\a_mB\end{bmatrix} $$ so you want to see that, for a row vector $y$, $yB$ is a linear combination of the rows of the rows of $B$; write $y=[y_1\ y_2\ \dots\ y_n]$, so, if $$ B=\begin{bmatrix}b_1\\b_2\\\vdots\\b_n\end{bmatrix} $$ we have $$ yB=y_1b_1+y_2b_2+\dots+y_nb_n $$ which is a linear combination of the rows of $B$.
This clearly implies that the row rank of $C$ is less than or equal to the row rank of $B$, because the row space of $C$ is a subspace of the row space of $B$.
Similarly, the columns of $C$ are linear combinations of the columns of $A$, because you can apply the same reasoning to $C^T=B^TA^T$ and the rows of $C^T$ are linear combinations of the rows of $A^T$.
Therefore the column rank of $C$ is less than or equal to the column rank of $A$, because the column space of $C$ is a subspace of the column space of $A$.