I have this question here which says the following.
Let $A$,$B$ be $3 \times 6$ matrices with the following properties.
$(i)$ For every b$\epsilon \mathbb{R}^3$, rank$(A)$ $=$ rank$([A|b])$
$(ii)$ dim$($Row$(B)$$)$$=2$.
Answer the following questions.
$(a)$ What is rank$(A)$?
My reasoning: I know that rank$(A)\leq \min(m,n)$ but I am not sure if that justifies in me saying that rank$(A)=3$. I want to say yes, but I am not certain.
$(b)$ What is nullity$(B)$?
My reasoning: rank$(B)+{}$nullity$(B)=$ # of columns.
Therefore, $2$+nullity$(B)$ $=6$
nullity$(B)$ $=6-2$
nullity$(B)$ $=4$
$(c)$ Is there a vector b$\epsilon \mathbb{R}^3$ such that $A$x=b is consistent?
My reasoning: Yes there is. Since rank$(A)$ $=$ rank$([A|b])$, there must be a consistent solution.
$(d)$ Is there a vector b$\epsilon \mathbb{R}^3$ such that $B$x=b is consistent?
My reasoning: Yes there is. Since rank$(B)$ $=2$, this means that rank$(B)$ $<6$ so the system is not only consistent, but has infinitely many solutions.
$(e)$ Is there a non-zero vector x $\epsilon \mathbb{R}^6$ such that Ax=$0$ and Bx=$0$ simultaneously?
I'm not sure about this one... Any guidance would be much appreciated!
Are my other answers reasonable? Thanks!
(a) The answer is correct but the argument is not quite there. From the given information we can say that every $b\in \mathbb{R}^3$ is in the column space of $\mathbb{R}^3$. Why we can conclude that the rank of $A$ is 3 from this fact?
(b) It is ok.
(c) The argument needs to be more precise.
(d) You need to connect the relationship between the rank and the consistency of system of equations.
(e) Use the fact that $\text{dim}(U+V)=\text{dim}(U)+\text{dim}(V)-\text{dim}(U\cap V)$.
For part (e) try this way. From the previous part we know that $\text{nullity}(A)=3$ and $\text{nullity}(B)=4$. Let $X=\{x_1,x_2,x_3\}$ and $Y=\{y_1,y_2,y_3,y_4\}$ be respectively be the basis of nullspace of $A$ and $b$.
We want to show that $\text{null}(A)\cap \text{null}{B}\neq \emptyset$. Assume otherwise and show that the assumption leads to the conclusion that $ X\cup Y$ is linearly independent, which is impossible (why?)