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?)

  • $\begingroup$ Okay so I fixed almost all of it. For part d, I know the rank is 2, which means after row reduction, there are two leading ones. Since it is a 3x6 matrix, that means there is 1 free variable and the system is consistent and there are infinitely many solutions. Is that right? I still can't get part e. I've never seen that theorem before. $\endgroup$ – Future Math person Dec 12 '18 at 1:17
  • $\begingroup$ Ah ok if you want to argue that way you need to say that you pick $b=0$. Since $B$ has rank 2, then $Bx=0$ has infinitely many solutions. Let me think if there is another way to solve it without that theorem. $\endgroup$ – user9077 Dec 12 '18 at 1:21
  • $\begingroup$ I add something for part (e) above. $\endgroup$ – user9077 Dec 12 '18 at 1:30
  • $\begingroup$ I get what you mostly did I am not sure what to do for the contradiction portion. When I took linear algebra 6 years ago, I don't think I ever covered the intersection of dimensions in any way. Can you elaborate on why that's needed? $\endgroup$ – Future Math person Dec 12 '18 at 2:49
  • $\begingroup$ $X\cup Y$ has 7 elements. So they must be linearly dependent. Hence there are $\alpha_i,\beta_i\in \mathbb{R}$ not all zero such that $$\alpha_1x_1+\cdots +\alpha_3x_3+\beta_1y_1+\cdots \beta_3y_3=0.$$ So $$\alpha_1x_1+\cdots+\alpha_3x_3=-(\beta_1y_1+\cdots+\beta_4y_4)\in \text{null}(A)\cap \text{null}(B)$$ $\endgroup$ – user9077 Dec 12 '18 at 2:53

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