$C$ is a $3\times5$ matrix and $\{s,t,u,v,w\}$ is a linearly independent set of vectors in $\Bbb R^5$. $Cu=0,Cv=0$, and $Cw=0$. Answer the following

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What are the possible values of the nullity of $C$? I said the nullity must be at least $3$, and the rank must be at most $2$ because the null space has at least a dimension of $3$.

What are the possible values for the rank of $C$? See above.

Is $Cx = b$ consistent for every choice of $b$ in $\Bbb R^3$? I want to say no because I'm not given any vectors in $C$ so I can't conclude whether or not it is consistent.

Suppose in addition that $\{Cs,Ct\}$ is a linearly independent set of vectors. What would your answer to the first and second questions be now?

Nullity must be $5$, therefore the rank must be $0$?

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You are on the right track, but you missed a few points.

(1) Note that the question is after all the possible values of $C$'s nullity, at least 3 is a bit inconcrete. For example the nullity cannot be 7 (as $C \in \def\M{\operatorname{Mat}}\M_{3,5}(\def\R{\mathbf R}\R)$, and hence $\def\nul{\operatorname{nullity}}0 \le \nul C \le 5$, together with your remark we have that $$ \nul C \in \{3,4,5\} $$

(2) Here you are correct, we must have $\def\r{\operatorname{rank}}\r C \in \{0,1,2\}$.

(3) Here you missed the track, note that we do not have that $Cs = Ct = 0$, but that $\{Cs, Ct\}$ is linear independent. Hence, the image of $C$ is at least two-dimensional (as it contains $Cs$ and $Ct$). This gives $\r C \ge 2$ and (1) and (2) imply $$ \nul C =3, \r C = 2$$