Find the nullity and rank of a 3x5 matrix C where three columns = 0? Tricky problem. So I have a 3x5 matrix C, where {s,t,u,v,w} (I'm assuming those are its columns) is a linearly independent set of vectors in R^5, and that Cu=0, Cv=0, Cw=0.
What is the rank and nullity of C?
I'm guessing with the latter part of the question that Cu, Cv, and Cw are linearly independent in R^3, meaning the other two (Cs and Ct) are free columns, making the nullity equal to 2 and thus making the rank equal to 3. This is a wild guess, so it's probably wrong.
 A: *

*The elements $\{s,t,u,v,w\}$ are not the columns of $C$, they are just linearly independent vectors in $\mathbb{R}^5$.

*The nullity of $C$ is the dimension of its nullspace, which is the subspace of $\mathbb{R}^5$ consisting of vectors $x$ satisfying $Cx=0$. You already have three linearly independent vectors in the nullspace of $C$, so the nullity is at least $3$.

*I am not sure you can say any more. If we let your five vectors be the standard basis vectors for $\mathbb{R}^5$, then both the matrices $$C=\begin{bmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{bmatrix}$$ and $$C=\begin{bmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&0&0&0\end{bmatrix}$$ send $(0,0,1,0,0)^\top$, $(0,0,0,1,0)^\top$, and $(0,0,0,0,1)^\top$ to zero, but they have different ranks and nullities.

A: The homogeneous system $CX$=$0$ has at least 3 linearly independent solutions viz. $u,v,w$ (which you wrongly took as its columns) thereby giving Nullity(C)  is at least 3. Using Rank(C)+Nullity(C)=5, we get Rank(C) is at most 2.
