I'm trying to find the column space of $\begin{bmatrix}a&0\\b&0\\c&0\end{bmatrix}$, which I think is $span\left(~\begin{bmatrix}a\\b\\c\end{bmatrix}~\begin{bmatrix}0\\0\\0\end{bmatrix}~\right)$. Since by definition a span needs to include the null vector, this is redundant. Would it also be correct to say that this is redundant because $\begin{bmatrix}0\\0\\0\end{bmatrix}$ is linearly dependent on $\begin{bmatrix}a\\b\\c\end{bmatrix}$ (can multiply it by a scalar of 0 to reach the null vector)? Or is that a misapplication of the concept of linear dependence?
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$\begingroup$If $\{\vec v_1,\vec v_2, \cdots, \vec v_n\}$ are linearly independent $c_1 \vec v_1+c_2 \vec v_2+\cdots+c_n \vec v_n= \vec 0$ iff $c_1=c_2=\cdots=c_n=0$. Considering $\vec v_n=\vec 0$, we can get $c_1 \vec v_1+c_2 \vec v_2+\cdots+c_n \vec v_n= \vec 0$ by setting $c_1=c_2=\cdots=c_{n-1}=0$ and taking any $c_n \neq 0$. So by definition, any set of vectors that contain the zero vector is linearly dependent.
$\endgroup$ $\begingroup$It is exactly as you say: in any vector space, the null vector belongs to the span of any vector.
$\endgroup$ 7 $\begingroup$If $S=\{v : v=(0,0)\}$ we will show that its linearly dependent. let suppose that S is Linearly independent such that $cv=0$.$c$ may not be a zero so $S$ is linearly dependent.
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