Given a matrix $A \in M_{m\times n}$ , Suppose the rows of the matrix $A$ are linearly dependent.
Prove/ Disprove: If we delete some of the columns of the matrix $A$ then the rows of the matrix $A$ will still be linearly dependent.
$\ $
My Attempt:
as $dim C(A) = dim R(A) = rank(A)$
as $dim C(A) = dim{\left(sp \{c_1, c_2, \ldots, c_n \} \right)}$ where $\{c_1, c_2, \ldots, c_n \}$ is the vector space of $A$'s columns,
and $dim R(A) = dim{\left(sp \{r_1, r_2, \ldots, r_m \} \right)}$ where $\{r_1, r_2, \ldots, r_m \}$ is the vector space of $A$'s rows.
$A$ rows are linearly dependent, hence $$rank(A) < n$$
That means – $nullity(A) > 0$.
As Linear maps: $\mathbb{R}^n \to \mathbb{R}^m$ are equivalent to $m\times n$ matrices – then even if we delete some columns of the matrix $A$, then because $nullity(A) > 0$, deleting some columns won't make $nullity(A) =0$ – thus the matrix $A'$ (the matrix $A$ after deleting some columns) won't be a full rank matrix – and specifically there will be still be linearly dependent rows
Is that correct?
Best Answer
If the rows of an $m \times n$-matrix $A$ are linearly dependent, then (by definition) we can write: $$\sum_{i=1}^m \alpha_i\vec A_{i*} = \vec o_n \tag{$*$}$$ where not all coefficients $\alpha_i$ are zero; $\vec A_{i*}$ denotes the $i$-th row of $A$ (as a vector in $\mathbb{R}^{n}$) and $\vec o_n$ is the zero vector in $\mathbb{R}^{n}$. But this implies that for the $j$-th coordinate, we must have: $$\sum_{i=1}^m \alpha_i A_{ij} = 0 \tag{$\star$}$$ This holds for all coordinates of the rows, so for every $j$ with $1\le j \le n$.
If you delete $k$ columns ($k<n$), we still have $(\star)$ for all the remaining columns so $(*)$ still holds, but in $\mathbb{R}^{n-k}$ instead of $\mathbb{R}^{n}$.