Prove: if the rows of $A$ are linearly dependent, then if we delete some of $A$ columns - $A$ rows will still be linearly dependent,

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?

• You're using the letter $A$ to mean two totally different things! You need to edit this, maybe saying the matrix is $A$ and using "$N$" for the other As. Dec 17 '18 at 14:51
• @DavidC.Ullrich I don't understand what do you refer by "the other as" Dec 17 '18 at 14:55
• I think it's just a weird english construction that's making it hard to read. "Suppose $A$ rows" I think means "Suppose the rows of $A$". However, I don't understand the proof, and in particular it is not true that deleting columns does not reduce the nullity. Dec 17 '18 at 14:58
• If the rows of $A$ are linearly dependent, then $\text{rank}(A)<m$ Dec 17 '18 at 15:00
• @Callus - fixed it Dec 17 '18 at 15:05

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.
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 ($$k), we still have $$(\star)$$ for all the remaining columns so $$(*)$$ still holds, but in $$\mathbb{R}^{n-k}$$ instead of $$\mathbb{R}^{n}$$.