# Linearly independent columns of a matrix product

Given $\mathbf{A} = \mathbf{B}\mathbf{C}$, with $\mathbf{B} \in \mathbb{R}^{m\times n}$ and $\mathbf{C} \in \mathbb{R}^{n\times p}$. Say we know that the columns of $\mathbf{A}$ are linearly independent. Does this also imply that the columns of $\mathbf{B}$ and $\mathbf{C}$ are linearly independent? How can I go about proving this?

We know $\text{Ker}(\mathbf{BC}) = \left\{\mathbf{0}\right\}$, so $\mathbf{x} = \mathbf{0}$ is the only vector satisfying $$\mathbf{BCx} = \mathbf{0}$$

From this: $$\mathbf{BC}(\mathbf{0}) = \mathbf{B}\mathbf(0) = \mathbf{0}$$

but this only shows that $\mathbf{0}$ is in the null space of $\mathbf{B}$, not that it's the only member of the nullspace.

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do you have any idea about determinants? – Praphulla Koushik Oct 14 '13 at 4:51
@PraphullaKoushik wouldn't determinants only be helpful if these were square matrices? – user100704 Oct 14 '13 at 4:53
Oh, yes yes.. I did not realize given matrices are not square matrices – Praphulla Koushik Oct 14 '13 at 4:55

It is not necessarily true that the columns of $B$ are linearly independent. For example, $$\pmatrix{1 & 0\cr 0 & 1\cr} = \pmatrix{1 & 0 & 0\cr 0 & 1 & 0\cr} \pmatrix{1 & 0\cr 0 & 1\cr 0 & 0\cr}$$ On the other hand, it is true that the columns of $C$ are linearly independent, because ${\rm Ker}(C) \subseteq {\rm Ker}(BC)$.
Thank you. Can you elaborate on the last statement? I can kind of see how to prove it. Suppose $\mathbf{x} \in \text{Ker}(\mathbf{C})$. $\mathbf{Cx} = \mathbf{0}$ so we have $\mathbf{BCx} = \mathbf{0}$, so x will be in $\text{Ker}(\mathbf{BC})$. But how do we prove there isn't an element of $\Ker(\mathbf{C})$ that isn't in $\text{Ker}(\mathbf{BC})$? By simple contradiction? – user100704 Oct 14 '13 at 5:15
Huh? You just proved that: if $x \in {\rm Ker} C$, then $x \in {\rm Ker}(BC)$. – Robert Israel Oct 14 '13 at 5:33
The column vector of $C$ must be linear independent. You might try to show that if $A = BC$ are just mappings and $A$ is injective, then so is $C$. However, $B$ might not be injective.