# If $Col(A)$ and $Col(B)$ are linearly independent, then $AB = 0$

What does $Col(A)$ and $Col(B)$ are linearly independent mean?

It is from an old linear algebra book and I don't see this used very often, does it mean a basis of $Col(A)$ and a basis of $Col(B)$ are linearly independent?

Assume the above definiton, then it means $Col(A)$ and $Col(B)$ are linearly independent iff $Col(A) \cap Col(B) = \{0\}$.

Just from this how could I show $AB = 0$?

Edit: I guess we need $A$ and $B$ to be symmetric, from the counter example $\begin{bmatrix} 1 & 2 \\ 1 &2 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 &1 \end{bmatrix}$.

• It probably means that the columns of A (resp. B) form a linearly set of vectors.
– Ned
Apr 1 '17 at 18:25
• @Ned you sure mean a linearly independent set of vectors? Apr 1 '17 at 18:33
• yes, thanks, forgot the word independent
– Ned
Apr 1 '17 at 18:38

First, an example: \begin{align} \mathbf{A} \mathbf{B} &= \mathbf{0} \\[5pt] \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)\, \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) &= \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \end{align} Certainly $\mathbf{A} \mathbf{B} = 0$.
Look at the matrix-vector multiplication $$\mathbf{A} \mathbf{B} x \overset{?}{=} \mathbf{0}$$ Define $$y = \mathbf{B} x$$ When does $$\mathbf{A} y = \mathbf{0}?$$ When the vector $y$ is not in the column space of $\mathbf{A}$. Yet the vector $y$ is in the column space of $\mathbf{B}$ by construction: $$y = \mathbf{B} x = x_{1} \mathbf{B}_{1} + x_{2} \mathbf{B}_{2} + \dots$$ It is assembled using the column vectors of $\mathbf{B}$.
The proof demonstrates that if the matrix product is $\mathbf{0}$, the multiplicand matrices must be linearly independent.
As noted by @Nan Li, it does not show the reverse direction, that the product of regular matrices is necessarily $\mathbf{0}$.
• What if $A=\begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix}$ and $B=\begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}$. Then $\mathcal C(A)=\text{span}\{\begin{pmatrix}1\\1 \end{pmatrix}\}$ and $\mathcal C(B)=\text{span}\{\begin{pmatrix}0\\1 \end{pmatrix}\}$, they are independent. However, $AB \neq 0$