Linear algebra: kernel, span, dimension? [closed]

Question

Let $A=\left( \begin{array}{cc}0 & 1\\ \end{array}\right)$ and $B=\left( \begin{array}{cc}1 & 0\\ \end{array}\right)$ be two $1\times 2$ matrices.

What is the dimension of $\operatorname{span}\ker(A) \cap \operatorname{span}\ker(B)$?

Answer 1

The kernel of $A$ is spanned by the vector $(1,0)$ while the kernel of $B$ is spanned by the vector $(0,1)$. So $\ker A$ is $1$-dimensional as well as $\ker B$.

Since $span\{(1,0)\}\cap span \{(0,1)\}=(0,0)$, the dimension of the intersection of the two subspaces is $0$.

Edit: let me explain how one computes the kernel for each $A$ and $B$. You want to find all column vectors $$ v = \left( \begin{array}{c} x \\ y \\ \end{array}\right) $$ so that $Av=0$ (this will give you the null space of $A$) and $Bv=0$ (this will give you the null space for $B$). Computing $Av$, we get $$ Av = \left( \begin{array}{cc} 0 & 1\end{array}\right) \left( \begin{array}{c} x \\ y \\ \end{array}\right) = 0 x + 1 y = y. $$ Since we need to set this equal to zero, we get $y=0$. So the set of all those vectors satisfying $Av=0$ are of the form $(x,0)$ where $x$ varies over the real numbers. Thus $$ \ker (A)= \{ (x,0): x\in \mathbb{R}\} = \{ x(1,0): x\in \mathbb{R}\} = span\{ (1,0)\}. $$

The same analysis works for $B$.

Answer 2

If $Av=0$ and $Bv=0$ then $\binom BAv=0$, where $\binom BA$ is the $2\times 2$ matrix with first line $B$ and second line $A$. But then $\binom BA=E$ the identity matrix. So $\binom BAv=Ev=v=0$.