Column Space and Nullspace

May 23, 2019

Union and Intersection of Subspaces

Let’s have abitrary two vector subspaces $S$ and $T$. The subspace $S\cup T$ is then the space that contains all vectors $v\in S$ and $w\in T$. But $S\cup T$ is not a vector space. Imagine $S$ and $T$ as two lines in $\mathbb R^2$ that go through origin, say $x-axis$ and $y-axis$. Then $v=\left[\begin{matrix} 1\\ 0 \end{matrix}\right], w=\left[\begin{matrix} 0\\ 1 \end{matrix}\right]$, but $v+w=\left[\begin{matrix} 1\\ 1 \end{matrix}\right]$ is neither in $S$ nor $T$.

But $S\cap T$ is a vector space. $S\cap T = \min(S,T,S\cap T)$, the new space will be always smaller than the old ones, therefore the new space will always be vector spaces. If $S=x-axis,T=y-axis$, $S\cap T=\mathbf{0}$, zero vector is a vector space. This is true for whatever vector subspaces. More formally, let $t,u\in S\cap T$, as $t,u\in S\cap T$, $t,u\in S$ and $t,u\in T$, as $S$ and $T$ are both vector spaces, we can perform any linear combinations on $t,u$, and thus $S\cap T$ is a vector space because $t,u$ are abitrary vectors in $S\cap T$

Column Spaces Continued

Let $A=\left[\begin{matrix} 1 & 1 & 2\\ 2 & 1 & 3\\ 3 & 1 & 4\\ 4 & 1 & 5 \end{matrix}\right]$ and we want to solve the system of equations $A\mathbf{x}=\mathbf{b}$. For what $\mathbf b$ there’s a solution to this system? There’s not always an answer to a system of equations, like in two-equation system, there’s no answer to $2x+2y=0, 2x+2y=1$. To solve $A\mathbf{x}=\mathbf{b}$, $\mathbf b$ must be a linear combination of the columns of $A$. Remember a vector/matrix on the right of $A$ is manipulating its columns. Namely, $\mathbf b$ must be in $A$’s column space(by definition of column space).

A preview of independence: here the third column of $A$ is equal to the sum of the first one and the second one. We say this column is dependent, and this column will not play a role in contrusting the column space of $A$(you can image this column lies on the plane spanned by $\left[\begin{matrix} 1\\ 2\\ 3\\ 4\\ \end{matrix}\right]$ and $\left[\begin{matrix} 1\\ 1\\ 1\\ 1 \end{matrix}\right]$). And thus the $C(A)$ is plane in $\mathbb R^4$.

Nullspace

The definition of nullspace of a matrix $A$ is the collection of all vectors $\mathbf x$ s.t. $A\mathbf x=\mathbf 0$. Let $A$ is same as above, the equation will be:

a simple case of $\mathbb x$ is just make it all zeros. But here it can be more than just zeros. We can also solve the equation with $\mathbb x= \left[\begin{matrix} 1\\ 1\\ -1 \end{matrix}\right]$ and $\mathbb x=\left[\begin{matrix} 2\\ 2\\ -2 \end{matrix}\right]$, in fact, $c\left[\begin{matrix} 1\\ 1\\ -1 \end{matrix}\right]$ will work. And this is the nullspace of $A$, a line in $\mathbb R^3$. We often write $N(A)$ as the nullspace of $A$.

Note that nullspace’s dimension is dependent on the row size of $A$. $A$ is $4\times 3$ here. But the column space is related to column size which is 4.

Other values of b

Why we call the solution to $A\mathbf x=\mathbf b$ when $\mathbf b=\mathbf 0$ as a space? Let’s make $\mathbf b=\left[\begin{matrix} 1\\ 2\\ 3\\ 4\\ \end{matrix}\right]$. A solution will be $\mathbf x=\left[\begin{matrix} 1\\ 0\\ 0 \end{matrix}\right]$, and we can also do $\mathbf x=\left[\begin{matrix} 0\\ -1\\ 1 \end{matrix}\right]$ but we cannot do all zeros. The point we cannot make $\mathbf x$ all zeros make these solutions not a space. The solutions to this $\mathbf b$ is a plane that does not go through origin in $\mathbb R^3$.