The Nullspace of A
The nullspace $N(A)$ consists of all solutions to $Ax=0$. These vectors $x$ are in $R^n$.
The solution vectors $x$ have $n$ components. They are vectors in $R^n$, so the nullspace is a subspace of $R^n$. The $C(A)$ is a subspace of $R^m$.
The $N(A)$ contains all the combinations of the special solutions to $Ax=0$.
Example $x+2y+3z=0$ comes from the 1 by 3 matrix $A=$
$$\begin{bmatrix}1&2&3\\end{bmatrix}$$
. Then $Ax=0$ produces a plane. All vectors on the plane are perpendicular to $(1, 2, 3)$.The plane is the nullspace of $A$. Set two free variables $y$ and $z$ to $0$ and $1$.
$$\begin{bmatrix}1&2&3\\end{bmatrix}$$
$$\begin{bmatrix}1\2\3\\end{bmatrix}$$
$$=0$$
has two special solutions $s_1$=
$$\begin{bmatrix}-2\1\0\\end{bmatrix}$$
and $s_2$=
$$\begin{bmatrix}-3\0\1\\end{bmatrix}$$
.
Those vectors $s_1$ and $s_2$ lie on the plane $x+2y+3z=0$. All vectors on the plane are combinations of $s_1$ and $s_2$.
We notice the last two components are free, because they were assigned $0$ or $1$.
The two key steps of this section:
- Reducing A to its row echelon form $R$.
- Finding the special solutons to $Ax=0$
Pivot Columns and Free Columns
Let’s continue to find something to make this system more stable. $C=$
$$\begin{bmatrix}1&2&2&4\3&8&6&16\\end{bmatrix}$$
becomes $U=$
$$\begin{bmatrix}1&2&2&4\0&2&0&4\\end{bmatrix}$$
. In $U$, columns 1 and 2 are pivot columns and columns 3 and 4 are free columns.
From this view, we can find elimination doesn’t change solutions — This is also the nullspace of $U$. To think about this, I will make a consideration of equations, because the value at the right side is always 0, so the elimination will not affect the solutions of the nullspace.
The Reduced Row Echelon Form R
$U(A)$ is a very easy way to get the special solutions to $Us=0$, however, not the easiest way. Now let’s see the Reduced Row Echelon Form $R$ ($rref(A)$).
How to make a $rref(A)$?
- Produce zeros above the pivots. Use pivot rows to eliminate upward in $R$.
- Produce ones in the pivots. Divide the whole pivot row by its pivot.
It’s same with $U(A)$, $N(A)$ also does not change the zero vector on the right side, another word, $N(A)=N(U)=N(R)$. This nullspace becomes the easiest to see when we reach the reduced row echelon form $R=rref(A)$. And through the operations of $N(R)$, we can easy get a very beautiful attribution, The pivot columns of R contain $I$.
Have an example:
$U=$
$$\begin{bmatrix}1&2&2&4\0&2&0&4\\end{bmatrix}$$
becomes $R=$
$$\begin{bmatrix}1&0&2&0\0&1&0&2\\end{bmatrix}$$
.
To get the special solutions, we can very easily value the free components (columns) $1$ or $0$ to get $s_1=(-2,0,1,0)$ and $s_2=(0, -2, 0, 1)$. Thanks to the rref operations, we can get the value at the pivot components very easily, just assign it a ‘minus’ to the free components which were given the $1$ in the nullspace. There is an example to help understand this process:
Pivot Varibles and Free Variables in the Echelon Matrix
$p$ means pivot columns while $f$ means free columns
$A=$
$$\begin{bmatrix}p&p&f&p&f\|&|&|&|&|\|&|&|&|&|\|&|&|&|&|\\end{bmatrix}$$
$R=$
$$\begin{bmatrix}1&0&a&0&c\0&1&b&0&d\0&0&0&1&e\0&0&0&0&0\\end{bmatrix}$$
$s_1=$
$$\begin{bmatrix}-a\-b\1\0\0\\end{bmatrix}$$
$s_2=$
$$\begin{bmatrix}-c\-d\0\-e\1\\end{bmatrix}$$
.
Now let’s talk something more interesting. About the dimension.
See the $N(R)$ below, think about what are the column space and the nullspace for this matrix $R$?
$R=$
$$\begin{bmatrix}1&0&x&x&x&0&x\0&1&x&x&x&0&x\0&0&0&0&0&1&x\0&0&0&0&0&0&0\\end{bmatrix}$$
The columns of $R$ have four components (rows) so they lie in $R^4$. The column space $C(R)$ consists of all vectors of the form $(b_1, b_2, b_3, 0)$. Counting the pivots leads to an extremely important theorem. Suppose $A$ has more columns than rows. With $n>m$ there is at least one free variable. The system $Ax=0$ has at least one special solution and this solution is not zero!
For a Short Wide matrix ($n>m$), in its nullspce always has nonzero vectors. And there must be at least $n-m$ free variables, since the number of pivots cannot exceed $m$. Here is the point: When there is a free variable, it can be set to $1$. Then the equation $Ax=0$ has at least a line of nonzero solutions.
About the dimension of nullspace
Nullspace is a subspace. Its dimension is the number of free variables.
The Rank of a Matrix
Sometimes we need to get the true size of a linear system, some rows may be a combination of other rows and becomes zero in the $U$ or the $rref$, and we don’’t want to count them. So there comes the Rank, the true size of $A$.
Here is an example:
$A=$
$$\begin{bmatrix}1&1&2&4\1&2&2&5\1&3&2&6\\end{bmatrix}$$
and $R=$
$$\begin{bmatrix}1&0&2&3\0&1&0&1\0&0&0&0\\end{bmatrix}$$
The first two columns of $A$ are $(1, 1, 1)$ and $(2, 2, 2)$, going in different directions, which will be pivot columns. What we need to be attention, column 3 and column4 is free column,
$Column 3 =2(column 1) + 0(column 2)$ —-> $s_1=(-2, -0, 1, 0)$
$Column 4 =3(column 1) + 1(column 2)$ —-> $s_2=(-3, -1, 0, 1)$
Actually, Every free column is a combination of earlier pivot columns. It is the special solutions $s$ that tell us those combinations.
Rank One
Matrices of rank one have only one pivot. The column space of a rank one matrix is ‘one-dimensional’.
For example,
$A=$
$$\begin{bmatrix}1&3&10\2&6&20\3&9&30\\end{bmatrix}$$
—->
$R=$
$$\begin{bmatrix}1&3&10\0&0&0\0&0&0\\end{bmatrix}$$
Here all columns are on the line through $u= (1, 2, 3)$. The columns of $A$ is $u, 3u, 10u$. Put those numbers into the row $v^T=$
$$\begin{bmatrix}1&3&10\\end{bmatrix}$$
and you have the special rank one form $A = uv^T$
$A = column times row = uv^T$
$$\begin{bmatrix}1&3&10\2&6&20\3&9&30\\end{bmatrix}$$
$=$
$$\begin{bmatrix}1\2\3\\end{bmatrix}$$
$$\begin{bmatrix}1&3&10\\end{bmatrix}$$
With rank one, $Ax=0$ is easy to understand. That equation $u(v^Tx)=0$ leads us to $u^Tx = 0$. All vectors $x$ in the nullspace must be othogonal to $v$ in the row space. This is the geometry when $r=1$: row space = line, nullspace = perpendicular plane.
$A$ and $U$ and $R$ also have r independent columns (pivot columns).
The rank r is the ‘dimension‘ of the column space. It is also the dimension of the row space. There is a very great thing: ##
The Dimension of the nullspace = $n-r$ ##
Very important!