Linear Algebra Lecture 7

Linear Algebra Lecture 7

1. Computing the nullspace

2. Pivot variables and free variables

3. Special solutions

4. rref(A) = R

---

Example 1

Compution the nullspace for a matrix A, find all the solutions to $Ax=0$

$A = \begin{bmatrix} 1&2&2&2 \\ 2& 4&6&8\\3&6&8&10 \end{bmatrix}$

Eliminations
While doing elimination, will not changing the null space, solving $Ax = 0$ by elimination.

$A = \begin{bmatrix} 1&2&2&2 \\ 2& 4&6&8\\3&6&8&10 \end{bmatrix} \to \begin{bmatrix} 1&2&2&2 \\ 0& 0&2&4\\0&0&2&4 \end{bmatrix} \to\begin{bmatrix} 1&2&2&2 \\ 0& 0&2&4\\0&0&0&0 \end{bmatrix}$

Eliminate the matrix into echelon form, we can find two pivot.
The number of pivots is the rank of the matrix.

Pivot variables and free variables

The critical step is separating out the pivot variables(主变量). The columns with pivots are called pivot columns and the other columns are called free columns.

Pivot columns are column 1 and 3, free columns are column 2 and 4.

For these free columns can assign any number freely to those variables, $x_2$ and $x_4$. And then can solve the equations for $x_1$ and $x_3$.

$\begin{cases}x_1+2x_2+2x_3+2x_4=0\\2x_3+4x_4=0\end{cases}$

Special solutions

Choose 1 and 0 for free variables, get the special solutions.

$x=c_1\begin{bmatrix} -2\\1\\0\\0 \end{bmatrix} + c_2\begin{bmatrix} 2\\0\\-2\\1 \end{bmatrix}$

These are the two special solutions, the special numbers I gave to the free variables, the values 0 and 1 for the free variables.
Once have special solutions we could take any multiple of that and it’s in the null space.

The null space contains exactly all the combinations of the special solutions. There’s one special solutions for every free variable.

In this example, the rank equals the number of pivot numbers $r=2$, number of free variables equals $n-r=4-2$.

---

Reduced row echelon form

In reduced row echelon form, has zeros above and below the pivots.

$\begin{bmatrix} 1&2&2&2 \\ 0& 0&2&4\\0&0&0&0 \end{bmatrix} \to \begin{bmatrix} 1&2&0&-2 \\ 0& 0&2&4\\0&0&0&0 \end{bmatrix}$

Make all the pivot to 1.

$\begin{bmatrix} 1&2&0&-2 \\ 0& 0&2&4\\0&0&0&0 \end{bmatrix} \to \begin{bmatrix} 1&2&0&-2 \\ 0& 0&1&2\\0&0&0&0 \end{bmatrix}$

$R=rref(A)$ reduced row echelon form in Matlab.

RREF got identity in pivot rows and columns.

$\begin{bmatrix} 1&0&2&-2 \\ 0& 1&0&2\\0&0&0&0 \end{bmatrix} \to R=\begin{bmatrix} I & F \\ 0&0 \end{bmatrix}$

When we calculate $Rx=0$

$\begin{bmatrix} I&F\end{bmatrix}\begin{bmatrix} x_{pivot} \\ x_{free}\end{bmatrix} = 0 \to x_{pivot} = - Fx_{free}$

We choose identity for free variables, so the pivot variables are $-F$.

---

Example 2

$A = \begin{bmatrix} 1&2&3 \\ 2& 4&6\\2&6&8 \\ 2 &8&10 \end{bmatrix} \to \begin{bmatrix} 1&2&3 \\ 0& 0&0\\0&2&2 \\ 0&4&4 \end{bmatrix} \to \begin{bmatrix} 1&2&3 \\ 0& 2&2\\0&0&0 \\ 0&0&0 \end{bmatrix} \to \begin{bmatrix} 1&0&1 \\ 0& 1&1\\0&0&0 \\ 0&0&0 \end{bmatrix} \to R$

$x=c\begin{bmatrix} -1\\-1\\1 \end{bmatrix} =\begin{bmatrix}-F\\I \end{bmatrix} = N(A)$