If v and w are in a vector space S , every combination cv + dw must be in S (S is $R^2$)
A subspace of S must satisfy two requirements: ( If v and ware vectors in the subspace and c is an scalar)
(i) v + w is in the subspace
(ii) cv is in the subspace
In other words, the set of vectors is "closed" under addition v + w and multiplication cv (and dw)
Notice: Ever subspace contains the zero vector
example 1
Find all subspace in
look at the picture above,if I multiply the vector on the dash line by zero.Then I will get (0,0),which is not in the dash line obviously.
So the subspaces of
-
$R^2$ counts as a subspace of itself - any line through zero (0,0)
- zero point
When we are trying to solve
DEFINITION The column space consists of all linear combinations of the column .
The combinations are all possible vectors Ax. They fill the column space
To solve
example
Here is a 3 by 2 matrix A. The column space of all combinations of the two columns flls up a plane in
DEFINITION : The row space of a matrx is the subspace of
The row space of A is
The rows of an
example
Describe the row space of A. $$ A= \left[\begin{array}{cc} 1&4\ 2&7\ 3&5\ \end{array} \right] $$
The row space of A is spanned by the three rows of A.This row space
is all of
The nullspace
Pretend A is a
The m by n matrix A can be square or rectangular. The right hand side is
1、For invertible matrices this is the only solution.
2、For other matrices, not invertible, there are nonzero solutions to
Check that the solution to
If
Then
So
So
If the right-hand side b is not equal to 0, namely
NO
example
solve
$$
\left[\begin{array}{cc}
1&2&2&2\
2&4&6&8\
3&6&8&10\
\end{array} \right]
$$
We can apply elimination to the solve linear equations
column two
$\left[\begin{array}{cc}2\0\0\\end{array} \right]$ and column four $\left[\begin{array}{cc}2\4\0\\end{array} \right]$
are free columns .
For the fee variables
First
Second
So the
$\left[\begin{array}{cc} -2\ 1\ 0\ 0\ \end{array} \right]$ and $\left[\begin{array}{cc} 2\ 0\ -2\ 1\\end{array} \right]$
are the special solutions to
The null space
($c_1$ , $c_2$ are any real number)
question1:how many special solutions are there?
Each of free variables corresponds to a special solution
question2:How many free variables are there ?
for a matrix , that's M rows,N columns with rank R. Then it has (N-R) free variables
N columns mean N variables
The rank R is the number of pivot .
In this case, R is 2 , N is 4 , So there are 2 free variables.
I still take the matrix above for example
These steps bring us to the best matrix R
1. Produce zeros above te pivots. Use pivot rows to eliminate upward in R
2. Produce ones in the pivots. Divide the whole pivot row by its pivot.
The right-hand side R is the reduced row echelon form of A
The pivot columns of R contain identity matrix I ,so the number under and over the pivot must be zero,just like the third column in matrix R $\left[\begin{array}{cc} 0\ 1\ 0\ \end{array} \right]$
$$ Rx= \left[\begin{array}{cc} 1&2&0&-2\ 0&0&1&2\ 0&0&0&0\ \end{array} \right] \left[\begin{array}{cc} x_1\ x_2\ x_3\ x_4\ \end{array} \right] \xrightarrow{exchange \ col2 \ and \ col3 } \left[\begin{array}{cc} 1&0&2&-2\ 0&1&0&2\ 0&0&0&0\ \end{array} \right] \left[\begin{array}{cc} x_1\ x_3\ x_2\ x_4\ \end{array} \right] \ \Rightarrow{ } \left[\begin{array}{cc} I &F\ 0 & 0\ \end{array} \right] \left[\begin{array}{cc} x_1\ x_3\ \hline x_2\ x_4\ \end{array} \right] =0 \Rightarrow{ } I \left[\begin{array}{cc} x_1\ x_3\ \end{array} \right]
-F \left[\begin{array}{cc} x_2\ x_4\ \end{array} \right] \ \Rightarrow{ } \left[\begin{array}{cc} 1&0\ 0&1\ \end{array} \right] \left[\begin{array}{cc} x_1\ x_3\ \end{array} \right]
\left[\begin{array}{cc} -2&2\ 0&-2\ \end{array} \right] \left[\begin{array}{cc} x_2\ x_4\ \end{array} \right] $$
In the example above , I make $I=\left[\begin{array}{cc} 1&0\ 0&1\ \end{array} \right]$ and $F=\left[\begin{array}{cc} 2&-2\ 0&2\ \end{array} \right]$
From reduced system
It's easy to prove that
So the solutions for
($c_1$ , $c_2$ are any real number)
example 2
$$ A=\left[\begin{array}{cc} 1&2&3\ 2&4&6\ 2&6&8\ 2&8&10\ \end{array} \right] \xrightarrow[ ]{ } \begin{matrix} \underbrace{ \left[\begin{array}{cc} 1&2&3\ 0&2&2\ 0&0&0\ 0&0&0\ \end{array} \right] } \ U\end{matrix} \xrightarrow[ ]{ } \begin{matrix} \underbrace{ \left[\begin{array}{cc} 1&0&1\ 0&1&1\ 0&0&0\ 0&0&0\ \end{array} \right] } \ R\end{matrix} \xrightarrow[]{F=\left[\begin{array}{cc}1\1\\end{array}\right]\ \ I=\left[\begin{array}{cc}1&0\0&1\\end{array}\right]} \left[\begin{array}{cc} I&F\ 0&0\ \end{array} \right] \ \\ N= \left[\begin{array}{cc} -F\ I\ \end{array} \right] \ x = \left[\begin{array}{cc} x_1\ x_2\ \hline x_3\ \end{array} \right]
\left[\begin{array}{cc} -1\ -1\ 0 \ \end{array} \right] $$
The solution of
conclusion
There are two methods to solve Ax=0
1、converted the Ax=0 to Ux = 0 through elimination ,and then compute the special solution by back substitition
2、converted the Ax=0 to Rx = 0 through elimination ,and then find the special solution by RN=0
first example
$$ A= \left[\begin{array}{cc} 1&3&0&2\ 0&0&1&4\ 1&3&1&6\ \end{array} \right] \left[\begin{array}{cc} x_1\ x_2\ x_3\ x_4\ \end{array} \right]
\left[\begin{array}{cc} 1\ 6\ 7\ \end{array} \right] \Rightarrow \begin{matrix} \underbrace{ \left[\begin{array}{cccc|c} 1&3&0&2&\mathbf{1}\ 0&0&1&4&\mathbf{6}\ 1&3&1&6&\mathbf{7}\ \end{array} \right] } \ augment \ matrix \end{matrix}
\left[\begin{array}{c|c} A&\mathbf{b}\ \end{array} \right] $$
After several elimination operations,
\left[\begin{array}{cccc|l} 1&3&0&2&\mathbf{b_1}\ 0&0&1&4&\mathbf{b_2}\ 1&3&1&6&\mathbf{b_3}\ \end{array} \right] \xrightarrow[]{} \left[\begin{array}{cccc|l} 1&3&0&2&\mathbf{b_1}\ 0&0&1&4&\mathbf{b_2}\ 0&0&0&0&\mathbf{b_3-b_2-b_1}\ \end{array} \right] $$
Only and if
First set two free variables
second we find out the solutions for Ax=0 is that $x_{nullspace} = c_2 \left[\begin{array}{cc} -3\1\0\0\\end{array} \right]+c_4\left[\begin{array}{cc}-2\0\-4\1\\end{array} \right]$
So the complete solution is that:
The complete solutions of
example
$$
A=
\left[\begin{array}{rr}
1&1\
1&2\
-2&-3\
\end{array} \right]
and
b=
\left[\begin{array}{rr}
b_1\
b_2\
b_3\
\end{array} \right]
$$
If
Supposing the equation is solvable.
This example has no free variables since
A is a typical example to explain full column rank.A has full column rank
Every column has a pivot. The rank is r = n. The matrix is tall and thin (m
(Row reduction puts I at the top, when A is reduced to R with rank n:)
\left[\begin{array}{cc} {n\ by\ n\ identity\ matrix} \ {m-n\ rows\ of\ zero} \end{array} \right] $$
conclusion
Every matrix A with full column rank (r = n) has all these properties:
1. All columns of A are pivot columns.
2. There are no free varables or special solutions.
3. The nullspace N(A) contains only the zero vector x = 0.
4. If Ax = b has a solution (it mght not) then it has only one solution
example $Ax=b$
$$
\left[\begin{array}{rrrr}
1&1&1\
1&2&-1\
\end{array} \right]
\left[\begin{array}{rrrr}
x_1\
x_2\
x_3\
\end{array} \right]
\left[\begin{array}{rrrr} 3\ 4\ \end{array} \right] $$
$$ \left[\begin{array}{rrr|r} 1&1&1&3\ 1&2&-1&4\ \end{array} \right] \xrightarrow[]{} \left[\begin{array}{rrr|r} 1&1&1&3\ 0&1&-2&1\ \end{array} \right] \xrightarrow[]{} \left[\begin{array}{rrr|r} 1&0&3&2\ 0&1&-2&1\ \end{array} \right]
\left[\begin{array}{r|r} R&d\ \end{array} \right] $$
we set free variable
From R ,we know that $x_{nullspace} = c_2 \left[\begin{array}{cc} -3\2\1\\end{array} \right]$
Complete solution $$ x=x_p+x_n= \left[\begin{array}{cc} 2\ 1\ 0\ \end{array} \right] + c_2 \left[\begin{array}{cc} -3\ 2\ 1\ \end{array} \right] $$
A matrix has full row rank
Every matrix A with full row rank (r = m) has all these properties:
- All rows have pivots, and R has no zero rows.
-
$Ax = b$ has a solution for every right side b. - The column space is the whole space
$R^m$ . - There are
$n - r = n - m$ special solutions in the nullspace of A.
| r=m=n | r=m<n | r=n<m | r<m,r<n | |
|---|---|---|---|---|
| shape | Square | Short and wide | Tall and thin | Not full rnk |
| solutions | one solutions | ∞ solutions | 0 or 1 solution | 0 or ∞ solutions |
| reduced R | $\left[\begin{array}{cc} I\\end{array} \right]$ | $\left[\begin{array}{cc} I&F\\end{array} \right]$ | $\left[\begin{array}{cc} I\0\\end{array} \right]$ | $\left[\begin{array}{cc} I&F\0&0\\end{array} \right]$ |
| linear dependent | The columns are certainly independent | The columns are certainly dependent , there are |
The columns of matrix A with full column rank are independent.Because there are n pivots and no free variables. Only |
|
| remark | invertible | example 4.1 |
DEFINITION : The sequence of vectors
There is a direct way to check if a sequence of vectors is dependent. Another way to describe linear dependence is this: "One vector is a combination of the other vector." see the example below
- If three vectors are not in the same plane, they are independent. No combination of
$v_1, v_2, v_3$ in Picture 1 gives zero except$0v_1 + 0v_2 + 0v_3$ . - If three vectors
$w_1, w_2, w_3$ are in the same plane,The combination$w_1 - w_2 + w_3$ is$(0, 0, 0)$ they are dependent.
How to fnd that solution to
$$
A=\left[\begin{array}{cc} 1&0&3\ 2&1&5\ 1&0&3\ \end{array} \right] \xrightarrow[ ]{ } \begin{matrix} \underbrace{ \left[\begin{array}{cc} 1&0&3\ 0&1&-1\ 0&0&0\
\end{array} \right] } \ R\end{matrix}
$$
The solution
summarization
After a sequence of elimination , A is converted to R(reduced row echelon form) , R is a
| m=n | m<n | n<m |
|---|---|---|
| The columns are certainly independent | The columns are certainly dependent , there are |
The columns of matrix A with full column rank are independent.Because there are n pivots and no free variables. Only |
DEFINITION A set of vectors spans a space if their linear combinations flll the space.
Example 1 $v_1=\left[\begin{array}{cc}1\0\\end{array} \right]$ and $v_1=\left[\begin{array}{cc}0\1\\end{array} \right]$ span the full two-dimensional space
Example 2 $v_1=\left[\begin{array}{cc}1\0\\end{array} \right]$ , $v_2=\left[\begin{array}{cc}0\1\\end{array} \right]$ , $v_3=\left[\begin{array}{cc}4\7\\end{array} \right]$ also span the full space
Example 3 $w_1=\left[\begin{array}{cc}1\1\\end{array} \right]$ and $w_2=\left[\begin{array}{cc}-1\-1\\end{array} \right]$ only span a line in
think of three vectors coming out fom
Two vectors can't span all of
DEFINITION: A basis for a vector space is a sequence of vectors with two properties:The basis vectors are linearly independent and they span the space.
The pivot columns of A are a basis for its column space,but not the pivot rows of its echelon for R.
The pivot rows of A are a basis for its row space. So are the pivot rows of its echelon for R.
example
Find bases for the column and row spaces of matrix A
$$
A=\left[\begin{array}{rrrrr}
1&1&-2&0&-1\
1&2&0&-4&1\
0&1&3&-3&2\
2&3&0&-2&0\
\end{array} \right]
\xrightarrow[ ]{ }
\left[\begin{array}{rrrrr}
1&1&-2&0&-1\
0&1&2&-4&2\
0&0&1&1&0\
0&0&0&0&0\
\end{array} \right]
=R
$$
And in fact,these rows still span the same space that the initial four rows did.Because when you do elimination,all that you're doing is recombining your rows by doing linear combinations of them.
So the basis for the row space is
We can no longer use first three columns of R as the basis for column space of A.Because when I did elimination ,I changeed the columns space. The column spaces of A and R are different. Their bases are different.But By the upper R,we keno that the privots are in first ,second and third columns. So the
Question Given four vectors in
First answer Make them the rows of A, and eliminate to find the nonzero rows of R.
$$
A=\left[\begin{array}{rrrrr}
1&1&-2&0&-1\
1&2&0&-4&1\
0&1&3&-3&2\
2&3&0&-2&0\
\end{array} \right]
\xrightarrow[ ]{ }
\left[\begin{array}{rrrrr}
\mathbf{1} &1&-2&0&-1\
0&\mathbf{1}&2&-4&2\
0&0&\mathbf{1}&1&0\
0&0&0&0&0\
\end{array} \right]
=R
$$
Second answer Put the five vectors into the columns of A. Eliminate to fnd the pivot columns (of A not R). Those pivot columns are a basis for the column space.
All bases for a vector space contain the same number ovectors,the number is the "dimension" of the space
Let's recall the example. The basis for the row space of is