Finding bases for the null space and row space of a matrix

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Christian Lawson-Perfect

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Christian Lawson-Perfect 5 years, 11 months ago

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Name	Status	Author	Last Modified
Finding bases for the null space and row space of a matrix	Doesn't work	Newcastle University Mathematics and Statistics	20/11/2019 14:50
John's copy of Finding bases for the null space and row space of a matrix	draft	John Steele	13/05/2019 04:25

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Question statement

The row-reduced echelon form $R$ of the following matrix $A$

$A=\var{testmatrix}$ is given by $R=\var{echmatrix}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

echelonform	list	List of 2 items
ech	list	Nested 6×5 list
echform	list	[ 1, 4, 5 ]
testmatrix	matrix	Matrix of size 6×5
record	list	List of 4 items
rref1	list	Nested 6×5 list
record_ops_matrix	list	Nested 7×6×5 list
record_ops_message	list	List of 7 items
rank	integer	3
maxint	integer	4
steps	number	6
rdmat	list	Nested 6×6 list
deter	number	-1373
nullity	integer	2
echmatrix	matrix	Matrix of size 6×5
trtestmatrix	matrix	Matrix of size 5×6
nbasis_column	list	[ 2, 3 ]
coordcol	list	List of 2 items
freecols	list	List of 2 items
rows	integer	6
columns	integer	5
column_basis	set	Set of 3 items
sum	number	10
null_basis	list	Nested 2×5 list
ec	matrix	Matrix of size 6×5
chosen	list	List of 5 items
student_cols_matrix	matrix	Matrix of size 6×3
not_chosen_col	matrix	matrix([2,0,0,0,0,0])
othercols	list	[ 2, 3 ]
c0	list	[ 1, 0, 0 ]
c1	list	[ 2, 0, 0 ]
colmark	list	[ 1, 0, 0, 1, 1 ]
z	list	[ 0, 0, 0, 0, 0 ]

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
echelon_form_matrix(rows,rank,columns,maxint)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `list`

[ [ [ 1, 1, 2, 0, 0 ], [ 0, 0, 0, 1, 0 ], [ 0, 0, 0, 0, 1 ], [ 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0 ] ], [ 0, 3, 4 ] ]

← Depends on:

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add
Part b) Gap-fill
1. Gap Choose several from a list
  Add
Part c) Choose several from a list
Part d) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Number entry
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5. Gap Gap 4. Number entry
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6. Gap Gap 5. Number entry
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Part e) Matrix entry

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Rank of $A=\;$Gap 0.

Nullity of $A=\;$Gap 1.

Advice

a) The following shows how $A$ is reduced to row-echelon form.

{solution(record_ops_matrix,record_ops_message)}

The rank of A is the number of columns of R with pivots, i.e. 3.

The nullity of A is the number of columns of R without pivots, i.e. 2.

b) Given the row-echelon form we see that the first three rows of $R$ form a basis for the row space of $A$.

c) One basis for the column-space of $A$ is given by the original columns which in the row-reduced echelon form have a pivot.

For example, $\{c_i:i \in \var{set(echform)}\}$

$\var{column_basis}$

There are other bases which you could have chosen and would also have been marked as correct.

d) If you look at the columns $c_{\var{othercols[0]}}$ and $c_{\var{othercols[1]}}$ we can simply read off these columns as combinations of the other columns by looking at the numbers above the pivot element in each of the columns $c_{\var{othercols[0]}}$ and $c_{\var{othercols[1]}}$.

e) Basis for the null-space.

The dimension of the null-space is $2$ and so we should have $2$ vectors in the basis.

The null space is given by $\operatorname{Null}(A)=\operatorname{Null}(R)= $ \[\left\{(x_1,x_2,x_3,x_4,x_5):\simplify{{ec}*vector(x_1,x_2,x_3,x_4,x_5)=vector(0,0,0,0,0,0)}\right\}\]

This gives three equations:

\begin{align}\simplify{{ec[0][0]}x_1+{ec[0][1]}x_2+{ec[0][2]}x_3+{ec[0][3]}x_4+{ec[0][4]}x_5}&=0\\\simplify{{ec[1][0]}x_1+{ec[1][1]}x_2+{ec[1][2]}x_3+{ec[1][3]}x_4+{ec[1][4]}x_5}&=0\\\simplify{{ec[2][0]}x_1+{ec[2][1]}x_2+{ec[2][2]}x_3+{ec[2][3]}x_4+{ec[2][4]}x_5}&=0 \end{align}

The free columns (without a pivot element) are $c_{\var{nbasis_column[0]}},c_{\var{nbasis_column[1]}}$

We obtain $2$ vectors which are linearly independent by:

1) Setting: $x_{\var{nbasis_column[0]}}=1,x_{\var{nbasis_column[1]}}=0$ gives $\var{rowvector(null_basis[0])}$ as a basis element for the null-space.

2) Setting: $x_{\var{nbasis_column[0]}}=0,x_{\var{nbasis_column[1]}}=1$ gives $\var{rowvector(null_basis[1])}$ as another basis element for the null-space.

Hence $\{\var{rowvector(null_basis[0])},\var{rowvector(null_basis[1])}\}$

is a basis for the null-space.

Note that this is just one possible basis, and if you input another correct basis for the null-space then it will be marked as correct.

Alternatively.

Note that on using the equations above, the variables $x_{\var{echform[0]}},\;x_{\var{echform[1]}}$ and $x_{\var{echform[2]}}$ can be expressed in terms of the variables $x_{\var{nbasis_column[0]}},x_{\var{nbasis_column[1]}}$.

If we do so we find that:

$ (x_1,x_2,x_3,x_4,x_5) \in \operatorname{Null}(A) \iff$ \[(x_1,x_2,x_3,x_4,x_5)= x_{\var{nbasis_column[0]}}\var{rowvector(null_basis[0])}+x_{\var{nbasis_column[1]}}\var{rowvector(null_basis[1])}.\]

As $ x_{\var{nbasis_column[0]}},\;x_{\var{nbasis_column[1]}}$ can take any values we see that the vectors

$\{\var{rowvector(null_basis[0])},\var{rowvector(null_basis[1])}\}$ form a basis as they span $\operatorname{Null}(A)$ and there are $2$ vectors which agrees with the nullity.

Give a worked solution to the whole question.

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Finding bases for the null space and row space of a matrix

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Christian Lawson-Perfect 5 years, 11 months ago

Christian Lawson-Perfect 5 years, 11 months ago

Newcastle University Mathematics and Statistics 9 years, 3 months ago

Error in variable testing condition

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Generated value: `list`

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Gap-fill

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Error

Finding bases for the null space and row space of a matrix

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 5 years, 11 months ago

Christian Lawson-Perfect 5 years, 11 months ago

Newcastle University Mathematics and Statistics 9 years, 3 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

Generated value: `list`