Register to reply

Null space and dimension

by ak123456
Tags: dimension, null, space
Share this thread:
ak123456
#1
Mar9-09, 06:03 PM
P: 51
1. The problem statement, all variables and given/known data
find a basis of the null space N(A) in R^5 of the matrix
A =
1 -2 2 3 -1
-3 6 -1 1 -7
2 -4 5 8 -4

in M3*5 (R) and hence determine the dimension


2. Relevant equations



3. The attempt at a solution
i found that
A=
1 -2 2 3 1
0 0 1/5 2/5 -2/5
0 0 0 0 0
by (r3-2r1)/5 for r2
(5r3-13r1-r2)/5 for r3
and Ax=0 for the null space
but i don't know how to continue , which one is the null space and how to determine the dimension(i guess the dimension is 1)
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
Mark44
#2
Mar9-09, 08:23 PM
Mentor
P: 21,215
Given A =
1 -2 2 3 -1
-3 6 -1 1 -7
2 -4 5 8 -4
you want to find the solution of the equation Ax = 0, where x is a vector in R^5 (i.e., a vector of the form (x1, x2, x3, x4, x5).

By row reduction, you ended up with a matrix A' that is equivalent to your first matrix (but not equal to it, as you imply by calling it A as well).

Keeping in mind that you are now trying to solve A'x = 0, what you have are two equations in five variables. The first row of your reduced matrix A' times x gets you
1*x1 - 2*x2 + 2*x3 + 3*x4 + 1*x5 = 0
The second row of A' times x gets you
1/5 * x3 + 2/5 * x4 - 2/5 *x5 = 0

Note that your matrix A' is not completely row reduced. There is a nonzero entry immediately above your first nonzero entry in the second row. It would be useful for you to replace the 2nd row by 5 times itself (to get rid of the fractions). It would also be useful for you to get rid of the first 2 in the first row, by replacing row 1 with itself plus -2 times row 2 (assuming that you got rid of the fractions in the second row first).

If you do these things you should end up with two equations: the first in x1, x2, x4, and x5, and the second equation in x3, x4, and x5.

In the first equation, isolate x1. In the second equation isolate x3.

You should end up with this:

x1 = something in x2, x4, and x5
x2 = x2
x3 = something in x4 and x5
x4 = x4
x5 = x5

These 5 equations describe every vector in the nullspace of your matrix A. Since there are three arbitrary constants, x2, x4, and x5, the dimension of the nullspace here is 3.

If you arrange the five equations just right, a basis for the nullspace will just about pop out at you.
ak123456
#3
Mar9-09, 09:16 PM
P: 51
Quote Quote by Mark44 View Post
Given A =
1 -2 2 3 -1
-3 6 -1 1 -7
2 -4 5 8 -4
you want to find the solution of the equation Ax = 0, where x is a vector in R^5 (i.e., a vector of the form (x1, x2, x3, x4, x5).

By row reduction, you ended up with a matrix A' that is equivalent to your first matrix (but not equal to it, as you imply by calling it A as well).

Keeping in mind that you are now trying to solve A'x = 0, what you have are two equations in five variables. The first row of your reduced matrix A' times x gets you
1*x1 - 2*x2 + 2*x3 + 3*x4 + 1*x5 = 0
The second row of A' times x gets you
1/5 * x3 + 2/5 * x4 - 2/5 *x5 = 0

Note that your matrix A' is not completely row reduced. There is a nonzero entry immediately above your first nonzero entry in the second row. It would be useful for you to replace the 2nd row by 5 times itself (to get rid of the fractions). It would also be useful for you to get rid of the first 2 in the first row, by replacing row 1 with itself plus -2 times row 2 (assuming that you got rid of the fractions in the second row first).

If you do these things you should end up with two equations: the first in x1, x2, x4, and x5, and the second equation in x3, x4, and x5.

In the first equation, isolate x1. In the second equation isolate x3.

You should end up with this:

x1 = something in x2, x4, and x5
x2 = x2
x3 = something in x4 and x5
x4 = x4
x5 = x5

These 5 equations describe every vector in the nullspace of your matrix A. Since there are three arbitrary constants, x2, x4, and x5, the dimension of the nullspace here is 3.

If you arrange the five equations just right, a basis for the nullspace will just about pop out at you.
thanks , you are so professional !


Register to reply

Related Discussions
Null space and Column Space Linear & Abstract Algebra 1
Basis for null space, row space, dimension Calculus & Beyond Homework 1
Column Space, Null Space Calculus & Beyond Homework 0
Dimension of Null Space Calculus & Beyond Homework 6
Null space Calculus & Beyond Homework 8