Question

linear algebra: basis for nullspace of A?

Suuppose A is a 4x5 matrix with rank 2 for which x1 = [1; 2; 3; 2; 1], x2=[1; -1; 0; 1; 2], x3 = [2; 1; 4; 4; 4] all satisfy Ax=0. Is the set S={x1, x2, x3} a basis for the nullspace of A? Justify your answer.

I learned a theorem in class that rank(A)+nullity(A)=n (the number of columns in A). So I tried row reducing S and got x1=0, x2=0, and x3, hence linearly independent. Does this mean that the nullity is 0? And so S is not a basis for the nullspace of A since 2+0 does not equal to 5? I am confused as to how to solve this question.

Answer 1

Try to reduce A to RREF

Answer 2

Check if x1, x2, x3 are linearly independent

Answer 3

I made a spelling error, yes x1, x2, and x3 were 0, and they were linearly independent.

Answer 4

And if they satisfy the null space condition, then they must be basis vectors...from Rank_nullity thm also u know the dimension shld be 3

Answer 5

so this is a 3x5 matrix then?

Answer 6

4x5..reduces to the null space dimensions

Answer 7

oh i see what you mean

Answer 8

@him1618 what do you mean the null space condition? Since it's linearly independent and hence x1=x2=x3=0, does it mean the nullity is zero? And thus they're not the basis? :S

Answer 9

@TuringTest yo

Answer 10

they are not all zero, you did the reduction wrong

Answer 11

Wht u tryis: Try to solve for Ax=0

The set of equations u get there..ull have to take some dummy variables
The no of these variables u take is the no of basis vectors in the soln set
This is the basis of the null space

Answer 12

if they were all reducible to zero the set would be linearly dependent, since the only solution to Ax=0 would be the trivial one

Answer 13

He should try the original method...then all theorems will make sense

Answer 14

\[\left[\begin{matrix}1 & 0&0 \\ 0 & 1&0\\ 0&0&1\\0&0&0\\0&0&0\end{matrix}\right]\]

This is what I got so I thought I got x1=0, x2=0, x3=0?

Answer 15

I imagined your matrix sideways :/

Answer 16

you mean x1,x2, and x3 are the column vectors?

Answer 17

So then the first 3 columns of ur original matrix compose the ans

Answer 18

that seems to simple... are you sure it's not sideways?

Answer 19

@TuringTest Yes, they're column vectors.. sorry for confusing you!

Answer 20

ok then him has stated the idea

Answer 21

See wht uve don is the linear independence test

Answer 22

If u reduce your matrrix...the columns where only leading ones remain correspond to the columns of the original which make up the reduced basis

Answer 23

get it?

Answer 24

Yes, so it means they're the basis, right?

Answer 25

yes

Answer 26

\(a\) possible basis

Answer 27

I see. Thanks a lot @him1618 and @TuringTest :)

Answer 28

welcome :)

Answer 29

A theater group made appearances in two cities. The hotel charge 500 before tax in the second city was higher than in the first. The tax in the first city was 6.5% , and the tax in the second city was 3.5% . The total hotel tax paid for the two cities was 317.50 . How much was the hotel charge in each city before tax?