Thread: Basis and Dimension of the solution space

I have a more simple problem, but my answer is wrong according to my book.

Find a basis for, and the dimension of, the solution space of Ax=0.

A is a matrix of rows:
1 2 -3
2 -1 4
4 3 -2

I have reduced this to row-echelon form and I get:
1 2 -3
0 1-2
0 0 0
Then, I know that the Rank(A) is 2.
columns=rank+nullity. Then, the nullity is 1.

Now, how can I have a nullity or dimension of the Nullspace to be 1, when I have two rows that are the basis. Namely, (1,2,-3) and (0,1,-2).

Thank you.

Ok, I made many mistakes. I know the answer now.

I end up with a reduced row echelon:
1 0 1
0 1 -2
0 0 0

(Numbers to the right of "x" represent sub-numbers)
x1=- x3
x2=2 x3
x3=t (all real)

Then, the basis is:
-1
2
1.

I have more questions. How do I know that I'm being asked to find a basis in this form rather than one for the row or column space?
Thank you for your patience.

Originally Posted by Jav

Now, how can I have a nullity or dimension of the Nullspace to be 1, when I have two rows that are the basis. Namely, (1,2,-3) and (0,1,-2).

You are confusing the null space with the row space. Take into account that is the set of such that Now, use the method we commented you.


P.S. Please, see rule #9 here

http://www.mathhelpforum.com/math-he...hp?do=vsarules

A basis of what? Not the kernel certainly, since and , not the 0 vector.
The kernel is given by the solutions to
] which can be got by row reducing the augmented matrix


Yes, that reduces to much what you have:

But now that is equivalent to the equations x+ 2y+ 3z= 0, y- 2z= 0. From the second equation, y= 2z and from the first, x+ 2y+ 3z= x+ 2(2z)+ 3z= x+ 7z= 0 from which x= -7z. That is, any vector in the kernel is of the form <x, y, z>= <-7z, 2z, z>= z<-7, 2, 1>. A basis for the kernel is the singleton set {<-7, 2, 1>}.