# Thread: Basis and Dimension of the solution space

1. ## 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.

2. ## Re: Vector Spaces

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.

3. ## Re: Vector Spaces

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 $N$ with the row space. Take into account that $N$ is the set of $(x,y,z)\in\mathbb{R}^3$ such that $\begin{Bmatrix} x+2y-3z=0\\y-2z=0\end{matrix}$ Now, use the method we commented you.

P.S. Please, see rule #9 here

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

4. ## Re: Vector Spaces

A basis of what? Not the kernel certainly, since $\begin{bmatrix}1 & 2 & -3 \\ 2 & -1 & 4 \\ 4 & 3 & -2\end{bmatrix}\begin{bmatrix}1 \\ 2 \\ -3 \end{bmatrix}= \begin{bmatrix}14 \\ -12 \\ 10\end{bmatrix}$ and $\begin{bmatrix}1 & 2 & -3 \\ 2 & -1 & 4 \\ 4 & 3 & -2\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ -2 \end{bmatrix}= \begin{bmatrix} 8 \\ 8 \\ 7\end{bmatrix}$, not the 0 vector.
The kernel is given by the solutions to
] $\begin{bmatrix}1 & 2 & -3 \\ 2 & -1 & 4 \\ 4 & 3 & -2\end{bmatrix}\begin{bmatrix}x \\ y\\ v \end{bmatrix}= \begin{bmatrix}0\\ 0 \\ 0\end{bmatrix}$ which can be got by row reducing the augmented matrix
$\begin{bmatrix}1 & 2 & -3 & 0 \\ 2 & -1 & 4 & 0 \\ 4 & 3 & -2 & 0\end{bmatrix}$

Yes, that reduces to much what you have:
$\begin{bmatrix}1 & 2 & 3 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$
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>}.

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# how to find a basis and dimension of the solution space of system equation

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