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I have row reduced a matrix to the following form: $$ A = \begin{bmatrix} 3 & -3 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

How do I compute the null space of this matrix?

Thank you.

Edit:

Specifically, how do you obtain the vectors.

If I let vector $v = \begin{bmatrix} x & y & z \end{bmatrix}^T$ be in the null space of $A$, I must solve the following equation: $$ 3x-3y+z=0$$

To find possible $v$, I thought to set $x,y$ and $z$ to $0$ at separate times. However, doing so gives me three vectors for the null space. However, I believe by the Rank-Nullity theorem the dimension of the null space must be 2.

I get the same vectors by this method as some that have been suggested, but I do not know which of the three to pick. Mathematica consistently chooses the same two vectors ($\begin{bmatrix} -1 & 0 & 3 \end{bmatrix}^T$ and $\begin{bmatrix} 1 & 1 & 0 \end{bmatrix}^T$).

audrow
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  • See http://math.stackexchange.com/a/1521354/265466 for a detailed description of how to read basis for the null space directly from the rref matrix. – amd Mar 02 '16 at 21:04

4 Answers4

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Let $\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ be an element of null space of $$ A = \begin{bmatrix} 3 & -3 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

We then have $$0= A\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} 3 & -3 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}3x_1-3x_2+x_3\\0\\0\end{bmatrix} $$

So, we want $\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ such that $3x_1-3x_2+x_3=0$.. Can you now find basis?

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It is obviously given by the vectors $v=(x,y,z)^T$ such that $3x-3y+z=0$, i.e. by the vectors $v$ such that $v\perp (3,-3,1)^T$. Two vectors on this plane are, for instance, $(1,1,0)^T$ and $(0,1,3)^T$: since they are linearly independent, they give a base of that null space.

Jack D'Aurizio
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    How do you obtain the two vectors? If I set x, y, and z to 0 at separate times I get three vectors for the null space. However, by Rank-Nullity theorem the dimension of the null space must be 2. – audrow Mar 02 '16 at 18:37
  • The dimension of the null space is clearly two, since a base of that space is given by two independent vectors. What is the purpose of setting $x=0,y=0,z=0$ at separate times? $v$ belongs to the null space of $A$ iff $Av=0$, that is the definition. – Jack D'Aurizio Mar 02 '16 at 18:42
  • At last, $(1,1,0)^T, (0,1,3)^T$ clearly satisfy $3x-3y+z=0$. You may recover them by imposing $z=0$ and $x=0$ at separate times, but when doing the same with $y=0$ you will get a linear combination of the previous two vectors. – Jack D'Aurizio Mar 02 '16 at 18:45
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Let $$X=\begin{bmatrix}x\\y\\z\end{bmatrix}\in\ker A$$

Then $AX=0$, i.e. $$\begin{bmatrix} 3 & -3 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$

Computing the product you get $3x-3y+z=0$ (the two ther lines give $0=0$). So the null space is a plane whose equation is $3x-3y+z=0$.

Note that we already knew that it was a plane, using the rank-nullity theorem.

Augustin
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I found this method particularly nice, which is of course equivalent to the ones stated by others and so it's simply a matter of personal taste.

First produce a $1$ in the $a_{11}$ entry by dividing by $3$. Second fill up the diagonal with $-1$. You get: $$ \begin{bmatrix} 1 & -1 & 1/3 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} $$

Now simply read of the kernel. It's the second and third column i.e.

$$ \begin{bmatrix} -1 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1/3 \\ 0 \\ -1 \end{bmatrix}$$

This can also be generalized to all kind of situations, i.e. all ranks and dimesions.

Maik Pickl
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