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Determine the null space of the matrix:$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$$

My try: $$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}_{R_2\rightarrow R_2-2R_1\\R_3\rightarrow R_3-R_1}$$ $$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}_{R_3\rightarrow 5R_3-2R_2}$$ $$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 0 \end{bmatrix}_{R_2\rightarrow \frac{R_2}{5}}$$ $$\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$

From this I got $$x-y=0\implies x=y\\y=0$$ $$(x,y,z)^T=(y,0,z)^T=y(1,0,0)^T+z(0,0,1)^T$$ So, $(1,0,0)^T$ and $(0,0,1)^T$ is the null space. Is this correct?

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  • $\begingroup$ No, the matrix has rank $2$, so the null space is $\{0\}$. There is no $z$. $\endgroup$ – egreg Aug 9 '18 at 20:59
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Recall that by definition the nullspace is the subspace of all vectors $\vec x$ such that $A\vec x=\vec 0$ and in that case we have ony the trivial solution $(x_1,x_2)=(0,0)$ then $Null(A)=\{\vec 0\}$.

Notably to solve $Ax=0$ we can proceed by RREF to obtain

$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}\to \begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}\to \begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$

that is

$$\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

that is $x_1=x_2=0$.

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  • $\begingroup$ So, is my method to find the null space correct? I reduced it to row echelon form. After that what am I supposed to do. $\endgroup$ – user572932 Aug 9 '18 at 21:04
  • $\begingroup$ @philip Yes your way is correct, we need to find the solutions for Ax=0 but your conclusion is wrong since $x$ has only 2 components. $\endgroup$ – user Aug 9 '18 at 21:07
  • $\begingroup$ @gimusi can you please show how to arrive to the solution using my method $\endgroup$ – user572932 Aug 9 '18 at 21:10
  • $\begingroup$ @philip Yes ok I thought it was enought, I'm going to add some more detail! $\endgroup$ – user Aug 9 '18 at 21:11
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Your method is correct to find that $x=y$ and $y=0$. But you have made an incorrect conclusion after that stage and you have made a mistake in the dimension of your vectors.

Firstly, see that your matrix acts on vectors in $\mathbb{R}^2$ to form vectors in $\mathbb{R}^3$ like so:

$$\left( \begin{matrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} x-y\\ 2x+3y\\ x+y \end{matrix} \right) $$

So you're searching for vectors $(x,y)^T$ in your null space.

If $x=y$ and $y=0$ then $x=y=0$, showing that your null space is just $\{(0,0)^T\}$

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Right Nullspace If vector $\alpha = [x,y]$ is in the right nullspace of A then $$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \\ \end{bmatrix}[x,y]^T = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ This gives $$x - y = 0$$ $$2x + 3y = 0$$ $$x + y = 0$$ First and third equation tells us that $x = y = -x$, i.e. the only solution is $[0,0]$. So, the right nullspace has dimension zero

Left Nullspace If vector $\alpha = [x,y,z]$ is in the left nullspace of A then $$\begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}$$ So we get $$x + 2y + z = 0$$ and $$-x + 3y + z = 0$$ First equation tells us $$x = -(2y+z)$$ and second one says $$x = 3y+z$$ So by equating $x=x$, we get $$-2y-z = 3y+z$$ that is $$5y = -2z$$ or $$y = - \frac{2}{5}z$$. Plugging this back into one of the two x equations, we get $$x= 3y + z = 3(- \frac{2}{5}z) + z = \frac{-6+5}{5}z = -\frac{1}{5}z$$ So, the Null space takes the form $$\begin{bmatrix} -\frac{1}{5} \\ - \frac{2}{5} \\ 1 \end{bmatrix} z$$

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  • $\begingroup$ That's the nullspace for $A^T$, that is the left nullspace. $\endgroup$ – user Aug 9 '18 at 21:18
  • $\begingroup$ Then, from this how can you conclude that Null space is $0$ $\endgroup$ – user572932 Aug 9 '18 at 21:20

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