Does the Null Space of a 2x3 Matrix Determine its Column Space?

[Drakkith]

Homework Statement

Let $A$ be a 2x3 matrix. If Nul($A$) is a line through the origin in ℝ³, then Col($A$) = ℝ². Explain why.
Hint: Think about the number of pivots in $A$.

Homework Equations

The Attempt at a Solution

So, Nul($A$) is the set of all solutions to the equation $Ax=0$.
Col($A$) is the set of all linear combinations of the columns of $A$.

If the Null Space forms a line in through the origin in ℝ³ that means... well I'm not sure what that means.

Looking at the hint, it appears there are at most 2 pivots in $A$ since there are only 2 entries in each vector. I'm not sure how this helps though.

The solution to the question says that since each row of $A$ has a pivot, the columns of $A$ span ℝ². But I'm having trouble understanding this. I know that Col($A$) = Span($A_1, A_2, A_3$) and that a set of vectors $V$ in ℝⁿ spans ℝⁿ if every vector in ℝⁿ is a linear combination of $V_1, V_2, ... V_p$.

I don't know where I'm having trouble. I feel like I have everything sitting right in front of me and just put it all together.

 

[fresh_42]

The easiest approach is always an example. At least this is what I did since I'm not used to pivot elements, which indicate some sort of row manipulation to achieve a certain final form, and my null space is called the kernel and the column space the image. We have
$$
\begin{bmatrix}a & b & c \\ d& e & f\end{bmatrix} \cdot \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}ax+by+cz \\ dx+ey+fz\end{bmatrix} \\
A : \mathbb{R}^3 \longrightarrow \mathbb{R}^2
$$
One way to look at it is to examine the dimensions. We know that a three dimensional vector space is mapped to a two dimensional. This means here, that we lose one dimension ($\dim \operatorname{Nul}(A) = 1$) plus we keep two ($\dim \operatorname{Col}(A) =2)$) and this has to be equal to the $3$ we started with.

What does pivot element mean here? I suppose it is some kind of manipulation of the matrix to achieve a certain form (row echelon form or so). Best would be $A' = \begin{bmatrix} 1& 0 & g \\ 0 & 1 & h\end{bmatrix}$. Can you get there? And if we then calculate $\operatorname{Nul}(A') = \{\vec{x}\,\vert \, A'\cdot \vec{x}= 0 \}$, we get the parametrized (by $z=(\vec{x}_3)$) form of the line that spans the null space.

 

[WWGD]

More formally, think of the Rank-Nullity theorem: Dimension of rank+ Dimension of nullspace = Number of columns.

 

[WWGD]

Homework Statement

Let $A$ be a 2x3 matrix. If Nul($A$) is a line through the origin in ℝ³, then Col($A$) = ℝ². Explain why.
Hint: Think about the number of pivots in $A$.

Homework Equations

The Attempt at a Solution

If the Null Space forms a line in through the origin in ℝ³ that means... well I'm not sure what that means.

.

I am not sure what type of answer you're looking for, but given a line has the form $y=mx$; $m$ a Real number, and your linear map is $L: \mathbb R^3 \rightarrow \mathbb R^2$ that $L(y)=0$ iff $y=mx$

The easiest approach is always an example. At least this is what I did since I'm not used to pivot elements, which indicate some sort of row manipulation to achieve a certain final form, and my null space is called the kernel and the column space the image. We have
$$
\begin{bmatrix}a & b & c \\ d& e & f\end{bmatrix} \cdot \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}ax+by+cz \\ dx+ey+fz\end{bmatrix} \\
A : \mathbb{R}^3 \longrightarrow \mathbb{R}^2
$$
.

Notice Fresheimer's matrix , that the zero set is the intersection of $ax+by+cz, dx+ey+fz$ , both of which are planes. Two planes intersect either empty ( if they are translates of each other) or at a line.

 

[Drakkith]

When it says that Nul($A$) is a line through the origin, does that mean that there's a single free variable in the nullspace?

 

[WWGD]

When it says that Nul($A$) is a line through the origin, does that mean that there's a single free variable in the nullspace?

Precisely.

 

[Drakkith]

Alright. Given that Nul($A$) has a free variable, then that means that $Ax= \begin{bmatrix} a&b& c\\ d &e &f \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix} = 0$

Multiplying gives us: $Ax= \begin{bmatrix} cz\\ fz \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$

In order for $Ax=0$ to be valid, $c$ and $f$ must be zero. So $A= \begin{bmatrix} a&b& 0\\ d &e &0 \end{bmatrix}$
Col($A$) = Span{$A_1, A_2, A_3$} and the columns of $A$ span ℝ².

But, isn't this true even if Nul($A$) isn't a line through the origin? I don't see the significance of that part of the question. If two of the three vectors already span all of ℝ², adding another vector wouldn't reduce the span, would it?

 

[fresh_42]

If you assume that the vector, which spans the null space and defines the line, i.e. its direction, already has the form $\begin{bmatrix}0\\0\\ z\end{bmatrix}$, then you made a choice considering the basis vectors, namely, that the line is the third basis vector and the other two which all together span $\mathbb{R}^3$ are perpendicular to it. There is of course a basis, which does this. But we started with another, arbitrary orthonormal basis, for which the matrix of $A$ is given by $\begin{bmatrix}a&b&c \\ d&e&f \end{bmatrix}$ and the basis vectors have the coordinates $\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\end{bmatrix}$. In this basis, we cannot know whether the null space is identical to the third basis vector - it might or might not be the case.
$$
A \cdot \vec{x} = \begin{bmatrix}a&b&c \\ d&e&f \end{bmatrix} \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} = 0 \\
\Longleftrightarrow \\
\begin{cases} ax+by+cz=0\\dx+ey+fz=0 \end{cases}
$$

The two equations each define a plane in three dimensional space. If we demand both to be true for the same choices of $x,y,z$ then this means we demand to be $(x,y,z)$ on both planes, i.e. the intersecting line - the one which spans the null space, as all points $(x,y,z)$ on the line are mapped by $A$ to the zero vector. Now row manipulation as those which lead to the row echelon form are nothing else than solving this system of linear equations by multiplying them or substracting them. Let's say we achieve the equations (and row echelon form)
$$
\begin{cases} x+gz=0\\y+hz=0 \end{cases}\\
\Longleftrightarrow \\
A' \cdot \vec{x} = \begin{bmatrix}1&0&g \\ 0&1&h \end{bmatrix} \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} = 0
$$
Then $\left\{\left. \begin{bmatrix}x \\y\\ z \end{bmatrix} \in \mathbb{R}^3 \, \right| \, A'\cdot \vec{x}= 0 \right\} = \left\{\left. \begin{bmatrix}x \\y\\ z \end{bmatrix} \in \mathbb{R}^3 \, \right| \, x=-gz\; , \;y=-hz \; \right\} = \left\{\left. \begin{bmatrix}-gz \\-hz \\ z \end{bmatrix} \in \mathbb{R}^3 \, \right| \, z \in \mathbb{R} \right\}$.

In physics the free variable $z$ is often abbreviated by $t$ as the (time) parameter of a walk along the line which here spans the null space. I'd say it is helpful to choose some numbers (small integers) for $A$ and draw the two planes given by the two equations, then see where they intersect, perform the calculations for the row echelon form and see whether the null space (intersection line) is given by the last equation as stated. E.g.
$$
A = \begin{bmatrix}-3 & 0 & -6 \\ 1 &4 &-2 \end{bmatrix}
$$
In this scenario, the null space isn't automatically spanned by a vector $\begin{bmatrix}0&0&z\end{bmatrix}$. This is only the case, if it turns out that $g=h=0$, but usually they aren't and the null space lies somehow diagonal according to the given basis. In the above, I didn't change the basis. (In general it could only be, that we need to shuffle the naming of $x,y,z$ in order to get the row echelon form above, which means a permutation of columns.)

 

[Drakkith]

I'm sorry Fresh, I'm a bit lost.

 

[fresh_42]

You can skip the first paragraph and start at the equations with $A\vec{x}=0$. In any case, try to do the example at the end. I think I've chosen easy numbers so that even a drawing should be possible, but at least calculations should be easy. An example often helps better than any explanations.

 

[Drakkith]

In any case, try to do the example at the end. I think I've chosen easy numbers so that even a drawing should be possible, but at least calculations should be easy. An example often helps better than any explanations.

I think I found the intersection of the planes as $(-2t, t, t)$
After that I don't know what to do.

 

[fresh_42]

Yes, I have this, too. It is the null space of $A$. And it is the intersection of, i.e. contained in both the planes given by the equations $-3x-6z=0$ and $x+4y-2z=0$. If you like, then you can draw them in a coordinate system like this and check the result.

What do you want to do more? You have found $g=2 \; , \;h=-1$ in the notation of post #8. It should only show you, that the null space isn't automatically of the form $\begin{bmatrix}0\\0\\z \end{bmatrix}$. To make it so, we would have to choose another basis, and the matrix of $A$ would have different numbers then.

To get the image of $A$, we simply have all points of the form $\left\{\left. A\cdot \vec{x} = \begin{bmatrix}-3x-6z \\ x+4y-2z\end{bmatrix} \right| x,y,z \in \mathbb{R}\right\}$. To get vectors which span it, we may choose any combinations of $(x,y,z)$, e.g. $(1,0,0)$ and $(0,1,0)$ which would transform by $A$ to the vectors $(-3,1)$ and $(0,4)$, resp., which span the image of $A$, a two dimensional space. This is not automatically embedded in $\mathbb{R}^3$ because we changed vector spaces by $A : U=\mathbb{R}^3 \longrightarrow V=\mathbb{R}^2$.

 

[Drakkith]

Okay. I think all that makes sense.
So how does all of this help with the original question?

 

[fresh_42]

Oops, I confused the threads me thinks. But for

I don't know where I'm having trouble. I feel like I have everything sitting right in front of me and just put it all together.

the answer now is yes, as the example has everything in front of you. The pivot elements are the ones in $A'$ or the non-zero entries which remain in $A$ when transformed into row echelon form. That is $A' = \begin{bmatrix}1&0&g \\ 0&1 &h \end{bmatrix}$, which translates to: The first two column vectors are linearly independent, so they span a two dimensional vector space, which is the image or column space of $A$. And vice versa there have to be two linear independent vectors and thus two pivot elements in diagonal places if the column space is, as given, two dimensional. That the null space is at least one dimensional is given by the fact that $A$ loses a dimension. The fact that it isn't of higher dimension (as given), means, that at least two linear independent column vectors have to remain in any row echelon form of $A$, which is the same as saying there are these two pivot elements diagonally placed.

 

[Drakkith]

Wait wait... if Nul($A$) is a line through the origin in ℝ³, then as we've shown above it has one free variable. Since $A$ has three columns, and one contains a free variable, the other 2 have to be pivot columns. Since there are two pivot columns, Col($A$) = ℝ².

Is that right?

 

[fresh_42]

Yes. This is one way to look at it. The transformation from $A$ to $A'$ is just to figure out which is which.