Understanding Null Space: A Layman's Explanation

[cragar]

Can someone give me a layman’s terms explanation of what a null space is .

 

[snipez90]

The way I thought of the null space when I first learned about it is that it is analogous to the zeros of a real function (i.e. the set of elements in the domain for which the value of the function is 0), but real functions aren't always nice. Now if you have a linear transformation T (which is a function with two very nice properties, namely additivity and T(av) = aT(v) for any scalar a and any vector v in V) from one vector space V to another vector space W, then the null space is simply the set of vectors in V for which the value of T at any particular vector in the set is 0.

But the null space is a subspace of V (you should prove this), and hence a vector space itself. Since you're working with these linear transformations with very nice properties over a very nice algebraic structure (the vector space), you get some nice theorems regarding the relationship between the linear transformation and the null space.

I'm sure that someone who likes algebra could give a better explanation. I'm pretty sure the null space has shown up often in theorems of functional analysis, but I can't remember how important they were. I know that some of the major theorems of functional analysis (Hahn Banach, Open Mapping, Bounded Inverse) don't really refer to the null space. But yeah this is more or less the extent of my knowledge about null spaces.

 

[Dickfore]

Null space is used with respect to a linear transformation. A linear transformation T is a mapping from one vector space U to another V (not necessarily identical) that has the following properties:
$$T: U \rightarrow V$$
1. $$T(\mathbf{x} \oplus_{u} \mathbf{y}) = T(\mathbf{x}) \oplus_{v} T(\mathbf{y}), \; \forall \mathbf{x}, \mathbf{y} \in U$$;

2. $$T(\lambda \odot_{u} \mathbf{x}) = \lambda \odot_{v} T(\mathbf{x})$$.

The null space of T is defined as:

$$\textrm{Null} T = \{\mathbf{x} : \mathbf{x} \in U, T(\mathbf{x}) = \mathbf{0}_{v} \}$$

 

[cragar]

Thanks for your answers

 

[Tac-Tics]

snipez90's reply is on the mark. Dickfore, those aren't really layman's terms :/

The definition is pretty easy to use.

Suppose you have a linear transformation T. The nullspace of T is the set of vectors v where T(v) = 0.

For many common operations, the nullspace is {0} (transformations, shears, non-zero scalings, reflections). The nullspace can never be empty, because it must always contain the 0 vector (why? because at the least, T(0) = 0).

When the nullspace contains more than just {0}, it means multiple vectors map to zero. As a consequence of this, the transformation is not invertible. The reason is because if v /= u, T(u) = 0, and T(v) = 0, you don't know whether T^-1(0) equals u or v. In effect, the transformation throws away information about its argument.

The simplest example of a nullspace greater than {0} is a projection transformation. Think of a 2D painting of a 3D landscape. That is effectively a projection -- a kind of linear transformation -- that maps points from the scene to points on a canvas. However, when this is done, a dimension is effectively "lost" in the painting. You can't tell if a mountain is small and near to the artist or if it is large and far away.

If you took a ruler from the artist's eye to a point on the mountain, from the artist's point of view, the ruler would look like a single point (because he's viewing the end of the ruler, not the side). This ruler embodies the nullspace -- an entire dimension maps to a single point (the zero point).

In this example, the nullspace would be a 1 dimensional subspace. Notice the math there: a 3D scene becomes a 2D painting and the resulting nullspace is 1D. 3 = 2 + 1. We have names for these different values. The space is 3D. The "rank" of the transformation (our projection) is 2. The "nullity" of the transformation is 1. The dimension of the space is always the sum of the rank and nullity.