Setting the record straight on rank, nullity, etc.

[Jamin2112]

Homework Statement

(Pictured)

Homework Equations

Some Wikipedia and Wolfram MathWorld definitions.

In linear algebra, a family of vectors is linearly independent if none of them can be written as a linear combination of finitely many other vectors in the collection.

The rank of a matrix A is the number of linearly independent rows or columns of A.

In linear algebra, the kernel or null space (also nullspace) of a matrix A is the set of all vectors x for which Ax = 0.

The nullity of a linear map of vector spaces is the dimension of its null space.

The Attempt at a Solution

I get a little confused with this stuff.

Say that we're looking at the first matrix. It can be row reduced as follows.

So it looks like we know that Ax = 0 whenever

x₃ = -1/2 x₁
and
x₃ = x₂.

In other words, any vector x of the form

, where x₃ is any real number, will solve Ax = 0. So the above picture is the basis is the null space of A; and since it has one vector, the rank of the null space, the nullity, is 1.

Tell me if I'm understanding this; and if I'm not understanding it, explain it to as if I were a 5-year-old.

 

[lanedance]

first your vector is not correct
x3 = -1/2 x1
x3 = x2
they lead to (start by setting x1=1)
(1, -1/2, -1/2)c

notice i used c as a scalar multplier, don't use a component, that is confusing

Now say you have a matrix A and want to find vectors x, such that Ax=0

Ax represents a set of linear equations, when you are row reducing you are just manipulating the linear equations whilst still solving them

now any vectors that satisfy Ax=0 are by definition in the nullspace of A.

As you have shown the set of vectors can be spanned by a single basis vector it has dimension 1, ie. the nullity of A is 1.

Geometrically the nullspace of A is the line through the origin with direction (1, -1/2, -1/2)

 

[HallsofIvy]

An n by n matrix maps vectors in R^n to vectors in R^n. If the m atrix is invertible, it maps R^n to all of R^n (the mapping is both "one-to-one" and "onto"). If it is not invertible, it maps R^n into some subspace of R^n. The "rank" of the matrix is the dimension of that subspace. Of course, that means the mapping is not "onto" and so cannot be "one-to-one".

A0= 0 for any matrix A so the one element of R^n we can be sure is in that subspace is 0. If A is not "one-to-one", more than on member of R^n (in fact, an entire subspace) is mapped to 0. That is the "nullspace" of A and its dimension is the "nullity". One can show that the sum of those two dimensions, rank+ nullity, must be n.
In the first case here,
$$\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 5 \\-1 & 0 & -2\end{bmatrix}$$

Finding the nullspace is the same as solving the equation
$$\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 5 \\-1 & 0 & -2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$$
You could solve that, of course, by setting up an "augemented" matrix and row reducing- but since the last column will be all "0"s, you don't really need that- just row reduce the matrix itself.
If you subract twice the first row from the second and add the first row to the third you get
$$\begin{bmatrix}1 & 1 & 1\\ 0 & -3 & 3 \\ 0 & 1 & -1\end{bmatrix}$$
Now, if you divide the second row by -3, then subtract that from the third you get
$$\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & -1 \\ 0 & 0 & 0\end{bmatrix}$$

We can interpret those as the equations x+ y+ z= 0, y- z= 0. From the second, z= y. Putting that into the first, x+ y+ y= x+ 2y= 0 so that x= -2y. That is, any vector in the nullspace is of the form <x, y, z>= <-2y, y, y>= y<-2, 1, 1>. The null space is 1 dimensional with basis {<-2, 1, 1>}.

We can also use that reduce matrix to say that any vector in the row space is of the form
$$\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & -1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}u \\ v \\ w\end{bmatrix}= \begin{bmatrix}u+ v+ w \\ v- w\\ 0\end{bmatrix}$$
That is, a vector <x, y, z> in the row space is of the form x= u+ v+ w, y= v- w, z= 0. The first two are "independent" which means that vectors in the row space are of the form <x, y, 0>= <x, 0, 0>+ <0, y, 0>= x<1, 0, 0>+ y<0, 1, 0>. It is two-dimensional with $\{<1, 0, 0>, <0, 1, 0>\}$ as basis.

Of course, rank+ nullity= 2+ 1= 3.

 

[Jamin2112]

first your vector is not correct
x3 = -1/2 x1
x3 = x2
they lead to (start by setting x1=1)
(1, -1/2, -1/2)c

Let me try this again.


If I reduce the matrix to [1 0 2; 0 1 -1; 0 0 0], then I know that Ax = 0 whenever

x₁ + 2x₂ = 0
and
x₂ = x₃.

So we let x₁ be an arbitrary constant c (Why?). Because c + 2x₂ = 0, x₂ is -1/2 of c, and because x₃ is x₂, x₃ is also 1/2 of c. We have that x = c(1 -1/2 -1/2)^T solves Ax = 0. In other words, I could do a loop in MATLAB like

for i = 1:10^6
x1 = rand;
x = [x1 -.5*x1 -.5*x1]';
Ax
end


and every Ax printed would be the zero vector.

The "basis" by definition is c(1 -1/2 -1/2)^T, and we'd say that (1 -1/2 -1/2)^T "spans" the "null space" of A. The basis has one vector; so its "nullity" is 1. The matrix has 3 columns; so its "rank" is 3 - 1 = 2.

I guess another thing that confuses me is "rank" versus "nullity." The rank is the number of linearly independent columns of a matrix; so to see whether there is non-trivial solution to Ax = 0 is to see whether all the vectors of A are linearly independent. It seems like we're using the same procedure for rank and nullity ... Can someone explain to me the subtle difference between the two?

 

[lanedance]

the two procedure are related as rank + nullity =n

if the only solution of Ax = 0 is x=0, then rank(A) = n