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# Dimension of the null space or nullity

Dimension of the Null Space or Nullity. Created by Sal Khan.

## Want to join the conversation?

• If we can have exercises for these concepts as well it will be great. As these concepts are difficult to master.
• why x2 is a free variable ? isnt x2 pivot as well ?
• When you're looking at free variables, pay attention to the pivot columns. The first and third columns hold a pivot. That leaves the second, fourth, and fifth as free variables.

Realistically, you could treat x2 as a non-free, but to do so would mean you'd have to make x1 free in order to satisfy the first row of the matrix. The general practice, though, is to treat the pivot, x1, as the non-free and x2 as the free.

When you solve an augmented matrix, it's solving a system of equations. If you were to place a vector of values into the matrix Sal used in an attempt to solve for values, you would still have to CHOOSE random values for x2, x4, and x5 to be able to solve for x1 and x3. That's where the notion of a "free" variable comes from. The math gives you the freedom to choose any value for those 3 variables and still come up with an answer for the pivots.
• I am confused. In a previous video ("null space and column space") we learned that the "basis" are the column vectors that correspond to the pivot entries in the rref(B). Which for our case would be the column vectors [1,1] and [2,3]. These TWO vectors should be the basis of the column space of B, which is clearly in R^2. How can the THREE vectors in this video be also a "basis"! (and these are in R^5?!). Or is it that a "basis of a column space" is different than a "basis of the null space", for the same matrix?
• In short, you are correct to say that 'a "basis of a column space" is different than a "basis of the null space", for the same matrix."

A basis is a a set of vectors related to a particular mathematical 'space' (specifically, to what is known as a vector space). A basis must:
1. be linearly independent and
2. span the space.
The 'space' could be, for example: R^2, R^3, R^4, ... , the column space of a matrix, the null space of a matrix, a subspace of any of these, and so on (there are many more examples; vector spaces appear everywhere in mathematics). When someone gives you a basis, you need to ask: "A basis of what space?"

In general, if I give you two different spaces, A and B, then a basis for A will be different to a basis for B.

Now, in general, the column space of a given matrix does not equal the null space of that matrix. Therefore, in general, a basis for the column space of the matrix will be different to a basis for the null space of that matrix.
• Hi sal i just wanna ask question about reducing echelon form,how to know if that matrix is been reduce completely???
• There are three conditions for a matrix to be in RREF
1) The first non-zero entry of a row must be a 1; this entry is called a pivot
2) The pivot for each row must to the right of all the pivots in any rows above.
3) Any columns that contain pivots must have zeros for all other entries except the pivot.

As a result of 2 and 3, any rows that contain all zeros must be at the bottom of the matrix.

Hope that helps.
• Didn't Sal say that if Nul B is not the zero vector, the B is linearly dependent?
Why are we proving that B is linearly INDEPENDENT so that it can form the basis for the null space?
• In this case, B is not linearly independent, what Sal is actually proving is that the set of solution vectors for NULL B (Bx=0) are linearly independent, which is a different thing.
• I've seen an example of a subspace in R^3 only have a dimension of 2. How is this visually possible? Doesn't a vector in R^3 mean it's in the third dimension?
(1 vote)
• The dimension only means the number of elements in the basis for the subspace. Think of a plane in ℝ³ that passes through the origin, how many vectors/elements does it take at minimum, such that when you take their span, you get a plane? Well, 2. And that tells us that the basis for a plane has 2 vectors in it. If the dimension is again, the number of elements/vectors in the basis, then the dimension of a plane is 2.
So even though the subspace of ℝ³ has dimension 2, the vectors that create that subspace still have 3 entries, in other words, they still live in ℝ³.
• What would the dimension of the nullspace be if the the only sloution to Ax = 0 is x = 0? Can you consider the null vector as a basis?
• Since the null space of A is the set of vectors x that satisfy Ax = 0, and you say that the only such x is the zero vector 0, so symbolically, N(A) = {0}.
The next part is to find the basis for N(A). This will be the set S of linearly-independent vector(s) that span N(A). If you think about it, only the null vector 0 can be in that set, as any non-null vector will span beyond N(A). Thus S = {0}.
The dimension of a null space is the number of elements in its basis. Since here the basis set S has one element, dim(N(A)) = 1. So, yes, you're right - the null vector 0 is the basis of the null space here.
• Is the null space the same thing as a solution space?
(1 vote)
• The null space of B is the solution space of the problem B x = 0.