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# Introduction to the null space of a matrix

Showing that the Null Space of a Matrix is a valid Subspace. Created by Sal Khan.

## Want to join the conversation?

• i still don't understand what a homogenous equation is • it's just an equation in the form
Ax=0
where A is a matrix, x is a vector written in matrix form, and 0 is a zero vector written in matrix form
nonhomogenous would be Ax=b, and b is a vector written in matrix form
• should be v_2 not v^2 • Do I understand correctly that if a null space is for a zero matrix, then all possible vectors are members of that null space, and if a null space is for a non-zero matrix, then the null subspace only consists of a zero vector? • Given a matrix A we can consider the vectors "x" such that Ax=0. We say that these vectors are in the null space of A. It only makes sense to talk about a null space associated to a given matrix. The zero matrix (the one whose only entries are 0) has the property that Ax=0 for any vector x which I think is what you meant.

For other matrices it is more complicated. For example, the identity matrix (with 1's on the diagonal) has the property that Ax=x so if Ax=0 then x=0 so the null space is just the zero vector. But what about the matrix whose rows are (1,0) and (0,0). If we multiply by (0,1) or any scaler multiple of that vector we get zero so the null space consists of all the vectors that look like (0,k) for some number k.

In general the nullspace of a matrix can be lots and lots of different things depending on the matrix. You will learn a lot more about it as you keep doing linear algebra.
• hey.. here somebody said in starting posts that The set { (0,0) , (1,1) } of vectors in R² contains the zero vector but it is certainly not closed under addition, therefore it's not a subspace. but by adding these two vectors i get (1,1) which belongs to r-2 .. so it ios not closed under addition? • It's not closed under addition because you need to cater for the possibility of adding one vector to itself. Here, if you add (1,1) and (1,1) you get (2,2) which is not in the set. Therefore the set is not closed under addition. Hope that helps.
• What's the difference between a subspace, a span, and a plane? • Let S be a set of vectors { v_1, ... ,v_n }.
Span:
The span of S are all the vectors that a linear combination of the vectors inside S can represent. If S is linearly independent, the vectors inside S span R^n. If one of the vectors is a multiple of the other, then S will span R^(n - 1).

Subspace:
Loosely speaking, a subspace is a "part" of R^n. It can be represented by the span of a set of vectors. Anything a set of vectors can span, is a subspace.

Plane:
A plane is a two dimensional space, just like the cartesian plane (xy plane).
• When proving that it is closed under addition, shouldn't we be adding just v1 + v2 rather than Av1 + Av2? • We should be checking that v1+v2 is in the nullspace. What it means to be in the nullspace is that A(v1+v2) should be the zero vector. But A(v1+v2)=Av1+Av2 (because matrix transformations are linear). Now if we assumed v1 and v2 are in the nullspace, we would have Av1=0 and Av2=0. So A(v1+v2)=Av1+Av2=0+0=0. So v1+v2 is indeed in the nullspace, so the nullspace is closed under vector addition.
• What is a zero vector?
And what is a subspace?
Thanks
(1 vote) • Isn't it redundant to say that the zero-vector must be a member of the subspace because doesn't the criteria 'closed under multiplication' cover this already? If the arbitrary number multiplied with the vector is zero then the vector is the zero vector.. • Yes you are correct, if you can show it is closed under scalar multiplication, then checking if it has a zero vector is redundant, due to the fact that 0*v*=0. However, there are many subsets that don't have the zero vector, so when trying to disprove a subset is a subspace, you can easily disprove it showing it doesn't have a zero vector (note that this technique of disproof doesn't always work, in which case you would have to check vector addition and scalar multiplication closures).
• Hi, got a question might sound a bit stupid...
To call some set a subspace, mustn't there be a reference? Here, doesn't mean that we are trying to prove if the vector X set is a subspace to the R^n range? Or to the matrix A?

BTW,
At , sal writes: AV1+AV2=0, but it does not mean V1+V2 is closed in vector set X, right? Same how to prove any c times v1 would also be closed in vector X set?
If answers are no, how can we say Vector X set is sub space?  