07 Null space and column space

Matrix vector products

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What we want to do now in this video is relate our notion of a matrix to everything we already know about vectors.

Let's define what is means when we take the product of our matrix A with some vector x. Our definition will only work if the vector x has the same number of components as A has columns.

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This matrix A also can be represented like this.

Now the interesting here is now the product Ax can be interpreted as a linear combination.

Introduction to the null space of matrix

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Before null space, let's review about the definition of subspace.

Conditions

1. Zero vector is a member of the subspace
2. Close under addition
3. Close under multiplication

Is N a subspace?

1. Does this contain zero vector? Yes

2. Close under addition? Yes

We know that $A\vec {v_1} and A\vec {v_2}$ are zero vectors. If you add them all, you can get zero vector!

3. Close under multiplication? Yes

When you have to find a null space of a matrix, your goal is to find the set of all x's th

Null space 2: Calculating the null space of a matrix

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We can make a RREF matrix and find the solution.

So the solution set is a linear combination of those two vectors.

The null space of A equals all the linear combinations of these vectors. It's span of two vectors!

So, the RREF of A times our vector x is equal to zero.

Null space3: Relation to linear independence

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Column space of a matrix

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Column space is all the linear combinations of column vectors. It is same with span!

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Dimension of the null space or nullity

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Null space in here is equal to a linear combination of these vectors.

$\vec {v_1}, \vec {v_2}, \vec {v_3}$ are basis for the null space B!!

• Dimension of subspace = # of elements in a basis for the subspace
• Nullity:, dim(N(B)) = 3 in this example!

The nullity of any matrix is equal to the number of free columns.

Dimension of the column space or rank

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Column 1, 2, 4 are pivot columns. They are clearly linearly independent.

Showing relation between basis cols and pivot cols

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Showing that the candidate basis does span C(A)

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