Linear Algebra Matrix Column Space

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[blazing fjord]

Linear Algebra Column Space

Linear Algebra Matrix Column Space

[topaz nacelle]

Well here I think you need to write the column space as a linear combination of the basis of the space

So you need to look which vectors in the columns are linearly dependent of each other

And create a basis for the space

For Null(A) you can use the rank theorem

[blazing fjord]

so vectors solve for what scalars make the vectors = 0?

[blazing fjord]

is this true for A?

(1,0,-1) and (-2,0,2) are linearly dependent

but other than that idk

[topaz nacelle]

Well here (1,0,-1) and (2,1,-2) are linearly dependent of (-2,0,2) and (0,3,0)

So a basis of the space becomes ((-2,0,2),(0,3,0))

They are the only two vectors that are not linearly dependent and that are generator of the column space

So you could say that Col(A)={a(-2,0,2)+b(0,3,0), a,b ∈ R}

or Col(A)=vectR((-2,0,2),(0,3,0)) which means the same thing

[topaz nacelle]

Well yes and no to test linear dépendance you indeed need to see if the linear combination of all the vectors equal to 0 implies the scalars equal to 0

Here I don’t think it’s necessary

for Null(A) we know it’s dimension is 1 because of the rank theorem

[blazing fjord]

i see

would using (1,0,-1) instead of (-2,0,2) also be valid

they seem to be interchangable to me

[topaz nacelle]

Yes it would

[blazing fjord]

okay i see

for b, k = 3, and for c, k = 4

[topaz nacelle]

Yep

[blazing fjord]

so in a matrix

row x column

is the column space dimension the same as the # of rows

and null space dimension the same as the # of columns

[topaz nacelle]

The number of linearly independent rows or columns, both are equivalent

[topaz nacelle]

For the null space you can use the rank theorem to determine its dimension

Or just solve AX=0

With X=(x,y,z,t) here