#*determine dimension of Nul A* Question 14
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Am I high or is the answer suppose to be that it has 3 dimensions for Nul because 3 free variables?
I was wondering if I was going to find the values of those free variables how would I do it?
Nullspace of A means to solve Ax = 0
dim(A) = rank(A) + nullity(A)
@tiny raft that seems helpful. However, rank is in the next chapter section and im not there yet to understanding this
@icy axle can you show me what you mean? step by step
The rank is equal to the number of pivots in the reduced row echelon form, and is the maximum number of linearly independent columns that can be chosen from the matrix.
so that would be 3?
for the pivots
i guess my next question is are col 5 and 6 independent or not? I know 2 isnt.
yeah
ig you mean for 14, theyre not multiples of each other
so i have my pivots (rank) and youre saying my number of linear independent columns is 5,
If you don't know rank yet, then you need to solve Ax = 0 and the number of variables in your solution set will tell you the dimension of Nul A
if i throw 0's in i guess that would give me a solution
they are free
you have infinite many solutions
it's just
x2, x5 and x6 can by anything
basically NulA is the vector space that is span by those 3 vectors
If you write it like:
x2[...] + x5[...] + x6[...]
it's basically the span
all linear combinations since x2, x5 and x6 are in R
so dimension would be also 3
of Nul A
let me right this out
in this example i was confused by the values they put into the vectors next to x2 and x4
{-2,1,0,0,0}
what i was trying to demonstrate here
x5 = 0 due to the system
-7x_5 = 0 -> x_5 = 0
also since x2 and x4 are the free variables we make every component in terms of these variables
𝔸dωn𝓲²s
𝔸dωn𝓲²s
that is spicy
𝔸dωn𝓲²s
And that's how the game works everytime, when doing these.
You solve some system and figure the dimension by counting the free variables
alright if im ever asked for the basis, I will know what to do. I will proatice this tomorrow morning.
dedication
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now do it lmao
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😂
sure!