Question:
If A is a 3 by 5 matrix, what information do you have about the nullspace of A?
Answer:
N(A) has dimension at least 2 and at most 5.
My trouble:
How do we know this?

- s3a

- schrodinger

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- anonymous

m>=r
m=3 so 0=

- s3a

You're using the notation where m is the number of rows and r is the rank, right?

- anonymous

yes
m=#rows
n=#columns
r=rank

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## More answers

- s3a

So m >= r always holds for the nullspace of a matrix?

- anonymous

rank is #pivot rows right?
can it be more than #of rows?

- anonymous

it is just a fact for every matrices

- anonymous

i mean you can't have more pivot 'rows' than rows

- s3a

No, it cannot. I now get the answer to that sub-question.

- s3a

I mean my sub-question not part (b). Let me look at 3(b) again

- s3a

Why is dim N(A) = n-r=5-r?

- s3a

Also is dim N(A) = dim ("the x variable vector in Ax = 0")?

- anonymous

i don't know whether you've heard the mit ocw lecture or not, but you should know dim N(A) = n -r

- anonymous

matrix is 3by 5 so n=5

- s3a

I haven't. Is N(A) = n - r something I am supposed to "just memorize"?

- s3a

dim N(A) = n - r that is

- anonymous

well dim N(A) is number of vectors in basis of nullspace of A

- anonymous

think about the basis of nullspace of A
think how you get those bases

- s3a

(where n and r are of the matrix A) (Sorry for the potentially dumb question, I'm kind of braindead at the moment but, like I said, I cannot take a break.)

- s3a

(My break will be when I sleep.)

- anonymous

can you think how you obtain the null space of A?

- anonymous

then it is clear cut that dim N(A) = n-r

- s3a

you find the vector x in Ax = 0

- s3a

?

- anonymous

yeah i'm asking do you know the exact procedures

- s3a

get A^(-1) and multiply both sides?

- s3a

(on the left)

- s3a

i mean left multiplication on each side of the equation

- anonymous

since there are r pivot columns, there are n-r free columns and there are n-r special solutions to the null space. And they form basis for nullspace. so dim N(A)=n-r

- anonymous

you can do that only when a is invertible

- anonymous

you use elimination process to compute nullspace right?

- anonymous

why don't you solve some examples to get dimN(A)=n-r
i think you know the definitions

- s3a

when a is invertible, like you said.
I'm not grasping something fundamental though: what is the nullspace of A? is it x in Ax = 0? In other words, what object's dimension is n - r?

- anonymous

nullspace(A) is a subspace in R^n, which a vector in it satisfies Ax=0

- anonymous

it is just whole solutions to Ax=0

- anonymous

solutions form a subspace, so they are called null'space'

- s3a

ok so it's a dimension of a vector space in which a vector x makes Ax = 0?

- anonymous

for example, when A is invertible, x is only 0 and the N(A)={0}

- anonymous

no nullspace is just a name of subspace

- anonymous

dim N(A) is what you are saying

- s3a

ya for two seconds, my brain was thinking about dim N(A).

- s3a

Will you be here in 1..75 hours?

- anonymous

you need to carry on the elimination yourself to see dim N(A) = n-r

- s3a

I have to go eat now.

- anonymous

i'm sorry

- s3a

(If you keep writing, I will come back and read what you said.)

- anonymous

i have to go :(

- s3a

o :(

- anonymous

i have a pdf about nullspace. do you want it?

- anonymous

http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/solving-ax-0-pivot-variables-special-solutions/
download the lecture summary for that lecture.
it has the just the right information for you.
good luck

- s3a

thanks and sry for leaving abruptbly

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