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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?
m>=r m=3 so 0=<r=<3 dim N(A) = n-r=5-r 2<=5-r<=5
You're using the notation where m is the number of rows and r is the rank, right?
yes m=#rows n=#columns r=rank
So m >= r always holds for the nullspace of a matrix?
rank is #pivot rows right? can it be more than #of rows?
it is just a fact for every matrices
i mean you can't have more pivot 'rows' than rows
No, it cannot. I now get the answer to that sub-question.
I mean my sub-question not part (b). Let me look at 3(b) again
Also is dim N(A) = dim ("the x variable vector in Ax = 0")?
i don't know whether you've heard the mit ocw lecture or not, but you should know dim N(A) = n -r
matrix is 3by 5 so n=5
I haven't. Is N(A) = n - r something I am supposed to "just memorize"?
well dim N(A) is number of vectors in basis of nullspace of A
think about the basis of nullspace of A think how you get those bases
(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.)
(My break will be when I sleep.)
can you think how you obtain the null space of A?
then it is clear cut that dim N(A) = n-r
you find the vector x in Ax = 0
yeah i'm asking do you know the exact procedures
get A^(-1) and multiply both sides?
i mean left multiplication on each side of the equation
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
you can do that only when a is invertible
you use elimination process to compute nullspace right?
why don't you solve some examples to get dimN(A)=n-r i think you know the definitions
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?
nullspace(A) is a subspace in R^n, which a vector in it satisfies Ax=0
it is just whole solutions to Ax=0
solutions form a subspace, so they are called null'space'
ok so it's a dimension of a vector space in which a vector x makes Ax = 0?
for example, when A is invertible, x is only 0 and the N(A)={0}
no nullspace is just a name of subspace
dim N(A) is what you are saying
ya for two seconds, my brain was thinking about dim N(A).
Will you be here in 1..75 hours?
you need to carry on the elimination yourself to see dim N(A) = n-r
(If you keep writing, I will come back and read what you said.)
i have a pdf about nullspace. do you want it?
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
thanks and sry for leaving abruptbly