s3a
  • s3a
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?
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
m>=r m=3 so 0=
s3a
  • s3a
You're using the notation where m is the number of rows and r is the rank, right?
anonymous
  • anonymous
yes m=#rows n=#columns r=rank

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s3a
  • s3a
So m >= r always holds for the nullspace of a matrix?
anonymous
  • anonymous
rank is #pivot rows right? can it be more than #of rows?
anonymous
  • anonymous
it is just a fact for every matrices
anonymous
  • anonymous
i mean you can't have more pivot 'rows' than rows
s3a
  • s3a
No, it cannot. I now get the answer to that sub-question.
s3a
  • s3a
I mean my sub-question not part (b). Let me look at 3(b) again
s3a
  • s3a
Why is dim N(A) = n-r=5-r?
s3a
  • s3a
Also is dim N(A) = dim ("the x variable vector in Ax = 0")?
anonymous
  • 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
  • anonymous
matrix is 3by 5 so n=5
s3a
  • s3a
I haven't. Is N(A) = n - r something I am supposed to "just memorize"?
s3a
  • s3a
dim N(A) = n - r that is
anonymous
  • anonymous
well dim N(A) is number of vectors in basis of nullspace of A
anonymous
  • anonymous
think about the basis of nullspace of A think how you get those bases
s3a
  • 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
  • s3a
(My break will be when I sleep.)
anonymous
  • anonymous
can you think how you obtain the null space of A?
anonymous
  • anonymous
then it is clear cut that dim N(A) = n-r
s3a
  • s3a
you find the vector x in Ax = 0
s3a
  • s3a
?
anonymous
  • anonymous
yeah i'm asking do you know the exact procedures
s3a
  • s3a
get A^(-1) and multiply both sides?
s3a
  • s3a
(on the left)
s3a
  • s3a
i mean left multiplication on each side of the equation
anonymous
  • 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
  • anonymous
you can do that only when a is invertible
anonymous
  • anonymous
you use elimination process to compute nullspace right?
anonymous
  • anonymous
why don't you solve some examples to get dimN(A)=n-r i think you know the definitions
s3a
  • 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
  • anonymous
nullspace(A) is a subspace in R^n, which a vector in it satisfies Ax=0
anonymous
  • anonymous
it is just whole solutions to Ax=0
anonymous
  • anonymous
solutions form a subspace, so they are called null'space'
s3a
  • s3a
ok so it's a dimension of a vector space in which a vector x makes Ax = 0?
anonymous
  • anonymous
for example, when A is invertible, x is only 0 and the N(A)={0}
anonymous
  • anonymous
no nullspace is just a name of subspace
anonymous
  • anonymous
dim N(A) is what you are saying
s3a
  • s3a
ya for two seconds, my brain was thinking about dim N(A).
s3a
  • s3a
Will you be here in 1..75 hours?
anonymous
  • anonymous
you need to carry on the elimination yourself to see dim N(A) = n-r
s3a
  • s3a
I have to go eat now.
anonymous
  • anonymous
i'm sorry
s3a
  • s3a
(If you keep writing, I will come back and read what you said.)
anonymous
  • anonymous
i have to go :(
s3a
  • s3a
o :(
anonymous
  • anonymous
i have a pdf about nullspace. do you want it?
anonymous
  • 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
  • s3a
thanks and sry for leaving abruptbly

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