Question

Suppose that a matrix A is formed by taking n vectors from R^m as its columns

(a) If those vectors are linearly independent, what is the rank of A and what is the relationship between m and n?

(b) If these vectors span R^m instead, what is the rank of A and what is the relationship between m and n?

(c) If these vectors form a basis for R^m, what is the relationship betweemn m and n then?

Answer 1

rank is the same thing as the dimension of space spanned by the row vectors

Answer 2

right?

Answer 3

hi?

Answer 4

earth to matrix girl

Answer 5

come in

Answer 6

sorry haha I was afk @518nad

Answer 7

@518nad , yes i believe that is right. But how does that definition come into play when those n vectors are linearly independent? how are m and n related?

Answer 8

n linearly dependant vectors from r^m

Answer 9

a single one of these n vectors will look like
n1=<c1,c2,c3,....,cm>
now if every single one of these n vectors are linearly independant at most u can select m linearly dependant n vectors

Answer 10

and what happens to te rank?

Answer 11

okay so rank is the same thing as the dimension of space spanned by the row vectors

Answer 12

if u pick 2 n vectors then u are gonna have m rows but only 2 colums, so at most it can be dim 2, in row land only 2 are needed

Answer 13

so the number of n vectors will keep determining the dimension, until n=m

Answer 14

Oh i see. Now what do u mean by "now if every single one of these n vectors are linearly independant at most u can select m linearly dependant n vectors" sorry if im having a hard time understanding, i'm still trying to grasp it

Answer 15

no worries take ur time

Answer 16

I know that Rank(A) = dimRS(A) = dimCS(A)

When it's saying the n vectors, i suppose that means columns? and so the rank would be dimCS(A)? and will the other one be that the relationship of n and m is thay they are both equal?

Answer 17

in part b) yeah n = m

Answer 18

as discussed above, when n =m, uve selected vectors in r^m space for each column, and theres m rows of them, and theyre all lin dependant filling out the r^m space,

Answer 19

theres m rows and m columns, the rows is obvious, shouldnt have restated that

Answer 20

okay.

in part b) would the rank be full rank if they all span one another?

Answer 21

i mean if the vectors span the R^m space?

Answer 22

yeah max rank possigble

Answer 23

okay. And for part c) since the vectors can form a basis, that means that they are linearly independent right. So would my answer be the same as what my answer for part a would be in terms of the relationship of n = m?

Answer 24

the relationship of n and m * ? sorry

Answer 25

in a n doesnt not have to equal m

Answer 26

so n can be greater than or equal to, less than or equal to m ?

Answer 27

in a the dim = n, until n=m

Answer 28

n cannot be greater than m in a because u cannot pick more linearly independent vectors than m, since the vectors n are of r^m

Answer 29

okay that makes more sense as to why it cannot be greater.

My professor gave us Thm 3.82. he said it would help out to think about this problem
(i) dimRS(A) + dimNS(A) = n
(ii) dimCS(A) + dinNS(A) = n
(iii) Rank(A) + dimNS(A) = n

when you mentioend that dimension = n until n = m, is that related to any of those above?

Answer 30

i meant to say, it would help out to be able to answer parts a - c. I'm not used to the abstractness. Especially in math!

Answer 31

whats the defn of dim ur prof gave u

Answer 32

actually he never said. The only thing i understand about dim is that there's 3d 2d 4d etc etc haha

Answer 33

ah i saw one on my notes it says
"The dimension of vector space V is the numbers of vectors in any basis."

Answer 34

what wud u say the dimRS is for this matrix
1.2.3
1.1.1

Answer 35

is taht R2?

Answer 36

yes

Answer 37

okay yes he did say that dimRS is the # of columns with pivots, but he never told us that beforehand. It kinda just slipped through

Answer 38

yes it makes sense

Answer 39

for example if dimRS(A) = m
For an m by m matrix, the null space is 0, because there will be no combination of the rows of A such that it will add to 0, other than the origin 0,0,0,0,,0 , which is considered dim 0

Answer 40

yes i see that

Answer 41

if dimRS(A) = n-1, then that means there are only n-1 independant rows, and u can imagine one set of linear combinations of those rows to give u the last row, so the null space would be just some n dimensional line vector

Answer 42

okay that makes sense. thanks for the visual idea. Its easier to discern that way.

Answer 43

unfortunately i have to sleep now . its already late where i live, but ill be back tomorrow, so thank you for the help tonight. ill get back on here to ask some more thigs if i dont understand still.

Answer 44

okay see ya :)