Question

Let U = Span({<1,0,0,0>,<0,1,0,0>}) and V = Span({<1,1,0,0>,<0,0,1,1>}) be subspaces of R^4.
a) Give a basis for U union V.
b) Give a basis for U intersection V.

Answer 1

what have you tried?

Answer 2

i know i've to calculate the union and the intersection of the 2 subspaces and then calculate the basis. Not sure how to calculate the union and intersection of subspaces.

Answer 3

if I said that W=Span({<1,0,0,0>,<0,1,0,0>,<1,1,0,0>,<0,0,1,1>})

could you find a basis for W?

Answer 4

the basis for W will be the set of linearly independent vectors in W

Answer 5

it would be the maximal linearly independent vectors in W...what would those be?

Answer 6

in this case <1,0,0,0>,<0,1,0,0> and <0,0,1,1>

Answer 7

that is your answer to part (a)

Answer 8

\[W=U\cup B\]

Answer 9

but how do i find the union given any set of vectors U and V?

Answer 10

\[U\cup V=\{w\in\mathbb{R}^4|w\in U\text{ or }w\in V\}\]

Answer 11

now that I look at it I don't believe there is a basis for the first part

Answer 12

the union of vector spaces is not always a vector space...the intersection is though

Answer 13

and intersection?
if i say the intersection of U and V is {<1,0,0,0>,<0,1,0,0>,<0,0,1,1>}...am i right in saying so?

Answer 14

the union of U and V does not include the vector <1,0,1,1>

Answer 15

so <1,0,0,0>,<0,1,0,0> and <0,0,1,1> cannot be a basis

Answer 16

do you see what I'm saying?

Answer 17

so, was i wrong in calculating a?

Answer 18

(a) has no answer since \(U\cup V\) is not a vector space. a basis is in reference to a vector space

Answer 19

from wiki..."A basis B of a vector space V over a field F is a linearly independent subset of V that spans V."

Answer 20

\[U\cap V\) is a vector space so we can find a basis for it

Answer 21

\[U\cap V\]

Answer 22

was the above the complete wording of the problem?

Answer 23

yes.

Answer 24

ok..so no answer for (a)

Answer 25

can you figure out \(U\cap V\)

Answer 26

\[U\cap V=\{w\in \mathbb{R}^4|w\in U \text{ and }w\in V\}\]

Answer 27

is it {<1,0,0,0>,<0,1,0,0>,<0,0,1,1>}?

Answer 28

no

too big

Answer 29

oops

Answer 30

give me a moment here

Answer 31

i'm sorry. is it {<1,0,0,0>,<0,1,0,0>,<1,1,0,0>}?

Answer 32

or null?

Answer 33

too big

is \(<1,0,0>\in V\)

Answer 34

no

Answer 35

should be zero space. right?

Answer 36

it is not the null set...the intersection of two vector spaces will contain the at least the zero vector

Answer 37

too small ;)

Answer 38

lol

Answer 39

so, b doesn't have a basis as well?

Answer 40

it does

Answer 41

zero vector itself?

Answer 42

more than that

Answer 43

you lost me here.

Answer 44

notice that the vector <1,1,0,0> is in both U and V

Answer 45

it's not.

Answer 46

U = Span({<1,0,0,0>,<0,1,0,0>}) and V = Span({<1,1,0,0>,<0,0,1,1>})

Answer 47

it is <1,0,0,0>+<0,1,0,0>=<1,1,0,0>

Answer 48

so \[<1,1,0,0>\in U\]

Answer 49

and <1,1,0,0> is a basis vector of V so it is in V

Answer 50

do you understand?

Answer 51

in that case coming back to (a), why does the union of U and V does not include the vector <1,0,1,1>? <1,0,0,0>+<0,0,1,1> = <1,0,1,1> right?

Answer 52

because <1,0,0,0> is in U and <0,0,1,1> is in V

Answer 53

they are not in the same set

Answer 54

but W contains both the vectors right

Answer 55

W contains the union of the two vector spaces but not the span of the combined vectors

Answer 56

remember \(U\cup V=\{w\in\mathbb{R}^4|w\in U\text{ or }w\in V\}\)

if <1,0,1,1> was in the union then either <1,0,1,1> is in U or <1,0,1,1> is in V. But it is not

Answer 57

so, W is not Span({<1,0,0,0>,<0,1,0,0>,<1,1,0,0>,<0,0,1,1>}) or is it just {Span({<1,0,0,0>,<0,1,0,0>}),Span({<1,1,0,0>,<0,0,1,1>})}?

Answer 58

\[W=Span({<1,0,0,0>,<0,1,0,0>})\cup Span({<1,1,0,0>,<0,0,1,1>})\]

which is not the same as

Span({<1,0,0,0>,<0,1,0,0>,<1,1,0,0>,<0,0,1,1>})

Answer 59

hmmm. ok. i get it now. so the intersection of U and V is just <1,1,0,0> which is the same as the basis. correct?

Answer 60

\[U\cap V=Span({<1,1,0,0>})\]

Answer 61

yeah, i meant that

Answer 62

and yes <1,1,0,0> is a basis vector

Answer 63

I'd ask your prof if the first part was a typo

Answer 64

great!! and if they ask to calculate orthonormal basis, how do i go about it?

Answer 65

you could use gram-schmit

Answer 66

or if it is just for this problem then just normalize <1,1,0,0>

Answer 67

I guess I spelled it wrong ...gram-schmidt

Answer 68

so, i calculate the basis and then run gram-schmit if it's the dimension is greater than 1?

Answer 69

yes...you could do that

Answer 70

ok. great. thank you so much.

Answer 71

np