Question

Linear Algebra

Answer 1

So if I remember right you
just write this (x1,6x1+1x2,9x1+8x2,4x1+2x2)^T
as
x1( __,__,__,__)^T+x2(__,__,__,__)^T
whatever two vectors you get in these blanks is the vectors that span your subspace.

Answer 2

\[(1 \cdot x_1 +0 \cdot x_2,6 \cdot x_1 +1 \cdot x_2, 9 \cdot x_1 +8 \cdot x_2, 4 \cdot x_1 + 2 \cdot x_2 )^T\]
=\[x_1( ?,?,?,?)^T+x_2(?,?,?,?)^T \]
figure out what ? marks go here

Answer 3

For the first ? we know we have in our main vector the first entry is 1x1+0x2
so
we know we have\[x_1( 1,?,?,?)^T+x_2(0,?,?,?)^T\]

Answer 4

see if you can finish the rest

Answer 5

ooh ok i got it
thanks

Answer 6

can you help me with one more
i can't figure out what the answer might be
also it can have more than one answers

Answer 7

it looks confusing

Answer 8

We must have:
u,v are in S then u+v is in S
u is in S and k is a real number then ku is in S
0 is in S

Answer 9

So for the first one can (0,0,0) ever be in that set?

Answer 10

hint the first entry is always -5 for each elements right?

Answer 11

so how can (0,0,0) ever be in that set?

Answer 12

what do you think? is (0,0,0) in the first set A?

Answer 13

when i said 0 had to be in S
0 could mean the 0 vector if we are working in R2 or R3 or Rmore than 3

Answer 14

please let me know if you don't know what I'm saying

Answer 15

yeah, im still confused

Answer 16

A
\[\{{(-5,y,z)|y,z \text{ are arbitrary numbers } } \}\]
This is our first set we are looking at

We must have the following in it for it to be a subspace of R3:

for every u and v in are set A we must have u+v is also in A
for every u in A and every scalar k in the real numbers we must also have ku in A
Last but not least we must have (0,0,0) in A.


All of our elements in A take on this form (-5,y,z) as said by the definition of the set A

Answer 17

We have 3 things we must have.
I'm looking at the easiest one first
which is,
is (0,0,0) in A?

Answer 18

Think about what the elements look like in A
They look like this (-5,y,z)
where the 2nd and 3rd entry can be any real number
but that first number is always -5

Answer 19

so A is not a subspace because it fails on the (0,0,0) check ?

Answer 20

right

Answer 21

now B
can 0+0+0 ever be 9?

Answer 22

nope

Answer 23

we say no so there is no way (0,0,0) is in B
so B is not a subspace

Answer 24

What do you think about C?

Answer 25

it is possible for C to be a subspace
it passes on (0,0,0)

Answer 26

ok now we need to look at property 2 or 3
let's look at propert 2 which says
for u in C and and scalar k
we have ku is in C
Suppose (a,b,c) is in C
Suppose k is a real number.
Since (a,b,c) is in C we have
-5a+8b-3c=0
Do we have
that (ak,bk,ck) also satisfies that equation.
So what I am asking is
do we have -5(ak)+8(bk)-3(ck)=0?

Answer 27

yes, i would think so

Answer 28

-5(1*1)+8(1*1)- 3(1*1) = 0
-5 + 8 - 3 = 0

Answer 29

yep because k(-5a+8b-3c)=0 if we factor the k
so that means k=0 or -5a+8b-3c=0
and we know the latter is true because we supposed it
no contradiction there
so we are good

now how about the first property

suppose (a1,a2,a3) and (b1,b2,b3) are in C
we want to see if (a1+b1, a2+b2,a3+b3) is also in C

Answer 30

that is we have the following are true
\[-5a_1+8a_2-3a_3=0 \text{ and } -5b_1+8b_2-3b_3=0\]
and we want to show the following:
\[-5(a_1+b_1)+8(a_2+b_2)-3(a_3+b_3)=0\]

Answer 31

that also checks out to be right

Answer 32

we could look at the last equation as this:
\[(-5a_1+8a_2-3a_3)+(-5b_1+8b_2-3b_3)=0+0=0\]
right
so C is indeed a subspace

Answer 33

now it looks like there may be multiple answers because it said which of the sets
it didn't say there was one answer
or i mean it isn't implied so we should check the other ones also

Answer 34

so set D
do we have (0,0,0)?

Answer 35

yes,
there is more than one answer

Answer 36

yes we do
now do you know how to check the other two properties?

Answer 37

yes, Thanks

Answer 38

tell me what you think about D when you are done analyzing it

Answer 39

D is also a subspace,
E is not a subspace
F is a subspace

Answer 40

that looks good
so you are saying C,D and F

Answer 41

yup