Question

show that the set S={(a,b,b,a+b)|a,b,c are element of real numbers} is a subspace of R^4(4-dimension).........plz plz

Answer 1

check for closure under addition and multiplication

Answer 2

\[\vec u=\{a,b,b,a+b\};a,b\in\mathbb R\]\[\vec v=\{x,y,y,x+y\};x,y\in\mathbb R\]\[\vec u+\vec v=\{a+x,b+y,b+y,a+x+b+y\}=\{p,q,q,p+q\}=\vec w;\vec w\in S\]so it's closed under addition

Answer 3

zero vectortest as well

Answer 4

I messed up closure under multiplication
I have heard that spaces like \(\mathbb R^4\) don't require the zero vector test because the zero vector is in all subs[paces of \(\mathbb R^n\) but that may be wrong...

Answer 5

if the subspace does not pass thru the origin; then it doesnt possess the zero vector

Answer 6

yeah, I didn't even word it right :/
and I can't seem to show closure under multiplication

Answer 7

i believe the top is a span, while the bottom is a proper subspace

Answer 8

is multiplication a dot product?

Answer 9

\[\vec u=\{a,b,b,a+b\};a,b\in\mathbb R\]\[\vec v=\{x,y,y,x+y\};x,y\in\mathbb R\]\[\vec u+\vec v=\{ax,by,by,ax+ay+bx+by\}\neq\{p,q,q,p+q\}\]I was just multiplying the respective entries...

Answer 10

\[ S=\{(a,b,b,a+b)~|~a,b,c \in R\}\]is this a typo by chance? or is c spose to be missing?

Answer 11

that should be a times in my last line...

Answer 12

ohhh sorry, it is supose to be like S={(a,b,b,a+c)}

Answer 13

ahh

Answer 14

<0,0,0,0+0> is the zero vector

Answer 15

\[\vec u=\{a,b,b,a+c\};a,b\in\mathbb R\]\[\vec v=\{x,y,y,x+z\};x,y\in\mathbb R\]\[\vec u+\vec v=\{a+x,b+y,b+y,a+x+(c+z)\}=\{p,q,q,p+r\}=\vec w;\vec w\in S\]\[\vec u\cdot\vec v=\{ax,by,by,ax+(cx+az+cz)\}=\{p,q,q,p+r\}=\vec w;\vec w\in S\]