Question

suppose that u and v are linearly independent vectors. show that 2u+3v and u+v are linearly independent

Answer 1

suppose that they are dependent and derive a contradiction

Answer 2

What do you mean?

Answer 3

assume there exists numbers \(a\) and \(b\) such that

\[a(2u+3v)+b(u+v)=0\]

then show that this leads so a contradiction with the fact that u and v are independent

Answer 4

with a,b not both zero

Answer 5

what do you mean by contradiction though? I have zero idea :\

Answer 6

get a result that is not possible.. [for example 0=1 is a contradiction]

Answer 7

take

\[a(2u+3v)+b(u+v)=0\]


show that a and b have to be zero

Answer 8

So like I just sub in random numbers to see?

Answer 9

no...
assume \(a,b\neq\)0 (or at least on of them is non-zero)
\[a(2u+3v)+b(u+v)=0\]

\[\Rightarrow (2a+b)u+(3a+b)v=0\]

since \(u,v\) are independent then \(2a+b=0\) and \(3a+b=0\)

now show that a,b are both zero

Answer 10

well b=-3a and -2a so wouldn't that mean it would have to be 0? is that what you're getting at?

Answer 11

subtract
\[\begin{array}{c}
2a+b=0\\
3a+b=0\\\hline
-a=0
\end{array}\]

thus a=0

if a=0 then \(2a+b=0\Rightarrow 2(0)+b=0\Rightarrow b=0\)

Answer 12

oh okay so that shows that they must be independent. Thanks!