If two linearly independent vectors are multiplied by 3 and 6, are they still linearly independent?

Question

Suppose $a$ and $b$ in a set are vectors that are linearly independent.

If you multiply vector $a$ by $3$ and and vector $b$ by $6$, are they still linearly independent?

And how do you prove this?

Answer 1

Unless one of the multipliers is zero, the multiplied vectors are still linearly independant. This follows quite directly from the definition of linear independance: Let $a,b$ be vectors and $u,v$ nonzero scalars. If $x,y$ are scalarss such that $xua+yub=0$ then, by linear independence of $a,b$, we have $xu=0$ and $yv=0$, hence $x=y=0$.

Answer 2

Suppose that vectors $a,b$ are linearly independed. Its mean that there is only trivial lineral combination $$ \alpha a +\beta b =0. $$ Multiply all by 6. $$ \alpha (6a) +\beta (6b) =0. $$ It means that $6a,6b$ also are linearly independed.

Answer 3

Multiplying vectors by a scalar only affects their module, but does not rotate them in the space we are working in (n-dimensional space).

This is easy to see as what you are doing when you multiply q*r (let q be a scalar and r a vector) is adding a number q of times the vector r. As all of them have the same direction the result of this operarion will be the a vector in the same direction which is larger or smaller than the original.

And as two vectors are lineary dependant when they have the same direction (we're talking of just two vectors, huh?) if they had at first a different direction they will, after the operation of multiplying by a scalar, have the same directions they had originally, which is the same as saying they are linearly independant.

This may be a more geometric than algebraic approach to the question, but maybe seeing things this way may help.