Show that $\{u_1,...,u_n\}$ are linearly independent.

Question

Let $T:U\rightarrow W$ be a linear transformation and assume that $\{T(u_1),...,T(u_n)\}$ are linearly independent vectors in $W$.

a) Show that $\{u_1,...,u_n\}$ are also independent.

b) Assume that $\{v_1,...,v_n\}$ are linearly independent vectors in $U$. Can you with absolute confidence say that $T(v_1),...,T(v_n)$ are linearly independent? why/why not?

So to show that some vectors are linearly independent, I need to show that the only solutions to the equation

$$\sum_{i=1}^n\lambda_iu_i=0$$

is the trivial solution $\lambda_i=0 \ \forall \ i.$

I'm confused by the transformation notation. $T$ is a transformation such that it takes a vector in $U$, applies some change and spits it out in $W$. How should I relate $T(u_i)$ to $u_i$? I don't understand how this is to be done.

Answer 1

It is easier to prove the contrapositive. The image of a linearly dependent set is linearly dependent. Hint: the $u_i$ satisfy $\sum \lambda_i u_i = 0$ for some non-trivial choice of $\lambda_i$. Applying $T$ to both sides, what are you allowed to do?

Answer 2

a) So

$$T(\sum_{i=1}^n\lambda_iu_i)=T(0)$$ so since $T$ is linear we have $$\sum_{i=1}^n\lambda_iT(u_i)=0$$

b) Say $Tx=0$ for each $x$, then all $T(u_i)=0$. Are they linear independent?

Answer 3

a) Assume the $u_i$ are linear dependent. Then there is a $c \ne 0$ with $$ \sum_{i=1}^n c_i u = 0 $$ due to $T$ being linear this means: $$ T\left( \sum_{i=0}^n c_i u \right) = T(0) = T(0 \cdot 0) = 0 \cdot T(0) = 0 $$ and $$ T\left(\sum_{i=0}^n c_i u\right) = \sum_{i=0}^n c_i T(u_i) $$ must vanish as well, which violates that the $T(u_i)$ are linear independent.

b) This depends on $\dim W$.

If $\dim W = n$ then the $u_i$ form a basis and $T$ would map the $u_i$ to $n$ linear independent vectors, so it must have $\DeclareMathOperator{\img}{img}\dim \ker = n - \dim \img T = n - n = 0$.

If the $T(v_i)$ were linear dependent, then there was a $c' \ne 0 $ with $$ \sum_{i=1}^n c'_i T(v_i) = 0 $$ but $$ \sum_{i=1}^n c'_i T(v_i) = T\left(\sum_{i=1}^n c'_i v_i \right) $$ as the $v_i$ are linear independent the last argument can only vanish for $c'=0$, so as $c' \ne 0$ it is not zero which violates that $T$ has only the null vector in its kernel. So the $T(v_i)$ are linear independent as well.

If $\dim W > n$, e.g. $\dim W = n+1$, then we could chose $n$ linear independent $v_i$, where $v_1 \not\in \langle u_i \rangle$ and there would be room for a $T$ which maps the $u_i$ to $n$ linear independent $T(u_i)$ but $T(v_1) = 0$. So the $T(v_i)$ would not form $n$ linear independent vectors.