Lin. dep.

**dwsmith** · September 26th 2011, 02:01 PM

Let $f:X\to Y$ be a linear transformation. If $x_1,\cdots , x_n$ are lin. dep., show that $f(x_1),\cdots , f(x_n)$ are lin. dep.

Since $x_1,\cdots , x_n$ are lin. dep., there exist an $x_i=\sum_{j\neq i}\lambda_j x_j$ where $i=1,\cdots ,n$ .

$f(x_i)=f\left(\sum_{j\neq i}\lambda_j x_j\right)=\sum_{j\neq i}\lambda_j f(x_j)$

$f(x_i)=\sum_{j\neq i}\lambda_j f(x_j)\Rightarrow 0=\left[\sum_{j\neq i}\lambda_j f(x_j)\right]-f(x_i)$

Which is a nontrivial sol. since not all $\lambda_i$ are equal to zero.

Correct?

**FernandoRevilla** · September 26th 2011, 07:22 PM

Originally Posted by dwsmith

Correct?

Yes, it is correct.

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