# Linear transformations and linear independance

• Mar 9th 2013, 09:44 AM
nabz09
Linear transformations and linear independance
Hi,

I was curious as to whether this statement was true or not, If it's true then can anyone show me a hint for the proof of it.

"If we have a linear transformation f:V -> V and if v(1),v(2),...,v(n) are linearly independant vectors in V, then does that imply that f(v(1)),f(v(2)),...f(v(n)) are also linearly independent in V. We are also told that f is injective, so f(v(k)) cannot equal 0, if v(k) is not equal to 0, since we know that with linear transformations f(0)=0."

v(i) means: v subscript i, for all i in [1,2,...,n],
• Mar 9th 2013, 02:08 PM
Nehushtan
Re: Linear transformations and linear independance
Suppose \$\displaystyle a_1f(v_1)+\cdots+a_nf(v_n)=0\$. As \$\displaystyle f\$ is a linear transformation, \$\displaystyle a_1f(v_1)+\cdots+a_nf(v_n)=f(a_1v_1+\cdots+a_nv_n)\$. Therefore \$\displaystyle f(a_1v_1+\cdots+a_nv_n)=0=f(0)\$. As \$\displaystyle f\$ is injective, this implies \$\displaystyle a_1v_1+\cdots+a_nv_n=0\$, and the linear independence of \$\displaystyle v_1,\ldots,v_n\$ implies \$\displaystyle a_1=\cdots=a_n=0\$
• Mar 9th 2013, 02:20 PM
HallsofIvy
Re: Linear transformations and linear independance
Of course, without the condition that the linear transformation is injective, this would be no longer true.