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I'm not really sure how to prove this but I know that I need to show that this map defines a norm on $X$, so that I must show the following:

(1) $||x|| = 0 \iff x=0$

(2) $||ax|| = |a| ||x||$ for every $a$ in $F$, $x \in X$

(3) $|| x + y || \leq ||x|| + ||y||$ for any $x,y \in X$

I know since $T$ is linear that $T(x+y) = T(x) + T(y)$ for all $x,y \in X$ and $T(\lambda x) = \lambda T(x)$ for all $\lambda \in F$. So that (2) is satisfied since $||T(\lambda x)|| = || \lambda T(x) || = |\lambda| ||T(x)||$. Not really sure how to go about (1) or (3). Thank you.

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    To make this question somewhat more self-contained, you could preface it by saying: "Let $X$ be a vector space, $Y$ a normed space with norm $\|\cdot\|$, and let $T:X\to Y$ be linear and injective." – 2017-01-17

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To avoid some confusion, I will write $\|x\|'=\|Tx\|$ for $x\in X$, then give some hints to show that $\|\cdot\|'$ is a norm on $X$.

(1): Suppose $\|x\|'=0$ for some $x\in X$. Then $\|Tx\|=0$, and thus $Tx=0$. What can you say now?

(2): You've already shown this.

(3): Suppose $x,y\in X$. Then we have $$ \|x+y\|'=\|T(x+y)\|=\|Tx+Ty\|. $$ From here it's pretty clear what to do.

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    $|| x+y ||' = || T(x+y) || = || Tx + Ty || \leq || Tx || + ||Ty || = ||x||' + ||y||'$ Is that right for (3) since the triangle inequality can be applied since its in the range space and $Y$ is a normed space so that the triangle inequality holds? – 2017-01-17
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    Then (1) follows since $Tx = 0$ and $T$ is injective we have that $x = 0$ – 2017-01-17
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    Very helpful by the way! Thank you. – 2017-01-17
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    @JohnSmith You got it. And you're welcome – 2017-01-17