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Here is my homework question,

Let $V$ and $W$ be vector spaces over a field $F$ and let $T$ from $V$ to $W$ be an isomorphism. Let $X$={$v_{1},v_{2},...,v_{n}$} be a subset of V, and recall that T(X)={$T(v_{1}),T(v_{2}),...,T(v_{n})$}.

a. Prove that if $X$ is linearly independent, then $T(X)$ is also linearly independent.

b. Prove that if $X$ spans $V$, then $T(X)$ spans $W$.

I am trying to solve a. but I don't know how to use isomorphism here... My work is.. Suppose X is linearly independent. Then for all $a_{1}v_{1},a_{2}v_{2},...,a_{n}v_{n}$ = 0. So, $a_{1},a_{2},...,a_{n}$ = 0. but I can't go further...

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By definition, isomoporhisms are injective (i.e. monomorphisms) and surjective (i.e. epimorphisms). – F M Apr 19 '12 at 5:24
@FernandoMartin Yes, I know that – yoodaniel Apr 19 '12 at 5:25
As for your work, if $X$ is linearly independent, then $a_1v_1,\dots,a_nv_n=0$ iff $a_1,\dots,a_n=0$, but $a_1v_1,\dots,a_nv_n=0$ certainly doesn't hold for all vectors (unless, of course, your space is the trivial vector space). – F M Apr 19 '12 at 5:26
Judging from this and your previous question, and the discussion that ensued, you really, really, really need to go learn the basic definitions. You need to know exactly what linearly independent means, to begin with. – Gerry Myerson Apr 19 '12 at 5:29
Maybe you do, but your writing "Then for all $a_1v_1,a_2v_2,\dots,a_nv_n=0$. So, $a_1,a_2,\dots,a_n=0$" suggests quite the opposite. – Gerry Myerson Apr 19 '12 at 5:41
up vote 1 down vote accepted

I understand that this is a homework question. However based on your previous question I am willing to give a little more help to help you get the ball rolling and basic ideas in these problems. I will help you out with $(a)$ because $(b)$ almost immediately follows from the definitions of $T$ being an isomorphism (in particular that it is surjective).

For $(a)$, you have an implication $p \implies q$, where $p$ and $q$ are respectively the statements

$p$: $X$ is linearly independent

$q$: $T(X)$ is linearly independent.

If you don't know about the contrapositive yet, it is a powerful way of proving statements. In your context it says that proving $p \implies q$ is equivalent to proving $\neg q \implies \neg p$ where the "$\neg$" symbol is logical negation.

So suppose we know $\neg q$. That is, we assume that the vectors in the collection $T(X)$ are linearly dependent. Now you will out the following details:

1) What does it mean for the vectors in $T(X)$ to be linearly dependent? You just need to apply the definition of linear dependence.

After you have written that down, using linearity of $T$ this implies that (.....)? You can fill in $(\ldots \ldots )$ here by looking at question 2) below:

2) Is there a vector in the kernel of $T$ now? Proceed to 3) below.

3) Using the fact that $T$ is an isomorphism can you conclude from here that $X$ is linearly dependent? Remember we started of with $\neg q$ and we want to prove $\neg p$, where $\neg p$ is the statement "$X$ is linearly dependent."

$\textbf{Edit:}$ Here is my response to your solution below: You cannot start with saying " suppose that $a_1 = a_2 = \ldots a_n = 0$." This is non-sensical. In fact in your proof the only thing you've done is showing that $T$ maps zero to zero. This is because you started of with saying that all the $a_i's$ were zero, so that $T$ applied to any linear combination involving these $a_i's$ will be zero anyway. Do you see this does not lead to a proof of what you want to prove?

What you need to do is this: Suppose we have a linear combination of the $T(v_i)'s$ that gives us zero. Viz. there are scalars $a_1 , a_2, a_3, \ldots a_n$ such that

$$a_1T(v_1) + \ldots a_nT(v_n) = 0.$$

Remember: You want to conclude from here, using information about $X$, that this linear combination can only be the trivial linear combination. Now by linearity of $T$ this means that the vector $a_1v_1 + \ldots a_nv_n$ is in the kernel of $T$. But then $T$ is an isomorphism so this means that $a_1v_1 + \ldots a_nv_n = 0$.

Do you see how to prove it correctly now?

Now because $X$ is linearly independent, this forces us to conclude that $a_1 = a_2 = \ldots = a_n = 0$.

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How can I post my solution? – yoodaniel Apr 19 '12 at 7:06
@yoodaniel That is also possible, but then again there can be many ways to solve a problem! Now post your solution as an answer and I will check it. – user38268 Apr 19 '12 at 7:07
Sorry I can't post answer yet because of the reputation. but here is my answer. $a_{1},...,a_{n}$= 0. So $a_{1}v_{1}+a_{2}v_{2}+...+a_{n}v_{n}$= 0. Then, $a_{1}v_{1}+a_{2}v_{2}+...+a_{n}v_{n}$ is an element in {0}. So, $a_{1}v_{1}+a_{2}v_{2}+...+a_{n}v_{n}$ is in Ker(T). T($a_{1}v_{1}+a_{2}v_{2}+...+a_{n}v_{n}$)= 0. So $a_{1}T(v_{1})+a_{2}T(v_{2})+...+a_{n}T(v_{n}) = 0$ . Thus $T(X)$ is a linearly independent set. – yoodaniel Apr 19 '12 at 7:19
Maybe I need to start opposite way... – yoodaniel Apr 19 '12 at 7:27
@yoodaniel Please see my response to your proof above. – user38268 Apr 19 '12 at 23:47

To expand on Martin's Hint:

Try to prove the following:

Let $V$ and $W$ be vector spaces over a field $\Bbb F$; let $T$ be a map from $V$ to $W$.

  1. The following are equivalent:

    • $T$ is injective.
    • $T$ takes each linearly independent set in $V$ to a linear independent set in $W$.
  2. The following are equivalent:

    • $T$ is surjective.
    • $T$ takes each spanning set of $V$ to a spanning set of $W$.
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ok I am on it.... – yoodaniel Apr 19 '12 at 5:33
I'd suggest rewriting, something like, "$T$ takes each linearly independent set...," lest someone think it's enough to show that $T$ takes some linearly independent set to a linearly independent set. – Gerry Myerson Apr 19 '12 at 5:46
@GerryMyerson I agree. I will make it clear. – user21436 Apr 19 '12 at 5:47

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