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These two are both very similar. The first is -Let $V$ and $W$ be finite dimensional vector spaces over a field $F$. Let $T:V\to W$ be a linear transformation. Suppose that $T$ is one-to-one. Show that there is a linear transformation $L:W\to V$ such that $LT=1v$? ----> So far, I have, If $T:V\to W$ then $B_v =\{v_1,\dots,v_n\}$ is a basis for $V$ and $B_w=\{w_1,\dots,w_m\}$ is basis for $W$. where $v=a_1v_1+\dots+a_nv_n$ and $T(v)=b_1w_1+\dots+b_mw_m$. confused where to go from there?

The second question is Let $T:V\to W$ and $L:W\to V$ be a linear transformation. Show 1.) $T$ is injective if $LT=1v$ and 2.) $T$ is surjective if $TL=1w$ For this one, all i know is that you have to show that $L$ is unique and that $L$ is linear.

Im really bad at linear transformations, so I could really use the guidance. Thanks!

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    $\begingroup$ This can help: drexel28.wordpress.com/2010/11/30/… (Note that the result is true for set-functions. You just need to check that the left-invese, defined as in the link, is indeed also linear transformation in your case) $\endgroup$ – user39280 Mar 8 '13 at 20:36
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You are looking for a left inverse, usually called a retraction. (Look at this.)

We need to define $L:W \to V$. For any $w$ in the image of $T$, so there is a $v \in V$ with $T(v) = w$, we can define $L(w) = v$. Why does that make sense? The transformation T is one-to-one, so the $v$ that we're using is unique.

What to do about $L(w)$ for $w \notin \operatorname{Im} T$? Hint: every $w = w_0 + w_1$ for some $w_0 \in \operatorname{Im} T$.

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  • $\begingroup$ Yes first part makes sense, but what does ImT mean? Isomorphism T? $\endgroup$ – kkkk Mar 8 '13 at 20:52
  • $\begingroup$ Im just not familiar with the Id_y in the link? $\endgroup$ – kkkk Mar 8 '13 at 20:56
  • $\begingroup$ Im T means "the image of T," which is the set of outputs of the function. $\endgroup$ – Sammy Black Mar 8 '13 at 22:49
  • $\begingroup$ Id_Y just means the identity function on Y. That is the function whose output is the same as input: Id_Y (y) = y for all y in Y. $\endgroup$ – Sammy Black Mar 8 '13 at 23:07
  • $\begingroup$ Im kinda confused on the last part there. However, if in fact T is one-to-one. The the nullity(T)=0 correct? So $\endgroup$ – kkkk Mar 8 '13 at 23:08

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