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The linear transformations $\def\R{\Bbb R}T_1: \R^4 \to\R^4$ and $T_2 : \R^4 \to\R^4$ are represented by the $4\times4$ square matrices $M_1$ and $M_2$. It is given that $\dim ( \operatorname{range of} T_1 ) = 3$, while nullity of $T_2 = 2$. It is also given that the null space of $T_2$ is a subspace of the range of $T_1$.

Prove that $\dim (\operatorname{range of} T_3 ) = 1$, where $T_3$ is the result of applying $T_1$ and then $T_2$ exactly in that order.

An attempt:

Essentially, the problem boils down to proving that the set $(Mv_1, Mv_2, Mv_3 )$ is not linearly independent, where $(v_1,v_2,v_3)$ are a basis for the range of $T_1$. One way of doing this is to assume that $(v_2,v_3)$ belong to the null space of $T_2$, which quickly gives us the answer. But isn't there any other method, apart from this and finding the product $M_2M_1$? Can't we prove that $(Mv_1,Mv_2,Mv_3)$ are linearly dependent without assuming that two out of $(v_1,v_2,v_3)$ belong to $N(T_2)$?

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    $\begingroup$ I've made the question typographically readable, but it still makes no sense. Linear transformations do not have a dimension (sub)space do; di you mean their rank instead (which is the dimension of their image subspace)? Linear transformation do not have subspaces (so $\ker(T_2)$ cannot be a subspace of $T_1$, did you mean the image of $T_1$ here? Nor do linear transformation have bases (again it is (sub)spaces that have them). $\endgroup$ May 29, 2017 at 5:02
  • $\begingroup$ Yes, everything you've said is correct. $\endgroup$
    – user440261
    May 29, 2017 at 5:37

1 Answer 1

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Yes. this result holds independently of how your basis vectors are chosen. You can prove this using (guess what...) the rank nullity theorem. The image (range) of the composed map of $T_1$ and $T_2$ is the same as the image of the restriction $T_2'$ of $T_2$ to the image $W$ of $T_1$; it is $T_2'$ that I'll apply the theorem to. It is given that $\dim(W)=3$ and that $\ker(T_2)\subseteq W$; therefore the $\ker(T_2')=\ker(T_2)\cap W=\ker(T_2)$ which it is given has dimension$~2$. Then by rank nullity the image of $T_2'$ has dimension $3-2=1$.

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  • $\begingroup$ What is T2'? And why does ker (T2') = ker ( T2) intersection W = ker (T2)? $\endgroup$
    – user440261
    May 29, 2017 at 12:26
  • $\begingroup$ Well $T_2'$ is the name I gave to the restriction mentioned immediately before the first occurrence of "$T_2'$". And the intersection there is one between a subset of $W$ and $W$, which is just that subset. $\endgroup$ May 29, 2017 at 12:50
  • $\begingroup$ What do you mean by restriction? $\endgroup$
    – user440261
    May 29, 2017 at 14:20
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    $\begingroup$ Restriction is the process of considering a given map only on a part (subset) of its domain. In linear algebra it is almost always about restricting to a subspace of the whole space. One just ignores what that map did on elements outside that subspace, but for those that are in the subspace one does not change what it is mapped to. So the restriction of $f:V\to W$ to a subspace $S$ of $V$ is $g:S\to W$ where $g(s)=f(s)$ for all $s\in S$. The operations is interesting because the restriction may have a different range and/or nullspace than the original map. $\endgroup$ May 29, 2017 at 15:07
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    $\begingroup$ The kernel of the restriction of a map $f$ to a subspace $W$ is the set of vectors $w\in W$ such that $f(w)=0$. Since $\ker(f)$ is the set of all vectors $v$ with $f(v)=0$, the kernel of the restriction is $\ker(f)\cap W$. $\endgroup$ May 29, 2017 at 20:54

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