# Number of linear maps is less or equal than the dimension of the vector space

## Problem

Let $V$ be a vector space of dimension $n$. Let $T_1,T_2,\dots,T_m$ be linear maps $V\rightarrow V$ such that $\dim R(T^{2}_{i})=\dim R(T_i)=1$ for all $i=1,2,\dots,m$, and $T_i\circ T_j$ is the zero map for all $i \neq j$. Prove that $m\leq n$.

## Thoughts

If I choose $v_i\in R(T_i),~v_i\neq 0~\forall~i=1,2,\dots,m$ and show that $T_i(v_i)=a_iv_i$ for some $a_i\neq 0$ and $T_i(v_j)=0~\forall~i\neq j$ . . .

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@YACP Oh, no. I am a very busy person. Honors, clubs, all that. I didn't have time. I'm sorry. I do appreciate your effort. Thank you. –  Yosef Qian May 1 '13 at 7:30
Don't do too much! ^_^ –  Trancot May 2 '13 at 22:30

## 2 Answers

There's no need to introduce orthogonality (which doesn't even make sense without an inner product).

Since $\dim R(T_i)=1$, there exists $m$ basis vectors $v_i \in R(T_i) \subseteq V$ for the respective ranges, i.e. $v_i$ is a basis of $R(T_i)$. Put $v_i = T_i(w_i)$. I affirm that these vectors are all linearly independent. If they were not, say with a linear dependence relation $a_1 v_1 + \ldots + a_m v_m = 0$ with $a_i \neq 0$, then applying $T_i$ gives $$0 = a_1 T_i(v_1) + \ldots + a_m T_i(v_m) = a_1 T_i(T_1(w_1)) + \ldots + a_m T_i(T_m(w_m)) = a_i T_i(v_i).$$

Since $a_i \neq 0$, we must have $T_i(v_i) = 0$, which implies that $R(T_i^2)=\{0\}$, a contradiction. Thus we have $m$ linearly independent vectors in a $n$-dimensional space; hence we must have $m \leq n$.

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Well done. You are right we don't need an inner product. –  Ross Millikan May 3 '13 at 0:47
@AlexP. Here here! –  Yosef Qian May 3 '13 at 2:43
@RossMillikan What is $w_i$ here? –  Yosef Qian May 5 '13 at 20:37
@YosefQian It's a vector in $V$ such that $T(w_i) = v_i$. –  A.P. May 6 '13 at 0:57

To have $\dim R(T^{2}_{i})=\dim R(T_i)=1$, each $T_i$ must map the whole space to a line and in fact map vectors along that line to that line (with a scaling factor allowed) and all vectors perpendicular to the line to $\vec 0$. Then the condition $T_i \circ T_j= 0$ forces the lines to be perpendicular. For if the line $T_j$ maps to is not perpendicular to the one $T_i$ maps to, the output vector of $T_j$ will have a component along the line $T_i$ maps to and the result will not be zero. The figure below shows what happens in $\Bbb R^2$ if the two axes are not perpendicular. As there are only $n$ perpendicular directions in $\Bbb R^n$ your are there.

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Is there a more "general" "picture" for dimensions for some arbitrary $n$? –  Trancot May 2 '13 at 23:45
@Trancot: The same will apply. I just can't draw it as well. You still need all the lines to be perpendicular. –  Ross Millikan May 2 '13 at 23:57
There's no notion of orthogonality in a general vector space. –  A.P. May 3 '13 at 0:39
@Alex P. Projections are orthogonal if their composition is the zero map are they not? –  Scott H. May 3 '13 at 0:47
@ScottH. (1) We're not necessarily dealing with projections. (2) What does orthogonal mean in a general vector space? –  A.P. May 3 '13 at 0:56