Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $T_i(v_j)=0~\forall~i\neq j$ . . .

share|improve this question
1  
@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 2

up vote 5 down vote accepted
+50

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 $m \leq n$.

share|improve this answer
    
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.enter image description here

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.