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.

The following is from Gowers's How to lose your fear of tensor products:

Because subscripts and superscripts are a nuisance in html, I shall now change notation, and imagine that we know the values of $f(s,t), f(u,v), f(w,x)$ and $f(y,z)$. If $s, u$ and $w$ are all multiples of a single vector, then some of this information is redundant, since $t, v$ and $x$ are not linearly independent (they live in ${\mathbb R}^2$). So let us suppose that no three of the first vectors are multiples and no three of the second are. It follows easily that we can take two pairs, without loss of generality $(s,t)$ and $(u,v)$, and assume that $s$ and $u$ are linearly independent, and that so are $t$ and $v$.

Here $f:{\mathbb R}^2\times{\mathbb R^2}\to X$ is a bilinear map, where $X$ is a vector space. Here are my questions:

  • What's the reasoning in bold? In ${\mathbb R}^2$, three vectors are always NOT linearly independent. Why does he mention "if $s, u$ and $w$ are all multiples of a single vector"?
  • What does it mean by "no three of the first vectors are multiples and no three of the second are"?
  • How can I deduce the last sentence?
share|improve this question
10  
Why down vote? Are my questions to stupid? Could anyone tell me how should I improve the questions? –  Jack Jul 12 '11 at 4:57
    
Does my answer below address the points of your question or would you like me to elaborate/discuss the Exercises that I have given (which are relevant to your question)? –  Amitesh Datta Jul 12 '11 at 7:47

1 Answer 1

up vote 3 down vote accepted

(1) If $s,u$ and $w$ are all multiples of a single vector $a$, then we can write $s=c_1a,u=c_2a$ and $w=c_3a$ for scalars $c_1,c_2$ and $c_3$. Since $t,v$ and $x$ are linearly dependent, we can write one of these vectors as a linear combination of the other two. (Exercise.) Let us assume, without loss of generality, that we can write $x=e_1t+e_2v$ for scalars $e_1$ and $e_2$. The bilinearity of $f$ implies that:

$f(w,x)=f(c_3a,x)=c_3f(a,x)=c_3f(a,e_1t+e_2v)=c_3\left[e_1f(a,t)+e_2f(a,v)\right]$.

Therefore, we know the value of $f(w,x)$ if we know the values of $f(a,t)$ and $f(a,v)$. Exercise: use the bilinearity of $f$ to show that (for $c_1,c_2\neq 0$) we know the values of $f(a,t)$ and $f(a,v)$ if we know the values of $f(s,t)$ and $f(u,v)$.

(2) I think he means that there does not exist a vector $a$ such that $s=c_1a, u=c_2a$ and $w=c_3a$ for scalars $c_1,c_2$ and $c_3$ and similarly there does not exist a vector $b$ such that $t=d_1b,v=d_2b$ and $x=d_3b$ for scalars $d_1,d_2$ and $d_3$.

The following exercises are relevant (two are already stated above but I will state them again (in a more general situation) for the convenience of location):

Exercise 1: Let $(v_1,\dots,v_n)$ be a linearly dependent tuple of vectors in a vector space $V$ for some positive integer $n$. If $v_1\neq 0$, prove that for some $1\leq i\leq n$, we can write $v_i$ as a linear combination of the tuple $(v_1,\dots,v_{i-1})$.

Exercise 2: Let $f:V\times V\to W$ be a bilinear map where $V$ and $W$ are vector spaces. Prove the following:

(a) $f(v_1,0)=0=f(0,v_2)$ for all vectors $v_1,v_2\in V$.

(b) Let $c\neq 0$. If we know the value of $f(cv_1,v_2)$ for vectors $v_1,v_2\in V$ and some scalar $c$, then we know the value of $f(v_1,v_2)$.

Exercise 3: Let $f:V\times V\to W$ be a bilinear map where $V$ and $W$ are vector spaces and let $(v_1,\dots,v_n)$ be a basis of $V$. Prove that if we know the values of $f(v_i,v_j)$ for all $1\leq i,j\leq n$, then we know the value of $f(a,b)$ for any pair of vectors $a,b\in V$.

Exercise 4: Let us maintain the notation of Exercise 3 and let $A$ be the $n\times n$ matrix with $f(v_i,v_j)$ as its $(i,j)$th entry. If $a,b\in V$, prove that $f(a,b)=a^{T}Ab$ where $a^{T}$ is the transpose of the column vector $a$ and $b$ is viewed as a column vector.

Exercise 5: Let $V$ be a vector space of dimension $n$ and let $a_1,\dots,a_n,a_{n+1},b_1,\dots,b_n,b_{n+1}$ be vectors in $V$. If all the $a_i$'s are multiples of a fixed vector, show that at least one of the values $f(a_i,b_i)$ for $1\leq i\leq n+1$ is redundant, i.e., can be deduced with a knowledge of the other $n$ values (plus possibly an understanding of basic properties of bilinear forms such as (a) of Exercise 2).

Challenge: Formulate conditions under which we can deduce the value of a bilinear map at a given ordered pair of vectors if we know the values of the bilinear map on a specific set of ordered pairs. (In other words, generalize Exercise 5 and think more about Exercise 3 and Exercise 4.)

I hope this helps!

share|improve this answer
    
Fair enough. Now I see. Thanks. –  Jack Jul 12 '11 at 15:08

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.