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.

I have something to discuss about mathematical understanding, and use the Linear Algebra as example.

Sometimes, I feel the intuitive understanding of definitions/theorems itself is very different from understanding its proofs. But for some others especially when it's generalized to abstract level ,or to general n-dimensions. Then I cannot use the 'intuitive' way to understand('see' the picture as 'intuition', for 2/3-spaces), but always the proof is indeed the understanding procedure('see' the algebraic expressions, but 'visible intuition' disappeared.)

Example:

  1. $^t(AB)={^t(B)}{^t(A)}$ , for this one, I cannot 'see' any intuition ,but the proof.$(AB)=\sum\limits_{j=1}^{n}{a_{ij}b_{jk}}=\sum\limits_{j=1}^{n}{b_{jk}a_{ij}}$ , to tranpose it, $^t(AB)=\sum\limits_{j=1}^{n}{b_{kj}a_{ji}}$ is exactly $^t(B){^t(A)}$

  2. Let $A_1, ...,A_r$ be generators of subspace $V$ of $R^n$, and $W$ be set of all vectors in $R^n$ which are perpendicular to $A_1, ..., A_r$, then W is perpendicular to each elments in $V$.

For this, I also cannot 'see' intuition of how it could do that in general space. But:

Proof: Let $w$ be any vector of $W$. Then, $w\cdot t_iA_i=t_i(w\cdot A_i)=0$, for $i = 1,...,r$, then, $w\cdot(\sum\limits_{i=1}^{r}{t_i A_i})=\sum\limits_{i=1}^{r}{t_i(w\cdot A_i)}=\sum\limits_{i=1}^{r}{0}=0$, as we know $\sum\limits_{i=1}^{r}{t_i A_i)}$ is a space generated by $A_i, i=1,...,r$, it concludes our proof.

The only idea I could 'see' is dot product has the distributive law and use the definition of linear combination, it'll be possible to prove its correctness.

share|improve this question
1  
I'm sorry but what exactly is your question then? –  Inquest Feb 17 '12 at 10:41
    
@Nunoxic I mean if I do this way for all def./theorems, is it the real understanding? or something else I need to do? Because sometimes when there's no 'intuitive image' in my mind, it's a little un-comfortable. –  Xingdong Feb 17 '12 at 10:45
2  
@Xingdong Perhaps it may be helpful to eliminate all these coordinates and look at the transpose of an operator on a vector space: If $A : V \longrightarrow V$ is an operator and $V$ equipped with a non-degenerate inner product then the transpose of $A$ is defined to be the unique operator $B$ such that $\langle Av, w \rangle = \langle v, Bw \rangle$ for all vectors $v,w \in V$. If I recall correctly one can prove that such a $B$ exists by the Riesz Representation Theorem for finite dimensions. –  user38268 Feb 17 '12 at 12:39
add comment

1 Answer

up vote 3 down vote accepted

Abstract proofs are, in an important sense, much better than visualizations. Strictly speaking, you cannot 'visualize' the truth of a statement in, say, two dimensions if it involves a quantifier of the form 'for all $v \in V$' where $V$ is an infinite set (e.g. a $\mathbb{R}$-vector space). What you really do is visualize what you take to be a representative example in your mind and verify the statement for that particular example in your mind. Granting that one can 'visually' prove a particular statement like this, you still have the issue of checking whether you have truly visualized every possible case of your universal quantifier. On the other hand, an abstract proof really does prove the statement, even if there is a universal quantification over an infinite set.

Of course, visualization is still very valuable, as it can provide insights that lead to conjectures, theorems, and/or proofs. Never underestimate the value of getting a good intuition for whatever field you are studying. However, visualization is only one of many ways to obtain intuition. (Carefully studying abstract proofs is another.)

As far as your specific examples go, sometimes you have to resort to more abstract visualization techniques for your intuition. As Benjamin Lin points out in the comments, you can visualize a linear map by writing $A : V \rightarrow W$ (indicating which vector spaces $A$ maps to and from). If we also write $B: U \rightarrow V$, then we can compose them to get $AB : U \rightarrow W$. Something you learn about transposes is that the transpose $A^t$ is a linear map $A^t : W^* \rightarrow V^*$, where $V^*$ denotes the dual vector space of $V$. (If you haven't seen this abstract formulation of linear algebra yet, that's okay. In matrix terms, the equivalent fact is that the transpose of an $m$-by-$n$ matrix is an $n$-by-$m$ matrix.) Similarly, we have $B^t : V^* \rightarrow U^*$ and $(AB)^t : W^* \rightarrow U^*$.

Using these maps, we conclude that not only would a formula $(AB)^t = A^t B^t$ be wrong, it is a nonsensical combination of symbols. The range of $B^t$ is $U^*$ and the domain of $A^t$ is $W^*$, so it is not possible to compose them. (Exercise: The relevant statement in matrix terms is that you cannot, in general, multiply $A^t B^t$ when $A$ and $B$ are matrices such that $AB$ is defined.) Looking at the the diagrams for the maps suggests the formula $(AB)^t = B^t A^t$ is worth studying, since at least $B^t A^t$ is defined and both sides of the formula are linear maps from $W^*$ to $U^*$. Of course, this does not prove the formula, but once you use this visual intuition to motivate what statement you are trying to prove, it is easy to prove this formula from the definitions.

share|improve this answer
add comment

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.