Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading Hoffman & Kunze's chapter on linear transformations, with a view towards understanding dual spaces. (I primarily want to read Calculus on Manifolds; in the first chapter of that book, Spivak marks one exercise concerned with dual spaces as "soon-to-be-important". I never studied dual spaces in my linear algebra course, and I want to be sure I follow his development.)

H&K state the theorem that the space of linear transformations $L(V, W)$ from $n$-dimensional $V$ to $m$ dimensional $W$ has dimension $nm$. I am having a hard time understanding their proof; they basically just juggle $\Sigma$-notation. I somewhat follow the steps, which loosely are as follows: you let $\mathcal{B} = \{\alpha_1, ... \alpha_n\}$ be a basis for $V$, and $\mathcal{B'} = \{\beta_1, ... \beta_m\}$ a basis for $W$. For $1 \leq p \leq m$, $1 \leq q \leq n$, define

$$E^{(p,q)} : V \to W : c_1\alpha_1 + ... + c_q\alpha_q ... +c_n\alpha_n \to c_q\beta_p;$$

that is, let

$$E^{(p,q)}(\alpha_i) = \left\{ \begin{array}{} 0, & i \neq q \\ \beta_p, & i = q \end{array} \right.$$

and put $E^{(p,q)}(\gamma) = \sum_{i=1}^n c_iE^{(p,q)}(\alpha_i).$ Fix some arbitrary l.t. $T: V \to W$ such that

$$T(\alpha_i) = \sum_{p=1}^m A_{(p,i)} \beta_p = A_{(1,i)}\beta_1 + A_{(2,i)}\beta_2 + ... +A_{(m,i)}\beta_m$$

for some weights $A_{(1,i)}, A_{(2,i)}, ... A_{(m,i)}, 1 \leq i \leq n$. Now consider

$$ T'(\gamma) := \sum_{p=1}^m\sum_{q=1}^nA_{(p,q)}E^{(p,q)}(\gamma).$$

On the one hand,

\begin{align} T'(\gamma) = \sum_{p=1}^m\sum_{q=1}^nA_{(p,q)}E^{(p,q)}(\gamma) &= \sum_{p=1}^m\sum_{q=1}^nA_{(p,q)}E^{(p,q)}(c_1\alpha_1 + ... + c_q\alpha_q ... +c_n\alpha_n)\\ & = \sum_{p=1}^m\sum_{q=1}^nA_{(p,q)}c_q \beta_p\\ & = \sum_{p=1}^m (\sum_{q=1}^nA_{(p,q)}c_q) \beta_p \end{align}

On the other hand,

$$ T(\gamma) = \sum_{i=1}^n c_iT(\alpha_i) = \sum_{i=1}^n c_i\sum_{p=1}^m A_{(p,i)}\beta_p = \sum_{p=1}^m(\sum_{i=1}^nc_iA_{(p,i)})\beta_p$$

so $T' = T$. Then you show that all the $E^{whatever}$ are linearly independent, and you are "done".

The problem with this is that I have no idea what's going on here. When I focus, I can make sense of the $\Sigma$-notation, but otherwise I don't follow the proof. So, I have to ask:

  • What is the intuition for why the theorem is true?
  • Can anybody make the proof given more intuitive? (For example, what are the $E^{whatever}$ really doing? How did H&K know that we'd have $T' = T$ (how did they think of $T'$)?
share|cite|improve this question
A vector space has dimension equal to the precise amount of real parameters one needs to uniquely specify an element of the space. Linear transformations $V\to W$ are in one-to-one correspondence with $m\times n$ matrices. Clearly the amount of real parameters to specify an $m\times n$ matrix is $nm$--the number of entries. This is the intuition. – Alex Youcis Mar 28 '13 at 3:05
@Alex I don't get it. (What do your 1st/3rd sentences mean?) – Chris Mar 28 '13 at 3:07
@AlexYoucis, I don't think that "definition" of dimension could pass not only a formal proof but even not an informal explanation for a newbie, in particular since it is not always clear, or even possible, to specify what parameter of what to specify abstract vectors in abstract vector spaces... – DonAntonio Mar 28 '13 at 3:14
The intuition I would give is that a map from $V$ to $W$ is completely determined by where it sends the $n$ basis vectors of $V$, and every linear map from $V$ to $W$ is a (unique) linear combination of maps which send $\alpha_i$ to $\beta_j$ and all other basis vectors to $0$. Since there are $nm$ pairs $(\alpha_i,\beta_j)$, the result follows. – Alex Becker Mar 28 '13 at 3:19
@DonAntonio The OP asked for intuition, I gave it to them. That "definition" is precisely the intuition for a vector space--it's just secretly a whole bunch of place holders (parameters) for real numbers. – Alex Youcis Mar 28 '13 at 3:32

Maybe matrices make it easier. Fix two bases $\{e_i\}$ for $V$ and $\{f_j\}$ for $W$. For every $T:V\longrightarrow W$, we denote $A_T$ the matrix of $T$ with respect to these bases. This is a well-defined linear (check) map from $L(V,W)$ to $M_{m\times n}(K)$.

As a linear operator is uniquely determined by its action on a basis, the latter is a bijection. As pointed out by Alex Becker, with this observation, you already have your dimension in a slightly informal way: $T$ is detemined by $\{T(e_1),\ldots,T(e_n)\}$, and each $T(e_i)$ is determined by its $m$ coefficients in $\{f_1,\ldots,f_m\}$. This makes $n\cdot m$ coeffiecients: your dimension.

Now with the isomorphism above: $$ \dim L(V,W)=\dim M_{m\times n}(K). $$ It is not hard to see that the matrices $E_{i,j}:=(\delta_{i,k}\delta_{j,l})$ (i.e. $1$ in $(i,j)$ position, $0$ elsewhere) for $1\leq i\leq m$ and $1\leq j\leq n$ constitute a basis of $M_{m\times n}(K)$. There are $mn$ of them. That's your dimension.

share|cite|improve this answer
I would guess that this is what @AlexYoucis meant to say. Thanks for the explanation. :) – Chris Mar 28 '13 at 3:20
@user1296727 Yes, I guess that's what he meant. Alex Becker's observation is very interesting too for the intuition. So I added a few words to somehow incorporate this. – 1015 Mar 28 '13 at 3:28
I just realized why you specified that the map is linear: otherwise, it doesn't follow that the dimension of the spaces is equal. – Chris Mar 28 '13 at 3:30
@user1296727 Correct. Two vector spaces have the same dimension if and only if they are isomorphic: i.e. there exists a linear bijection between them. Note that the inverse of the bijection is automatically linear. – 1015 Mar 28 '13 at 3:42

Your Answer


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.