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I am reading linear representations of finite groups by Serre and I want to understand what is meant by the following via an example:

Let $V$ be a vector space over the field $\mathbb{C}$ (unless stated otherwise) of complex numbers and let $GL(V)$ be the group of all automorphisms of $V$. An element $\alpha$ of $GL(V)$ is, by definition, a linear mapping of $V$ into $V$. When $V$ has finite dimension, say $n$, this implies we have a finite basis $\lbrace e_{i} \rbrace$ of $n$ elements with linear map $\alpha\colon V \rightarrow V$ defined by a square matrix $(\alpha_{ij})$ of order $n$, where the coefficients $\alpha_{ij}$ are complex numbers. Given a basis $\lbrace e_{i} \rbrace$ of $V$ and an automorphism $\alpha$ in $GL(V)$, then for every basis vector $\lbrace e_{i} \rbrace$ we have:

$$ \alpha(e_i)=\sum_{j=1}^n \alpha_{ij}e_j $$

An alternative definition to saying that $\alpha$ is an automorphism is to say that the determinant $\det(\alpha_{ij})$ is non zero. Therefore, the group $GL(V)$-- the general linear group on $V$ is thus isomorphic to the group of invertible (or non-singular) square matrices of order $n$.

I would appreciate if anyone can help.

EDIT: for example let $V=\mathbb{C}^3$ but I don't know how to proceed

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    $\begingroup$ What do you need help with? $\endgroup$ – Bill O'Haran Apr 15 at 13:42
  • $\begingroup$ I want an example so I can see what's going on. so for example if $V=\mathbb{R}^3$ then we have $\lbrace e_{1}, e_{2}, e_{3} \rbrace$. now what is $\alpha$ in this example?and how what is the matrix of $\alpha_{ij}$ $\endgroup$ – john Apr 15 at 13:43
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    $\begingroup$ @Javi Well, the field being $\mathbb{R}$ or $\mathbb{C}$ really does not matter here anyway. $\endgroup$ – Bill O'Haran Apr 15 at 13:48
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    $\begingroup$ @TheoBendit No because I am self teaching $\endgroup$ – john Apr 15 at 13:49
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    $\begingroup$ @BillO'Haran Agree, but the OP might be confusing basic notions, so I just wanted to make sure the problem was not related to the base field. $\endgroup$ – Javi Apr 15 at 13:49
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As I said in the comments, this is finding the matrix for a transformation given a basis, and is a standard thing in finite-dimensional linear algebra. I went hunting for some nice account for how to do this, but they all seemed to make what I see as a pedagogical error: they start the explanation in $\Bbb{R}^n$, which I think confuses the matter. Let's take a classic but different example.

Consider the vector space $V = P_2(\Bbb{C})$, the space of complex polynomials of degree $2$ or less. An arbitrary element of $V$ is a function $x \mapsto ax^2 + bx + c$, where $x$ is a complex variable. Addition and scalar multiplication are defined pointwise, and are exactly what one would expect.

One basis for this space is $(1, x, x^2)$; the constant function $1$, the identity function, and the squaring function. The map above can be written as a unique linear combination of these three functions: $$ax^2 + bx + c = a \cdot x^2 + b \cdot x + c \cdot 1.$$ As such, we can form the corresponding coordinate column vector in $\Bbb{C}^3$: $$\begin{bmatrix} a \\ b \\ c \end{bmatrix}.$$ This coordinate column vector is dependent on the particular basis. We could take a different basis $(x^2, x - 1, (x + 1)^2)$, and get a totally different coordinate vector for the same polynomial.

In this way, each basis represents a way to express the polynomials in $V$ as ordered triples of complex numbers in $\Bbb{C}^3$ (I'm trying to avoid using the word "vector" here). Indeed, any basis of any finite-dimensional vector space gives us a correspondence between the space and $n$-tuples of scalars. Different bases will produce different correspondences, but it shows that the spaces are structurally identical to powers of the scalar field.

Now, given a fixed basis for $V$, linear operators on $V$ can be seen to be linear operators on $\Bbb{C}^3$, mapping not the polynomials in $V$, but their coordinate vectors with respect to a given basis. As it turns out, such linear operators on $\Bbb{C}^3$ can always be uniquely envisioned as multiplication by a $3 \times 3$ complex matrix. This matrix is the matrix $(\alpha_{ij})$ referred to in the book.

To form this matrix, you compute the transformation on each basis vector individually, then writing the results as coordinate column vectors, with respect to the basis. The matrix whose columns are these coordinate vectors is the matrix in question. If $T$ is the operator, then it is the unique matrix $A$ that has the property that $$A[v]_B = [Tv]_B$$ where $[v]_B$ is the coordinate column vector of $v$ with respect to $B$.


Time for a concrete example. Consider the linear operator $D$, the differentiation map. It maps from $V$ to $V$. Fix the usual basis for $V$: $B = (1, x, x^2)$. We compute $$D(1) = 0 = 0 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \implies [D(1)]_B = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$ This is the first column of the matrix. Then, $$D(x) = 1 = 1 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \implies [D(x)]_B = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}.$$ This is the second column. Finally the third column: $$D(x^2) = 2x = 0 \cdot 1 + 2 \cdot x + 0 \cdot x^2 \implies [D(x^2)]_B = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}.$$ In total, the matrix for $D$ under $B$ is $$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix}.$$

In order to use it, consider differentiating a polynomial $(x + 1)^2$. First, we write it as a coordinate column vector: $$[(x + 1)^2]_B = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}.$$ Then, we multiply it to the above matrix: $$[D(x + 1)^2]_B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 0 \end{bmatrix}.$$ To confirm, we have that $D(x + 1)^2 = D(x^2 + 2x + 1) = 2x + 2$, which corresponds to the above calculation. In this way, we understand differentiation on $V$ as a simple matrix multiplication.

Note also that this matrix is not invertible, as there is no inverse to differentiation (indefinite integration doesn't count, due to the constant of integration)!

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    $\begingroup$ Sir, I cannot thank you enough. much appreciated +1 :)) $\endgroup$ – john Apr 15 at 14:46
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    $\begingroup$ The concrete example completely helped me understand what was going on. Salute to you. I mean it :) $\endgroup$ – john Apr 15 at 14:47
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    $\begingroup$ One quick question, where you assign the column vectors $[D(1)]_B$, shouldn't the second and third vectors be $[D(x)]_B$ and $[D(x^2)]_B$ where you've written the implies sign? probably a small typo when copying and pasting aha but I wouldn't want to confuse future readers! $\endgroup$ – john Apr 15 at 15:20
  • $\begingroup$ @john Ah yes, whoops! $\endgroup$ – Theo Bendit Apr 16 at 0:17
  • $\begingroup$ If possible, could you look at my recent question posted, hopefully giving an answer like the one above since it was very useful :) $\endgroup$ – john Apr 16 at 14:07

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