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I'd like to show that Any linear transformation between two finite-dimensional vector spaces can be represented by a matrix. I've seen a proof for linear transformation from and to $\mathbb{R}^n$ but I want to generalize it to any finite-dimensional vector space.

I'd also like to show that the composition of two linear tranformations can be written as a multipication of two matrices. I believe it should look like this: $$ [S\circ T]^B_D=[S]^C_D\cdot [T]^B_C $$ Where $T\rightarrow V:W, S\rightarrow U:V$ and $B,C,D$ are bases of $U,V,W$ respectively

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  • $\begingroup$ What is your question? $\endgroup$
    – copper.hat
    Commented Jun 30, 2020 at 5:40
  • $\begingroup$ How does one prove those theorems? $\endgroup$ Commented Jun 30, 2020 at 5:42
  • $\begingroup$ Pick a basis and show that the effect of the transformation on the space is the same as that of the corresponding matrix multiplication. It is sufficient to show this is true for elements of a basis. $\endgroup$
    – copper.hat
    Commented Jun 30, 2020 at 5:45

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Let the transformation $T$ be from $R^n \to R^m$. We will need bases for each of these spaces, let them be $B_n = \{e_{1n}, e_{2n}... e_{nn}\}$ and $B_m = \{e_{1m}, e_{2m}... e_{mm}\}$ respectively.

Now, any vector $v$ can be expressed as the following

$$v = \sum_1^na_ie_{in}$$

$$\implies T(v) = \sum_i^na_iT(e_{in})$$

To complete the matrix representation, we need to express each $T(e_{in})$ in the basis of the $m$-space

Hence, let $T(e_{in}) = \sum_{k=1}^mb_{ik}e_{km}$

Therefore

$$\implies T(v) = \sum_{i=1}^na_i\sum_{k=1}^mb_{ik}e_{km}$$

Now, we consider the matrix representation of $T$, we express $v$ as a column vector in $R^{n \times 1}$

$$v = \begin{bmatrix}a_1 \\ a_2 \\ . \\. \\. \\a_n\end{bmatrix}$$

Hence, $T(v)$ can be thought of as the sum of $m$ vectors in $R^{m \times 1}$, weighted by the $v$ column scalars. Therefore, we pre-multiply by the column wise representation of $T(e_{in})$ in the basis $B_m$, which is given by scalars $b_{ik}$ as defined above

$$[T] = \begin{bmatrix} b_{11} & b_{21} & b_{31} & ... & b_{n1} \\ b_{12} & b_{22} & b_{32} &...& b_{n2} \\ . & .& . \\ .&.&.&.\\b_{1m} & b_{2m} & b_{3m} &...&b_{nm} \end{bmatrix}$$

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    $\begingroup$ Thanks, I've seen the proof for $\mathbb{R}^n$ but as I stated, I'd like the proof for the general case. $\endgroup$ Commented Jun 30, 2020 at 6:00
  • $\begingroup$ proof of what? This is the proof of the general case? $\endgroup$ Commented Jun 30, 2020 at 6:02
  • $\begingroup$ any transformation $T\rightarrow V:W$ and not just from/to $\mathbb{R}^n$ (for some finite-dimensional vector space). Or is this the general proof since each vector space is isomorphic to $\mathbb{R}^n$? $\endgroup$ Commented Jun 30, 2020 at 6:05
  • $\begingroup$ Yeah, I mean, you could basically replace $R^{n}$ with $V$ and $R^m$ with $W$ in the above proof, I use none of their properties $\endgroup$ Commented Jun 30, 2020 at 6:07
  • $\begingroup$ Understood, thank you. What about showing the composition of two linear functions is a multipication of two matrices? $\endgroup$ Commented Jun 30, 2020 at 6:08

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