From S.L Linear Algebra:
Find the matrix associated with the following linear map:
$F: \mathbb{R}^n \rightarrow \mathbb{R}^n$ given by $F(X)=-X$
Considering that dimension of vector spaces are unknown, I've constructed a generalized method for the matrix associated with linear map given above.
Let $F: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a linear transformation given by $F(X)=-X$. Then $F(x_1,...,x_n)=-(x_1,...,x_n)$.
We shall find a $n \times n$ symmetric matrix $A=(a_{ij})$, such that $A(x_1,...,x_n)=-(x_1,...,x_n)$. With calculations, we can see that $\left ( (a_{11}x_1+...+a_{1j}x_{n}),...,(a_{i1}x_{1}+...+a_{ij}x_n) \right )=(-x_1,...,-x_n)$ and hence $\left ( ((1+a_{11})x_1+...+a_{1j}x_{n}),...,(a_{i1}x_{1}+...+(1+a_{ij})x_n) \right )=O$ ($O$ being a zero vector).
Considering that we are looking for basis of row space (image of linear transformation), all coefficients must be linearly independent within row vector, therefore kernel of all row vectors is trivial. We eventually get a $n \times n$ matrix with $a_{11},...,a_{ij}=-1$ (hence we get a "sparse" matrix which has $-1$ as diagonal elements):
$$\begin{pmatrix} -1_{11} & 0_{12} & ... & 0_{1j}\\ 0_{21} & -1_{22} & ... & 0_{2j}\\ ... & ... & ... & ... \\ 0_{i\,1} & 0_{i\,2} & ... &-1_{ij} \end{pmatrix}$$
I did index all elements of matrix to avoid confusion.
The generalized solution itself does seem quite confusing to me, it's much easier to find the matrix associated with linear transformation when dimension is known.
I'm looking for the basis of the matrix associated linear transformation above, which is the basis of row space of transformation (or the image).
Are there any mistakes in my solution? Is there a more simple and elegant way of finding a matrix associated with linear transformation where dimension of domain and codomain is unknown?
A matrix is determined by what it does to a basis. Let $\mathcal S=\{e_1,\dots,e_n\}$ be the standard basis. Then $Ae_i=-e_i, i=1,\dots, n$.
Thus $[A]_{\mathcal S}=\begin{pmatrix}-1&0&\dots&0\\0&-1&\dots&0\\\vdots&\ddots&\ddots&\vdots\\0&0&\dots&-1\end{pmatrix}$.