The Vector Space of Linear Maps

Definition Let $V$ and $W$ be vector spaces over the same field $\mathbb{F}$, where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, and let $b_1, \dots, b_n$ be a basis of the finite-dimensional space $V$. For any vectors $w_1, \dots, w_n$ in $W$, chosen freely and not necessarily distinct or independent, there is exactly one linear map $T: V \to W$ with $T(b_j) = w_j$ for $j = 1, \dots, n$. It is given by

The formula is well defined because every vector of $V$ has unique coordinates $x_1, \dots, x_n$ in the basis. It is linear because the coordinates of $ax + by$ are $a$ times those of $x$ plus $b$ times those of $y$. Any linear map with these basis images agrees with the formula, since linearity forces $T(\sum_j x_j b_j) = \sum_j x_j T(b_j)$. Both halves of the basis hypothesis matter. If the list spans $V$ but is dependent, prescribed images can contradict each other. If it is independent but does not span, such a map exists, and it is unique only when $W = \{0\}$.

Definition Let $V$ and $W$ be vector spaces over the same field $\mathbb{F}$. The set of all linear maps from $V$ to $W$ is written $\operatorname{Hom}(V, W)$. For $S, T \in \operatorname{Hom}(V, W)$ and $\lambda \in \mathbb{F}$, pointwise addition and scaling are defined at every $x \in V$ by

Both results are again linear. For the sum, linearity of each map and regrouping in $W$ give

and a similar computation works for $\lambda T$. With these operations $\operatorname{Hom}(V, W)$ is a vector space over $\mathbb{F}$. Its zero vector is the zero map, which sends every input to $0$, the negative of $T$ is the map $x \mapsto -T(x)$, and every other axiom holds because it holds in $W$ at each input. Two maps are equal when they agree at every input. No finite-dimensional hypothesis is needed. Composition is a different operation. It applies one map after another, needs the output space of the first map to be the input space of the second, and is not the addition of $\operatorname{Hom}(V, W)$.

Definition Let $V$ and $W$ be finite-dimensional vector spaces over $\mathbb{F}$ with bases $\mathcal{B} = (b_1, \dots, b_n)$ of $V$ and $\mathcal{C} = (c_1, \dots, c_m)$ of $W$. For $1 \le i \le m$ and $1 \le j \le n$, the matrix-unit map $E_{ij} \in \operatorname{Hom}(V, W)$ is the unique linear map with

where $\delta_{jk}$ is $1$ when $j = k$ and $0$ otherwise. So $E_{ij}$ sends $b_j$ to $c_i$ and every other basis vector of $V$ to $0$. On a vector with coordinates $x_1, \dots, x_n$ it gives

and its matrix $[E_{ij}]_{\mathcal{C} \leftarrow \mathcal{B}}$ has a single entry $1$, in row $i$ and column $j$, with zeros elsewhere. The maps depend on both chosen bases.

Definition Let $V$ and $W$ be finite-dimensional vector spaces over $\mathbb{F}$ with $\dim V = n$ and $\dim W = m$. The $mn$ matrix-unit maps $E_{ij}$ built from bases $b_1, \dots, b_n$ of $V$ and $c_1, \dots, c_m$ of $W$ form a basis of $\operatorname{Hom}(V, W)$, so

The coordinates of a map $T$ in this basis are the entries of its matrix. If $T(b_j) = \sum_{i} a_{ij} c_i$ for each $j$, then

To see that the maps span, apply both sides to $x = \sum_j x_j b_j$. Linearity gives $T(x) = \sum_{i,j} a_{ij} x_j c_i$, and the right side gives the same sum, so the two maps agree at every input. For independence, suppose $\sum_{i,j} a_{ij} E_{ij}$ is the zero map and evaluate it at $b_k$. Every term with $j \ne k$ vanishes, leaving $\sum_i a_{ik} c_i = 0$, and independence of the $c_i$ forces $a_{1k} = \cdots = a_{mk} = 0$. Doing this for each $k$ shows that every coefficient is zero. If $n = 0$ or $m = 0$, the only linear map is the zero map, and the formula correctly gives $0$.

Definition The rank of a linear map $T: V \to W$ is the dimension of its image $\operatorname{im} T = \{T(x) : x \in V\}$. A rank-one map is a linear map whose image is one-dimensional. Over $\mathbb{F}$, a linear map $T$ has rank one exactly when

for some nonzero vector $w \in W$ and some linear map $\varphi: V \to \mathbb{F}$ that is not the zero map. If the image is the line spanned by $w$, each $T(x)$ equals $\varphi(x) w$ for exactly one scalar $\varphi(x)$, and linearity of $T$ makes $\varphi$ linear. Conversely, a nonzero $\varphi$ takes every value in $\mathbb{F}$, so the image of such a map is the whole line spanned by $w$. In coordinates, an $m \times n$ matrix over $\mathbb{F}$ has rank one exactly when it equals a nonzero column $u \in \mathbb{F}^{m}$ times a nonzero row $r \in \mathbb{F}^{1 \times n}$. Each matrix-unit map $E_{ij}$ has rank one, since its image is the line spanned by $c_i$, and so does every nonzero multiple of it. Rank-one maps do not form a subspace, since the zero map has rank zero, and a sum can also raise the rank, as $E_{11} + E_{22}$ on $\mathbb{F}^2$ has rank two.