Let $V,U$ be $\mathbb F$-vector spaces, and $\varphi:V\to U$ be a linear map. Then the adjoint operator $\varphi^*:U^*\to V^*$ is defined by

$$\langle \varphi^*(f),v\rangle=\langle f,\varphi(v)\rangle,\quad \forall f\in U^*,v\in V.$$

A bilinear form $B:V\times U\to \mathbb F$ has matrix representation $G=(B(e_i,f_j))_{m\times n}$, where $\{e_i\}_{i=1}^m$ and $\{f_j\}_{j=1}^n$ are bases of $V$ and $U$, respectively, if $U,V$ are finite-dimensional. The rank of $B$ is defined as the rank of the matrix $G$, which is independent of the choice of bases. There exist bases $\{e_i\}_{i=1}^m$ and $\{f_j\}_{j=1}^n$ such that $B(e_i,f_j)=\delta_{ij}$ for $1\leq i,j\leq r$, and $B(e_i,f_j)=0$ otherwise, where $r=\mathrm{rank}(B)$.

That bilinear form $B$ is non-degenerate is equivalent to the following conditions:

• $\mathrm{rank}(B)=\dim(V)=\dim(U)$, when $V,U$ are finite-dimensional.
• Left and right kernels of $B$ are trivial, that is, $\{v\in V:B(v,u)=0,\forall u\in U\}=\{0\}$ and $\{u\in U:B(v,u)=0,\forall v\in V\}=\{0\}$.

A linear function is a linear map from $\mathbb F$-vector space to $\mathbb F$. So the co-dimension of it is always $1$.

For a non-zero linear function $f:V\to\mathbb F$, consider decomposition

$$V=\ker(f)\oplus \mathbb Fv_0,$$

by choosing a vector $v_0\in V$ such that $f(v_0)=1$. Then for any $v\in V$, we have

$$v=[v-f(v)v_0]+f(v)v_0\in \ker(f)\oplus \mathbb Fv_0.$$

Rank & Nullity

For any finite-dimensional vector space $V,U$, it holds

$$\dim (V+U)=\dim(V)+\dim(U)-\dim(V\cap U).$$

Compared with infinite-dimensional vector spaces, finite -dimensional vector spaces have some special properties based on the operation of dimension.

• Prove a subspace is equal to the whole space by showing that their dimensions are equal.

[611.4]

Matrix representation.

Rank.