One-to-one (injective) and onto (surjective) linear transformations are defined in a corresponding manner to functions: a linear transformation is one-to-one if it maps distinct vectors to distinct vectors, and onto if it covers the whole range.
Formally, a mapping $T: V \to W$ is one-to-one if, for all $u, v \in V$,
\(u \neq v \implies T(u) \neq T(v),\) or equivalently, \(T(u) = T(v) \implies u = v.\)
$T$ is onto if, for all $w \in W$, $\exists v \in V$ such that $T(v) = W.$
A linear mapping is called an isomorphism if it is one-to-one and onto, i.e. if it is a bijection.
Two vector spaces that have an isomorphism between them are called isomorphic; this is denoted by $V \cong W$.
Given a vector space $V$ with basis $B = {x_1, x_2, \cdots , x_n}$, any vector $v \in V$ can be expressed as $c_1v_1 + c_2v_2 + \cdots + c_nv_n$ for some collection of $c_i$.
These $c_i$ are called the coordinates of $v$, and the column vector consisting of the $c_i$ is called the coordinate vector of $v$ with respect to B, denoted by \([v]_ B = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}.\)
Note that the same vector has different coordinate vectors with respect to different bases.
1) Let $B = {v_1, v_2, \cdots , v_n}$ be a basis of a vector space $V$. If $u, v \in V$ and $c$ is a scalar, then
(i) $[v + u]_ B = [v]_ B + [u]_ B$, and
(ii) $[cu]_ B = c[u]_ B$.
Proof: Let $v = x_1v_1 + x_2v_2 + \cdots + x_nv_n$ and $u = y_1v_1 + y_2v_2 + \cdots + y_nv_n$.
This proves property (i).
Further,
This proves property (ii).
2) Let $B$ and $V$ be as above. If ${u_1, u_2, \cdots, u_k} \subseteq V$ is a linearly independent set of vectors, then ${[u_1]_ B, [u_2]_ B, \cdots, [u_n]_ B}$ is also linearly independent.
Proof: To prove this, suppose
This means that
If the set of coordinate vectors is linearly dependent, there exists scalars $c_i$ such that $c_1[u_1]_ B + c_2[u_2]_ B + \cdots + c_n[u_n]_ B = 0$. This means that
Now, consider the sum $c_1u_1 + c_2u_2 + \cdots + c_nu_n$
Taking the coefficients of each basis vector common, we get this to be equal to
$(c_1x_{11} + c_2v_{21} + \cdots + c_nv_{n1})v_1 + (c_1x_{12} + c_2x_{22} + \cdots + c_nv_{n2})v_2 + \cdots + (c_1x_{1n} + c_2x_{2n} + \cdots + c_nx_{nn})v_n$
From the equation in the preceding paragraph, we find the coefficient of each of the $v_i$ to be 0, making the sum 0.
Therefore, the $u_i$ are linearly dependent, which is a contradiction.
Therefore the $[u_i]_ B$ must also be linearly independent.
Note: This proof was not given in class (no proof was). If anyone knows a shorter or more elegant proof, please feel free to update it
If $V$ and $W$ are finite dimensional vector spaces with bases $B$ and $C$ respectively, where $B = {v_1, v_2, \cdots, v_n}$.
If $T: V \to W$ is a linear mapping, its matrix representation is the $m \times n$ matrix $A$ defined by $A = {[T(v_1)]_ C, [T(v_2)]_ C, \cdots, T[v_n]_ C}$.
The significance of this matrix is that for all $v \in V$,
Proof: Suppose $v$ can be expressed as $v = x_1v_1 + x_2v_2 + \cdots + x_nv_n$. Then $T(v) = x_1T(v_1) + x_2T(v_2) + \cdots + x_nT(v_n)$. Further, the right hand side of the given equation becomes
Consider the first column of $A$. It is equal to $T[v_1]_ C$.
By the definition of matrix multiplication, its first component is multiplied by $x_1$ and added to row 1 of the LHS;
its second component is multiplied by $x_1$ and added to row 2 of the LHS, and so on
Therefore the first component of the LHS column matrix is the first component of the vector $x_1[T(v_1)]_ C + x_1[T(v_2)]_ C + \cdots + x_1[T(v_n)]_ C$, which we have seen above is nothing but $[T(v)]_ C$.
Similarly, all other rows of the LHS are the other components of $[T(v)]_ C$, QED.
An inner product over a vector space $V(\mathbb{R})$ is an operation that assigns to every pair of vectors $u, v \in V$ a real number $(u,v)$ such that:
(i) $(u,v) = (v,u)$
(ii) $(u,v+w) = (u,v) + (u,w)$
(iii) $(cu,v) = c(u,v)$
(iv) $(u,u) \geq 0$ and $(u,u) = 0 \iff u = 0$.
Note: Strictly, this is the definition of the inner product restricted to the special case of vector spaces over the real numbers. A general definition exists for arbitrary vector spaces; this will be taught in the next class.
Other properties of the inner product (derivable from the above) are:
(v) $(u+v,w) = (u,w) + (v,w)$
(vi) $(u,cv) = c(u,v)$
(vii) $(u,0) = (0,u) = 0$.
Proof:
The proof is trivial, so we will not go into details. (v) and (vi) can be proved by switching $u$ and $v$ in (i) and switching $u$ and $v+w$ in (ii).
(vii) can be proved by substituting $v = w = 0$ in (ii).
Some functions on and properties of vectors defined in terms of the inner product are:
(a) The length or norm of a vector $v$, denoted by $\Vert v \Vert$, defined as $\Vert v \Vert = \sqrt{(v,v)}$.
(b) The distance between two vectors, denoted as $d(u,v)$, defined as $d(u,v) = \Vert u - v \Vert$.
(c) Orthogonality; two vectors $u$ and $v$ are said to be orthogonal if $(u,v) = 0$.