#### Date: 08 February 2018

• Let $V$ be a vector space over a field $k$ in which $2 \neq 0$ (say real or complex numbers).
• A Bilinear form $B(x, y)$ is a function $B : V \times V \to k$ which is linear in $x$ and $y$.
• We can associate to a Bilinear form $B$ a so called quadratic form $Q : V \to k$ defined by $Q(x) = B(x, x)$. Note that by definition a Quadratic form is associated to a Bilinear form.
• Given the Bilinear form $B$, we can define a symmetric Bilinear form $\bar{B}(x, y) = \frac{B(x, y) + B(y, x)}{2}$.
• If $\bar{Q}(x)$ is the quadratic form associated to $\bar{B}$, then it is easy to see that $\bar{Q} = Q$.
• Thus, every quadratic form $Q(x)$ is of the form $Q(x) = B(x, x)$ with $B$ symmetric.
• The symmetric Bilinear form $B$ with $Q(x) = B(x, x)$ is determined by $Q$, as can be shown by polarization.

### Bilinear forms and Linear operators

• Now assume that $V$ is a finite dimensional Euclidean space over $\mathbb{R}$.
• Given a linear transformation $L: V \to V$, we can define a Bilinear form $B(x, y) = \langle x, L(y) \rangle$.
• We see that all Bilinear forms are of this form.

#### Theorem

Let $B(x, y)$ be a Bilinear form on $V$. Then there exists a unique linear transformation $L$ such that $B(x, y) = \langle x, L(y) \rangle$ for all $x, y \in V$.

#### Proof

• Fix a Bilinear form $B(x, y)$ on $V$.
• For a fixed vector $y \in V$, we can define a linear functional $\theta_y: V \to \mathbb{R}$ by $\theta_y(x) = B(x, y)$.
• As in the note on duality, there is a unique vector $z_y$ such that for all $x \in V$, $\theta_y(x) = \langle x, z_y\rangle$.
• We define $L: V \to V$ by $L(y) = z_y$. It is easy to see that this is a linear transformation.
• Thus, $B(x, y) = \theta_y(x) = \langle x, z_y\rangle = \langle x, L(y) \rangle$ as required.

A linear transformation from a vector space $V$ to itself is also called a linear operator on $V$.

• Let $L: V\to V$ be a linear transformation. Let $B(x, y) = \langle x, L(y)\rangle$ as before.
• Then $(x, y) \mapsto B(y, x) = \langle y, L(x)\rangle$ is also a Bilinear form.
• Hence, by the above theorem, there exists a unique linear transformation $L^*$ such that $B(y, x) = \langle x, L^*(y)\rangle$.
• Using symmetry of the inner product, we can thus write $\langle x, L(y)\rangle = \langle L^*(x), y\rangle$ for all $x, y\in V$.
• $L^*$ is called the adjoint of $L$. We say $L$ is self-adjoint if $L = L^*$.
• As $B(x, y) = \langle x, L(y)\rangle$ and $B(y, x) = \langle x, L^*(y)$ for all $x, y\in V$, it follows that $B$ is symmetric if and only if $L= L^*$, i.e. $L$ is self-adjoint.
• In particular, any quadratic form $Q(x)$ is given by $Q(x) = \langle x, L(x)\rangle$ for a unique self-adjoint linear transformation $L: V \to V$.

### Matrices

• Let $v_1$, $v_2$, $\dots$, $v_n$ be an orthonormal basis for $V$ and $L: V \to V$ a linear transformation.
• Then the matrix $A$ of $L$ with respect to $v_1$, $v_2$, $\dots$, $v_n$ has entries $a_{ij} = \langle v_i, L(v_j)\rangle$.
• By definition, the entries $a_{ij}$ of $A$ are given by $L(v_j) = \sum\limits_{i=1}^n a_{ij}v_i$, i.e., the coefficients of $L(v_j)$ expressed in terms of the basis elements $v_i$.
• We take the inner product with $v_{i_0}$, for fixed $i_0$, to get $\langle v_{i_0}, L(v_j)\rangle = \sum\limits_{i=1}^n a_{ij}\langle v_{i_0}, v_i\rangle = a_{i_0j}$ as $v_1$, $v_2$, $\dots$, $v_n$ is orthonormal.
• It follows that $a_{ij} = B(x_i, x_j)$.
• As the adjoint $L^*$ corresponds to $(x, y) \mapsto B(y, x) = \langle y, L(x)\rangle$, it follows that the $(i, j)$-entry of the matrix of $L^*$ is $B(x_j, x_i)$, i.e., the matrix of $L^*$ is the transpose $A^T$.
• In particular, self-adjoint operators are those whose matrices are symmetric.