A bilinear form takes two vectors and produces a scalar from the underyling field.

The function is linear in addition in both arguments.

\(\phi (au+x, bv+y)=\phi (au,bv)+\phi (au,y)+\phi (x,bv)+\phi (x,y)\)

The function is also linear in multiplication in both arguments.

\(\phi (au+x, bv+y)=ab\phi (u,v)+a\phi (u,y)+b\phi (x,v)+\phi (x,y)\)

They can be represented as:

\(\phi (u,v)=v^TMu\)

Like bilinear forms, sesquilinear are linear in addition:

\(\phi (au+x, bv+y)=\phi (au,bv)+\phi (au,y)+\phi (x,bv)+\phi (x,y)\)

Sesqulinear forms however are only multiplictively linear in the second argument.

\(\phi (au+x, bv+y)=b\phi (au,v)+\phi (au,y)+b\phi (x,v)+\phi (x,y)\)

In the first argument they are "twisted"

\(\phi (au+x, bv+y)=\bar ab\phi (u,v)+\bar a\phi (u,y)+b\phi (x,v)+\phi (x,y)\)

For the real field, \(\bar b = b\) and so the sesqulinear form is the same as the bilinear form.

We can show the sesquilinear form as \(v^*Mu\)

\(f(M)=f([v_1,v_2])\)

We introduce \(e_i\), the element vector. This is \(0\) for all entries except for \(i\) where it is \(1\). Any vector can be shown as a sum of these vectors multiplied by a scalar.

\(f(M)=f([\sum^m_{i=1}a_{1i}e_i,\sum^m_{i=1}a_{2i}e_i])\)

\(f(M)=\sum_{k=1}^mf([a_{1k}e_k,\sum^m_{i=1}a_{2i}e_i])\)

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}f([a_{1k}e_k,a_{2i}e_i])\)

Because this in linear in scalars:

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}a_{1k}^*a_{2i}f([e_k,e_i])\)

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}a_{1k}^*a_{2i}e_k^*Me_i\)

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}a_{1k}^*a_{2i}e_k^*Me_i\)

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}a_{1k}^*a_{2i}e_k^*e_i\)

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}a_{1k}^*a_{2i}\delta_i^k\)

\(f(M)=\sum^m_{i=1}a_{1i}^*a_{2i}\)

For bilinear forms, the transformations which preserved metrics were:

\(P^T=P^{-1}\)

For sesquilinear they are different:

\(u^*Mv\)

\((Pu)^*M(Pv)\)

\(u^*P^*MPv\)

So we want the matrices where:

\(P^*MP=M\)

The unitary group is where \(M=I\)

\(P^*P=I\)

\(P^*=P^{-1}\)

We refer to these using \(U\) instead of \(P\).

\(U^*=U^{-1}\)

The unitary group depends on the dimension of the vector space, and the underlying field. So we can have:

\(U(n, R)\); and

\(U(n, C)\).

For the \(U(n, R)\) we have:

\(U^*=U^{-1}\)

\(U^T=U^{-1}\)

This is the condition for the orthogonal group, and so we would instead write \(O(n)\).

As a result, \(U(n)\) refers to \(U(n,C)\).

A matrix where \(M=M^*\)

For matrices over the real numbers, these are the same as symmetric matrices.

\(\phi (u,v)=u^*Mv\)

\((u^*Mv)^*=v^*M^*u=v^*Mu\)

\(\phi (u,v)=\overline {\phi (v,u)}\)

\((v^*Mv)^*=v^*M^*v=v^*Mv\)

So we have:

\((v^*Mv)^*=v^*Mv\)

Which is only satisfied for reals.

If \(A\) and \(B\) are Hermitian, \(AB\) is Hermitian if and only if \(AB\) commutes.

\((AB)^*=B^*A^*=BA\)

If it commutes then

\((AB)^*=AB\)

Hermitian matrices have real eigenvalues.

\(Hv=\lambda v\)

\(v^*Hv=\lambda v^*v\)

\(v^*Hv=\lambda\)

These are also known as anti-Hermitian matrices.

\(M^*=-M\)

Pauli matrices are \(2\times 2\) matrices which are unitary and hermitian.

That is, \(P^*=P^{-1}\).

And \(P^*=P\).

The matrices are:

\(\sigma_1 =\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}\)

\(\sigma_2 =\begin{bmatrix} 0&-i \\ i&0 \end{bmatrix}\)

\(\sigma_3 =\begin{bmatrix} 1&0 \\ 0&-1 \end{bmatrix}\)

The identity matrix is often considered alongside these as:

\(\sigma_0 =\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}\)

\(\sigma_i^2 =\sigma_i\sigma_i\)

\(\sigma_i^2 =\sigma_i\sigma_i^*\)

\(\sigma_i^2 =\sigma_i\sigma_i^{-1}\)

\(\sigma_i^2 =I\)

\(\det \sigma_i =-1\)

\(Tr (\sigma_i) =0\)

As the sum of eigenvalues is the trace, and the product is the determinant, the eigenvalues are \(1\) and \(-1\).

The matrix \(M\) is positive definite if for all non-zero vectors the scalar is positive.

\(v^TMv\)

We know that the outcome is a scalar, so:

\(v^TMv=(v^TMv)^T\)

\(v^TMv=v^TM^Tv\)

\(v^T(M-M^T)v=0\)

An inner product is a sesquilinear form with a positive-definite Hermitian matrix.

\(\langle u, v \rangle =u^*Hv\)

If we are using the real field this is the same as:

\(\langle u, v \rangle =u^THv\)

Where \(H\) is now a symmetric real matrix.

\(\langle v, v \rangle =v^*Hv\)

Always positive and real.

\(\langle u, v\rangle \langle v, u\rangle=|\langle u, v\rangle|^2\)

This states that:

\(|\langle u,v\rangle |^2 \le \langle u, u\rangle \dot \langle v, v\rangle\)

Consider the vectors \(u\) and \(v\). We construct a third vector \(u-\lambda v\). We know the length of any vector is non-negative. \(0\le \langle u-\lambda v, u-\lambda v\rangle\)

\(0\le \langle u, u\rangle+ \langle u, -\lambda v\rangle+\langle -\lambda v, u\rangle+ \langle -\lambda v, -\lambda v\rangle\)

\(0\le \langle u, u\rangle-\bar{\lambda }\langle u, v\rangle-\lambda { \langle v, u\rangle }+ \lambda \bar{\lambda }\langle v, v\rangle\)

We now look for a value of \(\lambda\) to simplify this equation.

\(\lambda = \dfrac{\langle u,v \rangle}{\langle v, v\rangle}\)

\(0\le \langle u, u\rangle-\dfrac{\langle v,u \rangle\langle u, v\rangle}{\langle v, v\rangle}-\dfrac{\langle u,v \rangle \langle v, u\rangle }{\langle v, v\rangle}+ \dfrac{\langle u,v \rangle}{\langle v, v\rangle}\dfrac{\langle v,u \rangle}{\langle v, v\rangle}\langle v, v\rangle\)

\(0\le \langle u, u\rangle-\dfrac{|\langle u,v \rangle|^2}{\langle v, v\rangle}\)

\(|\langle u,v \rangle|^2\ge \langle u, u\rangle\langle v, v\rangle\)

in inner product space, orthogonal projection

\(p_uv = \dfrac{<u,v>}{{v,v}}v\)

we then know that \(o=v-p_uv\) is orthogonal to \(u\).

if a set of vectors are all orthogonal, they form an orthogonal set if the set spans the vector space, it is an orthogonal basis.

can we form an orthogonal basis from a non-orthogonal basis? yes, using gram schmidt

we have \(x_1\), \(x_2\) \(x_3\) etc we want to make \(v_1\), \(v_2\) etc orthognal

\(v_1 = x_1\) \(v_2 = x_2 - p_{x_2}v_1\) \(v_3 = x_3 - p_{x_3}v_1 - p_{x_3}v_2v\)

The special unitary group, \(SU(n,F)\), is the subgroup of \(U(n,F)\) where the determinants are \(1\).

That is, \(|M|=1\)

The determinant of the unitary matrices is:

\(\det U^*=\det U^{-1}\)

\((\det U)^*=\dfrac{1}{\det U}\)

\((\det U)^*\det U = 1\)

\(||\det U||= 1\)

\(M^*M=MM^*\)

All symmetrix matrices are normal

All hermetitian matrices (inc subset symmetric) are normal

Normal matrix never defective