PX153 - K16 - diagonalization

o use similarity transformations to diagonalize matrices:

S = [\begin{matrix} ⋮ & ⋮ & ⋮ & ⋮ \\ {\vec{x}}^{1} & {\vec{x}}^{2} & {\vec{x}}^{3} & \dots & {\vec{x}}^{n} \\ ⋮ & ⋮ & ⋮ & ⋮ \end{matrix}]

make the $n^{t h}$ column the $n^{t h}$ eigenvector of $A$
transforming $A$ with $S$ will make $A^{'}$ , a diagonal matrices
if $n$ independent eigenvectors can be found , $A^{'}$ is diagonal, if $S$ has column vectors composed of the $n$ independent eigenvectors
for hermitian matrices, the eigenvectors are orthogonal:

({\vec{x}}^{i})^{†} ({\vec{x}}^{j}) = 0 f o r i \neq j

$S$ with columns as eigenvectors:

\begin{aligned} S^{†} & = [\begin{array}{c} \dots & ({\vec{x}}^{1})^{*} & \dots \\ \dots & ({\vec{x}}^{2})^{*} & \dots \\ ⋮ \\ \dots & ({\vec{x}}^{n})^{*} & \dots \end{array}] \\ S^{†} S & = I \end{aligned}

hence, $S$ is unitary
considering $S^{- 1} A S = S^{†} A S :$

S^{- 1} A S = λ_{n} I

the similarity transformation has made $A$ a diagonal matrix with eigenvalues in the diagonals
eg: diagonalize $A = [\begin{matrix} 3 & 2 \\ 2 & 0 \end{matrix}]$
- the matrix is hermitian: $A^{†} = A ⟹ S^{- 1} = S^{†}$
- for eigenvalues and eigenvectors:

\begin{aligned} det (A - I λ) & = 0 \\ λ^{2} - 3 λ - 4 & = 0 \\ (λ - 4) (λ + 1) & = 0 \\ [\begin{array}{c} 3 & 2 \\ 2 & 0 \end{array}] [\begin{array}{c} a \\ b \end{array}] & = λ [\begin{array}{c} a \\ b \end{array}] \\ f o r λ_{1} = 4 : \\ 3 a + 2 b & = 4 a \\ 2 a & = 4 b \\ ∴ a & = 2 b \\ ∴ {\vec{x}}^{1} & = \frac{1}{\sqrt{5}} [\begin{array}{c} 2 \\ 1 \end{array}] \\ f o r λ_{1} = - 1 : \\ 3 a + 2 b & = - a \\ 2 a & = - b \\ ∴ - 2 a & = b \\ ∴ {\vec{x}}^{2} & = \frac{1}{\sqrt{5}} [\begin{array}{c} 1 \\ - 2 \end{array}] \end{aligned}

- to diagonalize:

\begin{aligned} S & = [\begin{array}{c} {\vec{x}}^{1} & {\vec{x}}^{2} \\ ⋮ & ⋮ \end{array}] \\ S^{- 1} A S & = S^{†} A S \\ = \frac{1}{\sqrt{5}} \frac{1}{\sqrt{5}} [\begin{array}{c} 1 & - 2 \\ 2 & 1 \end{array}] [\begin{array}{c} 3 & 2 \\ 2 & 0 \end{array}] [\begin{array}{c} 1 & 2 \\ - 2 & 1 \end{array}] \\ = \frac{1}{5} [\begin{array}{c} - 5 & 0 \\ 0 & 20 \end{array}] \\ = [\begin{array}{c} - 1 & 0 \\ 0 & 4 \end{array}] \end{aligned}