We can use our Fourier transform algorithm to estimate the phase of an eigenvalue of a unitary matrix

Eigenvalues and Eigenvectors

💼 Case: let $A$ be an $m\times m$ matrix where $A_{ij}\in \mathbb C$

• We can write the eigen equation as, where $\vec v$ is an eigenvector and $\lambda$ eigenvalue

  $$ A\vec v=\lambda \vec v \qquad \Rightarrow \qquad (A-\lambda I)\vec v=0 $$

  now for $\vec v\ne \vec 0$ we need $\det (A-\lambda I)=0$ so that it is irreversible

• $\det (A-\lambda I)$ is a polynomial of degree $m$ with $m$ roots for $\lambda$ (some might be equal)

• If an eigenvalue $\lambda$ has multiple linearly independent eigenvectors we call it degenerate

💼 Case: lets look at spectral decomposition

• A linear operator $A$ on a vector space can be written in the form

  $$ A=\sum_j \lambda_j \ket{j}\bra{j} $$

  if there is an orthonormal basis $\{\ket{j}\}$ of eigenvectors of $A$. In that basis the matrix representation of $A$ is diagonal with diagonal entries $\lambda_j$

  🗒️ Note: all Hermitian operators $A=A^\dag$ on a finite complex vector space are diagonalisable

Diagonalisation of Linear Maps

💼 Case: consider a diagonalisable linear map $L:\mathbb C^n \to \mathbb C^n$

🧠 Remember: a linear map is a function $L$ between two spaces such that for any $\vec u,\vec v,\alpha$ we have

$$ L(\vec u+\vec v)=L(\vec u)+L(\vec v) \qquad L(\alpha \vec u )=\alpha L(\vec u) $$

🗒️ Note: we can represent linear maps as matrices (relation to above) $L(\vec x)\leftrightarrow A\vec x$

🤖 Algorithm: the following algorithm will diagonalise the linear map $L$ defined above

1. We find all the eigenvalues $\lambda_{1},\ldots ,\lambda_k$ where $1\le k \le n$ of A

   🗒️ Note: some $\lambda$ might be degenerate ie $\lambda_i=\lambda_j$ where $j\ne i$, which is why we defined $k$ not necessarily equal to $n$ (number of dimensions)

   • We need so specify the multiplicity $n_j \in \N$ of each $\lambda_j$ such that $\sum^k_{j=1}n_j=n$

2. For each eigenvalue of $\lambda_j$ we find $n_j$ orthonormal eigenvectors $\ket{v_{j,m}}$ where $1\le m\le n_j$

💎 Conclusion: the eigenvectors will form an orthonormal basis where $L$ is diagonal

💼 Case: we have a quantum operator $U$ that acts on $l\in \N$ qubits with eigenvalue $e^{2\pi i \phi~}$ associated with the eigenvector $\ket{u}$

🧽 Assume: we don’t know $U$, $\phi$ or $\ket{u}$ and $\phi$ can be expressed by exactly $t$ qubits