HOME

Date: <2024-10-19 Sat>

Singular Vector

Let \(u\) and \(v\) be a pair of singular vectors of \(A\) . i.e.

\begin{equation*} Av = \sigma u \end{equation*}

\(u^T A = \sigma v^T\)

Then,

\begin{align*} u_i \cdot u_j = 0 \textrm{ for } i \neq j \\ v_i \cdot v_j = 0 \textrm{ for } i \neq j \end{align*}

Caveat: Singular vectors corresponding to distinct singular values are orthogonal. But, if the singular values are non distinct (equal) then the singular vectors need not be othogonal. However, given a set of nonorthonal singular vectors, we can construct an orthogonal basis for that set. So, even for a set of singular values where not all values are distinct, we can create a corresponding set of orthogonal singular vectors.

Proof: Since \(v\) is eigenvector of \(A^TA\) which is a real symmetric matrix, its eigenvectors are orthogonal. Proof for othogonality of eigenvector for symmetric matrices follows from definition of self adjoint operator.

\begin{equation} \label{eqn:v1} v_1 = \underset{||v||=1}{\mathrm{argmax}} ||Av|| \tag{eqn:v1} \end{equation} \begin{equation*} v_2 = \underset{||v|| = 1, v \cdot v1 = 0}{\mathrm{argmax}} ||Av|| \end{equation*}

The first statement \eqref{eqn:v1} can proved by folowing rough reasoning: [Page 10]


Backlinks


You can send your feedback, queries here