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Eigendecomposition

We call \(\textbf{v}\) an eigenvector of a square matrix \(\textbf{A}\) if \(\textbf{v}\) is nonzero and for some scalar \(\lambda\) we have:

\(\textbf{A} \textbf{v} = \lambda \textbf{v}\)

We call \(\lambda\) the eigenvalue of \(\textbf{A}\).

A linear transformation acts on its eigenvectors like a scalar would. The eigenvalue is the factor by which the eigenvector is stretched (or squished) under the transformation.

Suppose that \(\textbf{A}\) has \(n\) linearly independent eigenvectors, then the eigendecomposition of \(\textbf{A}\) is:

\(\textbf{A} = \textbf{V} \textrm{diag}(\boldsymbol{\lambda}) \textbf{V}^{-1}\)

where each column of \(\textbf{V}\) is an eigenvector of \(\textbf{A}\) and the \(i\)-th component of \(\boldsymbol{\lambda}\) is the eigenvalue associated with the \(i\)-th column of \(\textbf{V}\).

When \(\textbf{A}\) is a real, symmetric matrix, then \(\textbf{A}\) can be decomposed as follows:

\(\textbf{A} = \textbf{Q} \boldsymbol{\Lambda} \textbf{Q}^T\)

where each column of \(\textbf{Q}\) is an eigenvector of \(\textbf{A}\) and \(\Lambda_{i,i}\) is the eigenvalue associated with the \(i\)-th column of \(\textbf{Q}\). We can then interpret the original transformation as scaling the space by \(\Lambda_{i,i}\) in the direction of the \(i\)-th column of \(\textbf{Q}\).

One application of the eigendecomposition is to compute \(\textbf{A}^n\) efficiently (https://math.stackexchange.com/questions/2628253/compute-the-100th-power-of-a-given-matrix).

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