# Spectral Theorem

In linear algebra, one is often interested in the *canonical forms* of a linear transformation. Given a particularly nice basis for the vector spaces in which one is working, the matrix of a linear transformation may also be particularly nice, revealing some information about how the transformation operates on the vector space.

The **spectral theorem** provides a sufficient criterion for the existence of a particular canonical form. Specifically, the spectral theorem states that if the inverse of \(M\) equals the transpose of \(M\), then \(M\) is *diagonalizable*: there exists an invertible matrix \(C\) such that \(C^{-1} MC \) is a *diagonal* matrix. Recall that a diagonal matrix is any matrix for which all entries off the main diagonal (the diagonal from top left to bottom right) are zero.

A matrix \(M\) with entries in \(\mathbb{R}\) is called

symmetricif \(M =M^{T}\). Thespectral theoremstates that any symmetric matrix is diagonalizable.

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## Motivation

Consider the matrix \[A:= \begin{pmatrix} 2 & 6 \\ 0 & -1 \end{pmatrix}.\] The eigenvalues of this matrix are \(\lambda_1, \lambda_2 = 2, -1\), with corresponding eigenvectors \(v_1 = (1,0)\) and \(v_2 = (-2,1)\). These eigenvectors \(\{v_1, v_2\}\) form a basis for \(\mathbb{R}^2\), with change of basis matrix \[C:= \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\] sending the standard basis vector \(e_1 = (1,0)\) to \(v_1\) and \(e_2 = (0,1)\) to \(v_2\). In the basis \(\{v_1, v_2\}\), the matrix for the linear transformation described by \(A\) is precisely \[C^{-1} AC = \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}.\] This *diagonal* matrix is easier to parse than \(A\); looking at it immediately tells one that \(A\) acts on \(\mathbb{R}^2\) by scaling two axes, one by a factor of \(2\) and one by a factor of \(-1\) (i.e., reflection).

A matrix \(M\) is called *diagonalizable* if there is a basis in which the linear transformation described by \(M\) has a diagonal matrix, i.e. a matrix whose entries off the main diagonal (the diagonal from top left to bottom right) are all zero. Equivalently, \(M\) is diagonalizable if and only if there exists an invertible matrix \(C\) such that \(C^{-1}M C\) is a diagonal matrix. Diagonalizable matrices are easier to work with than nondiagonalizable matrices, since they can be placed in this canonical diagonal form with a change of basis.

## Proof Of Spectral Theorem

**Cite as:**Spectral Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/spectral-theorem/