# Spectral Decomposition

Originally, spectral decomposition was developed for symmetric or self-adjoint matrices. Following tradition, we present this method for symmetric/self-adjoint matrices, and later expand it for arbitrary matrices.

A square matrix A is said to be unitary diagonalizable if there is a unitary matrix U such that $${\bf U}^{\ast} {\bf A}\,{\bf U} = {\bf \Lambda} ,$$ where Λ is a diagonal matrix and $${\bf U}^{\ast} = {\bf U}^{-1} .$$
A matrix A is said to be orthogonally diagonalizable if there is an orthogonal matrix P such that $${\bf P}^{\mathrm T} {\bf A}\,{\bf P} = {\bf \Lambda} ,$$ where Λ is a diagonal matrix and $${\bf P}^{\mathrm T} = {\bf P}^{-1} .$$
Theorem: Let A be a square n × n matrix.
1. Matrix A is orthogonally diaginalisable if and only if A is symmetric ($${\bf A} = {\bf A}^{\mathrm T}$$ ).
2. The matrix A is unitary diaginalisable if and only if A is normal ($${\bf A}\, {\bf A}^{\ast} = {\bf A}^{\ast} {\bf A}$$ ). ■

Example: The matrix
${\bf A} = \begin{bmatrix} 1&{\bf j}&0 \\ {\bf j}&1&0 \\ 0&0&1 \end{bmatrix}$
is symmetric, normal, but not self-adjoint. Another matrix
${\bf B} = \begin{bmatrix} 2&1 \\ -1&2 \end{bmatrix}$
is normal, but not self-adjoint. Therefore, both matrices are unitary diagonalizable but not orthogonally diagonalizable.
Example: Consider a symmetric matrix
${\bf A} = \begin{bmatrix} 2&1&1 \\ 1&2&1 \\ 1&1&2 \end{bmatrix}$
that has the characteristic polynomial $$\chi_{A} (\lambda ) = \det \left( \lambda {\bf I} - {\bf A} \right) = \left( \lambda -1 \right)^2 \left( \lambda -4 \right) .$$ Thus, the distinct eigenvalues of A are $$\lambda_1 =1,$$ which has geometrical multiplicity 2, and $$\lambda_3 =4.$$ The corresponding eigenvectors are
${\bf u}_1 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} , \quad {\bf u}_2 = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \qquad \mbox{and} \qquad {\bf u}_3 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} .$
The vectors $${\bf u}_1 , \ {\bf u}_2$$ form the basis for the two-dimensional eigenspace corresponding $$\lambda_1 =1 ,$$ while $${\bf u}_3$$ is the eigenvectors corresponding to $$\lambda_3 =4 .$$ Applying the Gram--Schmidt process to $${\bf u}_1 , \ {\bf u}_2$$ yields the following orthogonal basis:
${\bf v}_1 = {\bf u}_1 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \quad \mbox{and} \quad {\bf v}_2 = {\bf u}_2 - \frac{\langle {\bf u}_2 , {\bf v}_1 \rangle}{\| {\bf v}_1 \|^2} \, {\bf v}_1 = \frac{1}{2} \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$
because $$\langle {\bf u}_2 , {\bf u}_1 \rangle = -1$$ and $$\| {\bf v}_1 \|^2 =2 .$$ Normalizing these vectors, we obtain orthonormal basis:
${\bf q}_1 = \frac{{\bf v}_1}{\| {\bf v}_1 \|} = \frac{1}{\sqrt{2}} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} , \quad {\bf q}_2 = \frac{{\bf v}_2}{\| {\bf v}_2 \|} = \frac{1}{\sqrt{6}} \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} , \quad {\bf q}_3 = \frac{{\bf v}_3}{\| {\bf v}_3 \|} = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} .$
Finally, using $${\bf q}_1 ,\ {\bf q}_2 , \ {\bf q}_3$$ as column vectors, we obtain the unitary matrix
${\bf U} = \begin{bmatrix} \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0& \frac{-2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{bmatrix} ,$
which orthogonally diagonalizes A. As a check, we confirm
${\bf U}^{\mathrm T} {\bf A} \,{\bf U} = \begin{bmatrix} \frac{-1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{6}} & \frac{-2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{bmatrix} \, \begin{bmatrix} 2&1&1 \\ 1&2&1 \\ 1&1&2 \end{bmatrix} \, \begin{bmatrix} \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0& \frac{-2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{bmatrix} = \begin{bmatrix} 1 &0&0 \\ 0 &1&0 \\ 0 &0&4 \end{bmatrix} ,$
Theorem: Let A be a symmetric or normal square n × n matrix, with eigenvalues $$\lambda_1 , \ \lambda_2 , \ \ldots , \ \lambda_n$$ and corresponding eigenvectors $${\bf u}_1 , \ {\bf u}_2 , \ \ldots , \ {\bf u}_n .$$ Then
${\bf A} = \begin{bmatrix} \uparrow & \uparrow & \cdots & \uparrow \\ {\bf u}_1 & {\bf u}_2 & \cdots & {\bf u}_n \\ \downarrow & \downarrow & \cdots & \downarrow \end{bmatrix} \, \begin{bmatrix} \lambda_1 &&&0 \\ &\lambda_2 && \\ &&\ddots & \\ 0&&& \lambda_n \end{bmatrix} \, \begin{bmatrix} \longleftarrow & {\bf u}_1 & \longrightarrow \\ \longleftarrow & {\bf u}_2 & \longrightarrow & \\ \vdots & \\ \longleftarrow & {\bf u}_n & \longrightarrow \end{bmatrix} = \sum_{i=1}^n \lambda_i {\bf u}_i {\bf u}_i^{\ast} ,$
which is called spectral decomposition for a symmetric/ normal matrix A. ■

The term was cointed around 1905 by a German mathematician David Hilbert (1862--1943). A substantial part of Hilbert’s fame rests on a list of 23 research problems he enunciated in 1900 at the International Mathematical Congress in Paris. In his address, “The Problems of Mathematics,” he surveyed nearly all the mathematics of his day and endeavoured to set forth the problems he thought would be significant for mathematicians in the 20th century. Many of the problems have since been solved, and each solution was a noted event. He reduced geometry to a series of axioms and contributed substantially to the establishment of the formalistic foundations of mathematics. His work in 1909 on integral equations led to 20th-century research in functional analysis. His work also established the basis for his work on infinite-dimensional space, later called Hilbert space, a concept that is useful in mathematical analysis and quantum mechanics.

In 1895, Hilbert became Professor of Mathematics at the University of Göttingen, which was the 20th century global hub of renowned mathematicians. It was here that he enjoyed the company of notable mathematicians. Under Hilbert, Göttingen reached its peak as one of the great mathematical centres of the world. No one in recent years has surpassed his dual capacity, for seeing and overcoming the central difficulty of some major topic, and for propounding new problems of vital importance. On January 23, 1930, David Hilbert reached the mandatory retirement age of 68. Among the many honours bestowed upon him, he was made an "honorary citizen" of his native town of Königsberg (now Kaliningrad, Russia). He continued working as co-editor of Mathematische Annalen until 1939. The last years of Hilbert’s life and of many of his colleagues and students was overshadowed by the Nazi rule. Hilbert courageously spoke out against repression of Jewish mathematicians in Austria and Germany in mid 1930s. However, after mass evictions, several suicides, and assassinations, he eventually remained silent.

Denoting the rank one projection matrix $${\bf u}_i {\bf u}_i^{\ast}$$ by $${\bf E}_i = {\bf u}_i {\bf u}_i^{\ast} ,$$ we obtain spectral decomposition of A:
${\bf A} = \lambda_1 {\bf E}_1 + \lambda_2 {\bf E}_2 + \cdots + \lambda_n {\bf E}_n .$
This formular allows one to define a function of a square symmetric/self-adjoint matrix:
$f\left( {\bf A} \right) = f(\lambda_1 )\, {\bf E}_1 + f(\lambda_2 )\, {\bf E}_2 + \cdots + f(\lambda_n )\,{\bf E}_n$
because the matrices $${\bf E}_k , \quad k=1,2,\ldots n,$$ are projection matrices:
${\bf E}_i {\bf E}_j = \delta_{i,j} {\bf E}_i = \begin{cases} {\bf E}_i , & \mbox{ if } i=j, \\ {\bf 0} , & \mbox{ if } i \ne j , \end{cases} \qquad i,j =1,2,\ldots n.$
Example: Consider a self-adjoint 2-by-2 matrix
${\bf A} = \begin{bmatrix} 1 &2+{\bf j} \\ 2- {\bf j} & 5\end{bmatrix} ,$
where $${\bf j}^2 =-1 .$$ We have $${\bf S}^{\ast} {\bf A} {\bf S} = {\bf S}^{-1} {\bf A} {\bf S} = {\bf \Lambda}$$ for the matrices
${\bf S} = \begin{bmatrix} \left( 2+{\bf j} \right) / \sqrt{6} & \left( 2+{\bf j} \right) / \sqrt{30} \\ - 1/\sqrt{6} & 5\sqrt{30} \end{bmatrix} \quad \mbox{and} \quad {\bf \Lambda} = \begin{bmatrix} 0&0 \\ 0& 6 \end{bmatrix} .$
Here column vectors in matrix S are normalized eigenvectors corresponding eigenvalues $$\lambda_1 =0 \quad \mbox{and} \quad \lambda_2 =6 .$$

Therefore, the spectral decomposition of A becomes $${\bf A} = 0\,{\bf E}_1 + 6\,{\bf E}_2 ,$$ which is clearly matrix A itself.

In our case, projection matrices are

${\bf E}_1 = \frac{1}{6} \begin{bmatrix} 5 & -2 - {\bf j} \\ -2+{\bf j} & 1 \end{bmatrix} , \qquad {\bf E}_2 = \frac{1}{6} \begin{bmatrix} 1 &2+{\bf j} \\ 2- {\bf j} & 5\end{bmatrix} = \frac{1}{6}\, {\bf A} .$
It is easy to check that
${\bf E}_1^2 = {\bf E}_1 , \qquad {\bf E}_2^2 = {\bf E}_2 , \qquad \mbox{and} \qquad {\bf E}_1 {\bf E}_2 = {\bf 0} .$
The exponential matrix-function is
$e^{{\bf A}\,t} = {\bf E}_1 + e^{6t} \,{\bf E}_2 .$
Example: Consider a symmetric 3-by-3 matrix from the previous example
${\bf A} = \begin{bmatrix} 2&1&1 \\ 1&2&1 \\ 1&1&2 \end{bmatrix} .$
Its spectral decomposition is $${\bf A} = 1\,{\bf E}_1 + 1\,{\bf E}_2 + 4\,{\bf E}_3 ,$$ where the projection matrices $${\bf E}_i = {\bf q}_i {\bf q}_i^{\ast}$$ are obtained from the orthonormal eigenvectors:
\begin{align*} {\bf E}_1 &= \frac{1}{6} \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix} \left[ -1 \ -1 \ 2 \right] = \frac{1}{6} \begin{bmatrix} 1&1& -2 \\ 1&1& -2 \\ -2&-2& 4 \end{bmatrix} , \\ {\bf E}_2 &= \frac{1}{2} \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} \left[ -1 \ 1 \ 0 \right] = \frac{1}{2} \begin{bmatrix} 1&-1&0 \\ -1&1&0 \\ 0&0&0 \end{bmatrix} , \\ {\bf E}_3 &= \frac{1}{3} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \left[ 1 \ 1 \ 1 \right] = \frac{1}{3} \begin{bmatrix} 1&1& 1 \\ 1&1& 1 \\ 1&1& 1 \end{bmatrix} . \end{align*}
Indeed, these diagonalizable matrices satisfy the following relations:
${\bf E}_1^2 = {\bf E}_1 , \quad {\bf E}_2^2 = {\bf E}_2 , \quad {\bf E}_3^2 = {\bf E}_3 , \quad {\bf E}_1 {\bf E}_2 = {\bf 0} , \quad {\bf E}_1 {\bf E}_3 = {\bf 0} , \quad {\bf E}_3 {\bf E}_2 = {\bf 0} ,$
and they all have eigenvalues $$\lambda = 1, 0, 0 .$$ Using this spectral decomposition, we define two matrix-functions corresponding to $${\Phi}(\lambda ) = \cos \left( \sqrt{\lambda} \,t \right)$$ and $${\Psi}(\lambda ) = \frac{1}{\sqrt{\lambda}} \,\sin \left( \sqrt{\lambda} \,t \right)$$ that do not depend on the branch of the choisen square root:
\begin{align*} {\bf \Phi} (t) &= \cos \left( \sqrt{\bf A} \,t \right) = \cos t\, {\bf E}_1 + \cos t\, {\bf E}_2 + \cos (2t) \,{\bf E}_3 = \frac{\cos t}{3} \, \begin{bmatrix} 2&-1&-1 \\ -1&2&-1 \\ -1&-1& 2 \end{bmatrix} + \frac{\cos 2t}{3} \begin{bmatrix} 1&1& 1 \\ 1&1& 1 \\ 1&1& 1 \end{bmatrix} , \\ {\bf \Psi} (t) &= \frac{1}{\sqrt{\bf A}} \,\sin \left( \sqrt{\bf A} \,t \right) = \sin t\, {\bf E}_1 + \sin t\, {\bf E}_2 + \frac{\sin (2t)}{2} \,{\bf E}_3 = \frac{\sin t}{3} \, \begin{bmatrix} 2&-1&-1 \\ -1&2&-1 \\ -1&-1& 2 \end{bmatrix} + \frac{\sin 2t}{6} \begin{bmatrix} 1&1& 1 \\ 1&1& 1 \\ 1&1& 1 \end{bmatrix} . \end{align*}
These matrix-functions are solutions of the following initial value problems for the second order matrix differential equations:
\begin{align*} & \ddot{\bf \Phi}(t) + {\bf A}\,{\bf \Phi} (t) ={\bf 0} , \qquad {\bf \Phi}(0) = {\bf I}, \quad \dot{\bf \Phi}(0) = {\bf 0}, \\ &\ddot{\bf \Psi}(t) + {\bf A}\,{\bf \Phi} (t) ={\bf 0} , \qquad {\bf \Phi}(0) = {\bf 0}, \quad \dot{\bf \Psi}(0) = {\bf I} . \end{align*}
Since the given matrix A is positive definite, we can define four square roots:
\begin{align*} {\bf R}_1 &= {\bf E}_1 + {\bf E}_2 + 2\,{\bf E}_3 = \frac{1}{3} \begin{bmatrix} 4&1&1 \\ 1&4&1 \\ 1&1&4 \end{bmatrix} , \\ {\bf R}_2 &= {\bf E}_1 + {\bf E}_2 - 2\,{\bf E}_3 = \begin{bmatrix} 0&-1&-1 \\ -1&0&-1 \\ -1&-1&0 \end{bmatrix} , \\ {\bf R}_3 &= {\bf E}_1 - {\bf E}_2 + 2\,{\bf E}_3 = \frac{1}{3} \begin{bmatrix} 1&4&1 \\ 4&1&1 \\ 1&1&4 \end{bmatrix} , \\ {\bf R}_4 &= -{\bf E}_1 + {\bf E}_2 + 2\,{\bf E}_3 = \begin{bmatrix} 1&0&1 \\ 0&1&1 \\ 1&1&0 \end{bmatrix} , \end{align*}

and four others are just negative of these four; so total number of square roots is 8. Note that we cannot obtain $${\bf R}_3$$ and $${\bf R}_4$$ using neither Sylvester's method nor the Resolvent method because they are based on the minimal polynomial $$\psi (\lambda ) = (\lambda -1)(\lambda -4) .$$

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