$$

Find the spectral decomposition

Problem

Let

$$A=\left[\begin{array}{cccc}0& 0& 1& 1\\ 0& 0& 1& 1\\ 1& 1& 0& 0\\ 1& 1& 0& 0\end{array}\right].$$

Find the spectral decomposition of

$A$.

Thought

The spectral decomposition of

$A$ is a way to write

$A$ as

$$A=\sum _{j=1}^q{\mu }_j{P}_j$$

such that

• $P_i^2=P_i$ for any
  $i$,
• $P_i{P}_j=O$ for any
  $i$ and
  $j$, and
• $\sum _{j=1}^q{P}_j=I_n$.

Intuitively,

$P_j$'s are projection matrices whose ranges are orthogonal to each other and span the whole space.

In practice, we may first find the eigenvalues

${\lambda }_1,\dots ,{\lambda }_n$ and an orthonormal basis

$\{{\mathbf{u}}_1,\dots ,{\mathbf{u}}_n\}$ of

${\mathbb{R}}^n$. Then

$$\begin{array}{rl}A& ={\lambda }_1{\mathbf{u}}_1{\mathbf{u}}_1^{\mathrm{\top }}+\cdots +{\lambda }_n{\mathbf{u}}_n{\mathbf{u}}_n^{\mathrm{\top }}\\ & ={\mu }_1{P}_1+\cdots +{\mu }_q{P}_q,\end{array}$$

where

${\mu }_1,\dots ,{\mu }_q$ are the distinct eigenvalues of

$A$ and

$P_j$ is the sum of

${\mathbf{u}}_i{\mathbf{u}}_i^{\mathrm{\top }}$ for all

${\lambda }_i={\mu }_j$.

Sample answer

According to Find an orthonormal basis composed of eigenvectors, we already have

${\lambda }_1=2$,

${\lambda }_2=-2$,

${\lambda }_3={\lambda }_4=0$. Thus, we may set

${\mu }_1=2$,

${\mu }_2=-2$ and

${\mu }_3=0$ with

$q=3$.

The corresponding eigenvectors are

${\mathbf{u}}_1$,

${\mathbf{u}}_2$,

${\mathbf{u}}_3$, and

${\mathbf{u}}_4$. As

${\mathbf{u}}_3$ and

${\mathbf{u}}_4$ both correspond to the eigenvalue

${\mu }_3=0$, we will add them together to form a projection matrix onto

$\ker (A-0I)$. In summary, we have

$$\begin{array}{rl}A& ={\lambda }_1{\mathbf{u}}_1{\mathbf{u}}_1^{\mathrm{\top }}+{\lambda }_2{\mathbf{u}}_2{\mathbf{u}}_2^{\mathrm{\top }}+{\lambda }_3{\mathbf{u}}_3{\mathbf{u}}_3^{\mathrm{\top }}+{\lambda }_4{\mathbf{u}}_4{\mathbf{u}}_4^{\mathrm{\top }}\\ & ={\mu }_1{\mathbf{u}}_1{\mathbf{u}}_1^{\mathrm{\top }}+{\mu }_2{\mathbf{u}}_2{\mathbf{u}}_2^{\mathrm{\top }}+{\lambda }_3({\mathbf{u}}_3{\mathbf{u}}_3^{\mathrm{\top }}+{\mathbf{u}}_4{\mathbf{u}}_4^{\mathrm{\top }})\\ & =2P_1+(-2)P_2+0P_3,\end{array}$$

where

$$P_1=\frac{1}{4}\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 1& 1& 1\\ 1& 1& 1& 1\\ 1& 1& 1& 1\end{array}\right],P_2=\frac{1}{4}\left[\begin{array}{cccc}1& 1& -1& -1\\ 1& 1& -1& -1\\ -1& -1& 1& 1\\ -1& -1& 1& 1\end{array}\right],$$

and

$$P_3=\frac{1}{2}\left[\begin{array}{cccc}1& -1& 0& 0\\ -1& 1& 0& 0\\ 0& 0& 1& -1\\ 0& 0& -1& 1\end{array}\right],$$

which is the spectral decomposition of

$A$.

This note can be found at Course website > Learning resources.