EigenValues and EigenVectors

Eigenvalues and eigenvectors of matrices are needed for some of the methods such as Principal Component Analysis (PCA), Principal Component Regression (PCR), and assessment of the input of collinearity.

Eigenvalues and Eigenvectors

For a real, symmetric matrix $A_{n\times n}$ there exists a set of $n$ scalars $\lambda_i$, and $n$ non-zero vectors $Z_i\,\,(i=1,2,\cdots,n)$ such that

\begin{align*}
AZ_i &=\lambda_i\,Z_i\\
AZ_i – \lambda_i\, Z_i &=0\\
\Rightarrow (A-\lambda_i \,I)Z_i &=0
\end{align*}

The $\lambda_i$ are the $n$ eigenvalues (characteristic roots or latent root) of the matrix $A$ and the $Z_i$ are the corresponding (column) eigenvectors (characteristic vectors or latent vectors).

There are non-zero solutions to $(A-\lambda_i\,I)=0$ only if the matrix ($A-\lambda_i\,I$) is less than full rank (only if the determinant of $(A-\lambda_i\,I)$ is zero). $\lambda_i$ are obtained by solving the general determinantal equation $|A-\lambda\,I|=0$.

The determinant of $(A-\lambda\,I)$ is an $n$th degree polynomial in $\lambda$. Solving this equation gives the $n$ values of $\lambda$, which are not necessarily distinct. Each value of $\lambda$ is used in equation $(A-\lambda_i\,I)Z_i=0$ to find the companion eigenvectors $Z_i$.

When the eigenvalues are distinct, the vector solution to $(A-\lambda_i\,I)Z_i=0$ is unique except for an arbitrary scale factor and sign. By convention, each eigenvector is defined to be the solution vector scaled to have unit length; that is, $Z_i’Z_i=1$. Furthermore, the eigenvectors are mutually orthogonal; ($Z_i’Z_i=0$ when $i\ne j$).

When the eigenvalues are not distinct, there is an additional degree of arbitrariness in defining the subsets of vectors corresponding to each subset of non-distinct eigenvalues.

Example: Let the matrix $A=\begin{bmatrix}10&3\\3 & 8\end{bmatrix}$.

The eigenvalues of $A$ can be found by $|A-\lambda\,I|=0$. Therefore,

\begin{align*}
|A-\lambda\, I|&=\Big|\begin{matrix}10-\lambda & 3\\ 3& 8-\lambda\end{matrix}\Big|\\
\Rightarrow (10-\lambda)(8-\lambda)-9 &= \lambda^2 -18\lambda+71 =0
\end{align*}

By Quadratic formula, $\lambda_1 = 12.16228$ and $\lambda_2=5.83772$, arbitrarily ordered from largest to smallest. Thus the matrix of eigenvalues of $A$ is

$$L=\begin{bmatrix}12.16228 & 0 \\ 0 & 5.83772\end{bmatrix}$$

The eigenvectors corresponding to $\lambda_1=12.16228$ are obtained by solving

$(A-\lambda_2\,I)Z_i=0$ for the element of $Z_i$;

\begin{align*}
(A-12.16228I)\begin{bmatrix}Z_{11}\\Z_{21}\end{bmatrix} &=0\\
\left(\begin{bmatrix}10&3\\3&8\end{bmatrix}-\begin{bmatrix}12.162281&0\\0&12.162281\end{bmatrix}\right)\begin{bmatrix}Z_{11}\\Z_{21}\end{bmatrix}&=0\\
\begin{bmatrix}-2.162276 & 3\\ 3 & -4.162276\end{bmatrix}\begin{bmatrix}Z_{11}\\Z_{21}\end{bmatrix}&=0
\end{align*}

Arbitrary setting $Z_{11}=1$ and solving for $Z_{11}$, using first equation gives $Z_{21}=0.720759$. Thus the vector $Z_1’=\begin{bmatrix}1 & 0.72759\end{bmatrix}$ statisfy first equation.

$Length(Z_1)=\sqrt{Z_1’Z_1}=\sqrt{1.5194935}=1.232677$, where $Z’=0.999997$.

\begin{align*}
Z_1 &=\begin{bmatrix} 0.81124&0.58471\end{bmatrix}\\
Z_2 &=\begin{bmatrix}-0.58471&0.81124\end{bmatrix}
\end{align*}

The elements of $Z_2$ are found in the same manner. Thus the matrix of eigenvectors for $A$ is

$$Z=\begin{bmatrix}0.81124 &-0.58471\\0.8471&0.81124\end{bmatrix}$$

Note that matrix $A$ is of rank two because both eigenvalues are non-zero. The decomposition of $A$ into two orthogonal matrices each of rank one.

\begin{align*}
A &=A_1+A_2\\
A_1 &=\lambda_1Z_1Z_1′ = 12.16228 \begin{bmatrix}0.81124\\0.58471\end{bmatrix}\begin{bmatrix}0.81124 & 0.58471\end{bmatrix}\\
&= \begin{bmatrix}8.0042 & 5.7691\\ 5.7691&4.1581\end{bmatrix}\\
A_2 &= \lambda_2Z_2Z_2′ = \begin{bmatrix}1.9958 & -2.7691\\-2.7691&3.8419\end{bmatrix}
\end{align*}

EigenValues and EigenVectors

Thus the sum of eigenvalues $\lambda_1+\lambda_2=18$ is $trace(A)$. Thus sum of the eigenvalues for any square symmetric matrix is equal to the trace of the matrix. The trace of each of the component rank $-1$ matrix is equal to its eigenvalue. $trace(A_1)=\lambda_1$ and $trace(A_2)=\lambda_2$.

Computer MCQs Test Online

R and Data Analysis

Leave a Comment

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Discover more from Statistics for Data Analyst

Subscribe now to keep reading and get access to the full archive.

Continue reading

Scroll to Top