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Normalization and Decomposition of Eigenvectors

In linear algebra, an eigenvector is a special part of vectors containing a system of linear equations. Eigenvalues and eigenvectors feature prominently in the analysis of linear transformations, such as in the field of stability analysis, atomic orbitals, matrix diagonalisation, vibration analysis and many more. In this section, we will study the eigenvectors definition, vectors, normalization and decomposition of eigenvectors.

Formal Definition of Eigenvector

A nonzero vector is mapped by a given linear transformation of a vector space onto a vector that is the product of a scalar multiplied by the original vector. The eigenvector of a square matrix is defined as a non-vector by which a given matrix is multiplied and is equal to a scalar multiple of that vector.

Explanation

Suppose that A is an n x n square matrix and if v is a non-zero vector, then the product of matrix A and vector v is defined as produced of a scalar quantity λ, and the given vector, such that,

Av = λv

Where v is the eigenvector, and λ is the scalar quantity termed as the eigenvalue associated with the given matrix A.

Normalized Eigenvector

In problems related to finding eigenvectors, we often come across computation of normalized eigenvectors. A normalized eigenvector is an eigenvector having a unit length.

It can be found by simply dividing each component of the vector by the length of the vector, the vector is converted into the vector of length one.

The formula for finding the length of the vector:

\(\begin{array}{l}X = \begin{bmatrix} x_{1}\\ x_{2}\\ .\\ .\\ x_{n} \end{bmatrix}\\\end{array} \)
\(\begin{array}{l}L =\sqrt{x_{1}^{2}+x_{2}^{2}+…+x_{n}^{2}}\end{array} \)

For example, the given eigenvector is

\(\begin{array}{l}\begin{bmatrix} 1\\ -5\\ -1 \end{bmatrix}\end{array} \)

Here, 

\(\begin{array}{l}L =\sqrt{1^{2}+(-5)^{2}+1^{2}}\end{array} \)

L = 3√3

Its normalized form is represented by:

\(\begin{array}{l}\begin{bmatrix} \frac{1}{3\sqrt{3}}\\ \frac{-5}{3\sqrt{3}}\\ \frac{-1}{3\sqrt{3}} \end{bmatrix}\end{array} \)

Eigenvector Decomposition

The decomposition of a square matrix A into eigenvalues and eigenvectors is known as eigendecomposition. The decomposition of any square matrix into eigenvalues and eigenvectors is always possible as long as the matrix consisting of the eigenvectors of the given matrix is a square matrix, also explained in the eigendecomposition theorem.

As we are aware of the fact that a matrix represents a system of linear equations, the matrix can be worked out in order to determine its eigenvalues as well as eigenvectors. The determination of eigenvectors can be done only after the computation of eigenvalues. The whole process of determining eigenvectors is known as eigenvalue decomposition. It is also termed as eigendecomposition.

Solved Examples on Eigenvector Decomposition

Example 1: Show the process of eigenvector decomposition of matrix 

\(\begin{array}{l}\mathbf{A} = \begin{bmatrix} 1 & 0 \\ 1 & 3 \\ \end{bmatrix}\end{array} \)

Solution:

The 2 × 2 real matrix 

\(\begin{array}{l}\mathbf{A} = \begin{bmatrix} 1 & 0 \\ 1 & 3 \\ \end{bmatrix}\end{array} \)
may be decomposed into a diagonal matrix through the multiplication of a non-singular matrix
\(\begin{array}{l}{\displaystyle \mathbf {B} ={\begin{bmatrix}a&b\\c&d\end{bmatrix}}\in \mathbb {R} ^{2\times 2}.}\\\end{array} \)

Then

\(\begin{array}{l}\begin{bmatrix} a &b \\ c&d \end{bmatrix}^{-1}\begin{bmatrix} 1 &0 \\ 1&3 \end{bmatrix}\begin{bmatrix} a &b \\ c&d \end{bmatrix}=\begin{bmatrix} x &0 \\ 0&y \end{bmatrix}\\\end{array} \)
 
\(\begin{array}{l}\text{for some real diagonal matrix}\ \begin{bmatrix} x &0 \\ 0&y \end{bmatrix}\\\end{array} \)

Multiplying both sides of the equation on the left by B, we get 

\(\begin{array}{l}{\displaystyle {\begin{bmatrix}1&0\\1&3\end{bmatrix}}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}={\begin{bmatrix}a&b\\c&d\end{bmatrix}}{\begin{bmatrix}x&0\\0&y\end{bmatrix}}.}\\\end{array} \)

The above equation can be decomposed into two simultaneous equations.

Factoring out the eigenvalues x and y: 

\(\begin{array}{l}\begin{cases} \begin{bmatrix} 1 & 0\\ 1 & 3 \end{bmatrix} \begin{bmatrix} a \\ c \end{bmatrix} = x\begin{bmatrix} a \\ c \end{bmatrix} \\ \begin{bmatrix} 1 & 0\\ 1 & 3 \end{bmatrix} \begin{bmatrix} b \\ d \end{bmatrix} = y\begin{bmatrix} b \\ d \end{bmatrix} \end{cases}\\\end{array} \)

Letting: 

\(\begin{array}{l}{\displaystyle {\overrightarrow {a}}={\begin{bmatrix}a\\c\end{bmatrix}},\quad {\overrightarrow {b}}={\begin{bmatrix}b\\d\end{bmatrix}},}\end{array} \)
 this gives us two vector equations:

\(\begin{array}{l}\begin{cases} A \overrightarrow{a} = x \overrightarrow{a} \\ A \overrightarrow{b} = y \overrightarrow{b} \end{cases}\\\end{array} \)

And can be represented by a single vector equation involving two solutions as eigenvalues: 

\(\begin{array}{l}\mathbf{A} \mathbf{u} = \lambda \mathbf{u}\\\end{array} \)

Where, λ represents the two eigenvalues x and y, and u represents the vectors a and b.

Shifting λu to the left-hand side and factoring u out,

\(\begin{array}{l}{\displaystyle (\mathbf {A} -\lambda \mathbf {I} )\mathbf {u} =\mathbf {0} }\end{array} \)

Since B is non-singular, it is essential that u is non-zero. Therefore, 

\(\begin{array}{l}{\displaystyle \det(\mathbf {A} -\lambda \mathbf {I} )=0}\\\end{array} \)

Thus,

\(\begin{array}{l}{\displaystyle (1-\lambda )(3-\lambda )=0}\end{array} \)
gives us the solutions of the eigenvalues for the matrix A as λ = 1 or λ = 3, and the resulting diagonal matrix from the eigendecomposition of A is thus
\(\begin{array}{l}\begin{bmatrix} 1 &0 \\ 0&3 \end{bmatrix}.\\\end{array} \)

Putting the solutions back into the above simultaneous equations,

\(\begin{array}{l}\begin{cases} \begin{bmatrix} 1 & 0 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} a \\ c \end{bmatrix} = 1\begin{bmatrix} a \\ c \end{bmatrix} \\ \begin{bmatrix} 1 & 0\\ 1 & 3 \end{bmatrix} \begin{bmatrix} b \\ d \end{bmatrix} = 3\begin{bmatrix} b \\ d \end{bmatrix} \end{cases}\\\end{array} \)

Solving the equations, we get

\(\begin{array}{l}{\displaystyle a=-2c\quad {\text{and}}\quad b=0,\qquad [c,d]\in \mathbb {R} .}\\\end{array} \)

Thus, the matrix B required for the eigendecomposition of A is

\(\begin{array}{l}{\displaystyle \mathbf {B} ={\begin{bmatrix}-2c&0\\c&d\end{bmatrix}},\qquad [c,d]\in \mathbb {R} ,}\\\end{array} \)

That is,

\(\begin{array}{l}{\displaystyle {\begin{bmatrix}-2c&0\\c&d\end{bmatrix}}^{-1}{\begin{bmatrix}1&0\\1&3\end{bmatrix}}{\begin{bmatrix}-2c&0\\c&d\end{bmatrix}}={\begin{bmatrix}1&0\\0&3\end{bmatrix}},\qquad [c,d]\in \mathbb {R} }\end{array} \)

Related Links:

Eigenvectors of a Matrix

Eigenvalues and Eigenvectors Problems

Frequently Asked Questions

Q1

What do you mean by eigenvector decomposition?

Eigenvector decomposition is the method of decomposition of a square matrix A into eigenvalues and eigenvectors.

Q2

Give two applications of eigenvectors.

Eigenvectors are used in quantum mechanics. Eigenvector decomposition is used to solve linear equations of first order.

Q3

What is the eigenvalue of a singular matrix?

A singular matrix has a 0 eigenvalue.

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