When studying eigenvalues of a matrix, we often compute them one by one, starting with the eigenvalue that has the largest absolute value. However, repeatedly applying the power method to the same matrix will cause the iterations to converge to the eigenvalue that has already been found. The deflation method solves this problem by applying a transformation that replaces the computed eigenvalue with zero.
Eigenvalues of a Matrix: The Idea Behind the Deflation Method
Consider a real square matrix \( A \) of order \( n \):
\[
A=
\begin{pmatrix}
a_{11} & a_{12} & \dots & a_{1n}\\
a_{21} & a_{22} & \dots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \dots & a_{nn}
\end{pmatrix}.
\]
Suppose that the matrix \( A \) has \( n \) linearly independent eigenvectors and that its eigenvalues are real and arranged in descending order of their absolute values:
\[
|\lambda_1|>|\lambda_2|>\dots>|\lambda_n|.
\]
This ordering makes it possible to find the eigenvalues sequentially using the power method. The complete set of eigenvalues of a matrix is called its spectrum.
Let the right eigenvectors of the matrix \( A \) be denoted by
\[
X_1,X_2,\dots,X_n.
\]
For each of them, the following equality holds:
\[
A\cdot X_j=\lambda_j\cdot X_j,\qquad j=1,2,\dots,n.
\]
The eigenvalue \( \lambda_1 \) with the largest absolute value and its corresponding eigenvector \( X_1 \) can be found using the power method. If this method is applied to the original matrix again, it will once more approximate \( \lambda_1 \). Therefore, before finding \( \lambda_2 \), we need to construct a transformed matrix whose spectrum contains zero instead of \( \lambda_1 \).
For a nonsymmetric matrix, left eigenvectors are considered together with the right eigenvectors:
\[
Y_1,Y_2,\dots,Y_n.
\]
They are eigenvectors of the transpose of \( A \):
\[
A^{T}\cdot Y_j=\lambda_j\cdot Y_j.
\]
After transposing this equality, we obtain
\[
Y_j^{T}\cdot A=\lambda_j\cdot Y_j^{T}.
\]
The right and left eigenvectors are normalized so that the biorthogonality condition holds:
\[
Y_i^{T}\cdot X_j=
\begin{cases}
1, & i=j,\\
0, & i\ne j.
\end{cases}
\]
In particular,
\[
Y_1^{T}\cdot X_1=1,
\]
while for the other right eigenvectors,
\[
Y_1^{T}\cdot X_j=0,\qquad j=2,3,\dots,n.
\]
These relationships make it possible to replace the eigenvalue \( \lambda_1 \) with zero without changing the rest of the matrix spectrum.
Matrix Transformation: Replacing the First Eigenvalue with Zero
Suppose that the eigenvalue \( \lambda_1 \), the right eigenvector \( X_1 \), and the left eigenvector \( Y_1 \) have already been found. We construct a new matrix:
\[
A_1=A-\lambda_1\cdot X_1\cdot Y_1^{T}.
\]
The matrix
\[
\lambda_1\cdot X_1\cdot Y_1^{T}
\]
is the part of the spectral decomposition associated with the eigenvalue \( \lambda_1 \). It is formed by multiplying a column vector by a row vector, so it has rank one.
Let the eigenvectors have the following form:
\[
X_1=
\begin{pmatrix}
x_{11}\\
x_{21}\\
\vdots\\
x_{n1}
\end{pmatrix},
\qquad
Y_1^{T}=
\begin{pmatrix}
y_{11} & y_{21} & \dots & y_{n1}
\end{pmatrix}.
\]
Their product is then
\[
X_1\cdot Y_1^{T}=
\begin{pmatrix}
x_{11}\cdot y_{11} & x_{11}\cdot y_{21} & \dots & x_{11}\cdot y_{n1}\\
x_{21}\cdot y_{11} & x_{21}\cdot y_{21} & \dots & x_{21}\cdot y_{n1}\\
\vdots & \vdots & \ddots & \vdots\\
x_{n1}\cdot y_{11} & x_{n1}\cdot y_{21} & \dots & x_{n1}\cdot y_{n1}
\end{pmatrix}.
\]
The elements of the transformed matrix \( A_1 \) are calculated using the formula
\[
a_{ij}^{(1)}=a_{ij}-\lambda_1\cdot x_{i1}\cdot y_{j1},\qquad i,j=1,2,\dots,n.
\]
Let us show that after this transformation, the vector \( X_1 \) is associated with the eigenvalue zero:
\[
A_1\cdot X_1=
\left(A-\lambda_1\cdot X_1\cdot Y_1^{T}\right)\cdot X_1=
A\cdot X_1-\lambda_1\cdot X_1\cdot\left(Y_1^{T}\cdot X_1\right).
\]
Since
\[
A\cdot X_1=\lambda_1\cdot X_1
\]
and
\[
Y_1^{T}\cdot X_1=1,
\]
we obtain
\[
A_1\cdot X_1=
\lambda_1\cdot X_1-\lambda_1\cdot X_1\cdot 1=0.
\]
Therefore,
\[
A_1\cdot X_1=0\cdot X_1.
\]
Thus, the vector \( X_1 \) remains an eigenvector of the matrix \( A_1 \), but it now corresponds to the eigenvalue zero.
The formula above applies to the general case. If the matrix \( A \) is symmetric, its left and right eigenvectors are the same. In addition, they can be chosen to be orthonormal:
\[
X_i^{T}\cdot X_j=
\begin{cases}
1, & i=j,\\
0, & i\ne j.
\end{cases}
\]
Therefore, for a symmetric matrix, the transformation takes a simpler form:
\[
A_1=A-\lambda_1\cdot X_1\cdot X_1^{T}.
\]
Spectrum Preservation: Why the Other Eigenvalues Do Not Change
Now let us check how this transformation affects the eigenvectors \( X_2,X_3,\dots,X_n \). Consider an eigenvector \( X_j \), where
\[
j=2,3,\dots,n.
\]
Multiply the matrix \( A_1 \) by this vector:
\[
A_1\cdot X_j=
\left(A-\lambda_1\cdot X_1\cdot Y_1^{T}\right)\cdot X_j=
A\cdot X_j-\lambda_1\cdot X_1\cdot\left(Y_1^{T}\cdot X_j\right).
\]
For the vector \( X_j \), the following equality holds:
\[
A\cdot X_j=\lambda_j\cdot X_j.
\]
At the same time, the biorthogonality condition gives
\[
Y_1^{T}\cdot X_j=0.
\]
Therefore,
\[
A_1\cdot X_j=
\lambda_j\cdot X_j-\lambda_1\cdot X_1\cdot 0=
\lambda_j\cdot X_j.
\]
Thus, the eigenvalues
\[
\lambda_2,\lambda_3,\dots,\lambda_n
\]
and their corresponding eigenvectors remain unchanged. The spectrum of the matrix \( A_1 \) therefore consists of
\[
0,\lambda_2,\lambda_3,\dots,\lambda_n.
\]
Therefore, the transformation changes only the first eigenvalue, while the rest of the matrix spectrum is preserved.
Sequential Transformation: Moving to the Next Eigenvalues
Since the eigenvalues are arranged in descending order of their absolute values, the eigenvalue of the matrix \( A_1 \) with the largest absolute value is \( \lambda_2 \). Therefore, applying the power method to the transformed matrix makes it possible to find the eigenvalue of the original matrix with the second-largest absolute value and its corresponding eigenvector.
After finding \( \lambda_2 \), we construct the next matrix:
\[
A_2=A_1-\lambda_2\cdot X_2\cdot Y_2^{T}.
\]
Its spectrum is
\[
0,0,\lambda_3,\dots,\lambda_n.
\]
The eigenvalue of the matrix \( A_2 \) with the largest absolute value is now \( \lambda_3 \). After finding it, we can construct the matrix
\[
A_3=A_2-\lambda_3\cdot X_3\cdot Y_3^{T}
\]
and continue the computation.
In the general case, after sequentially replacing the first \( k \) eigenvalues with zeros, we obtain the matrix
\[
A_k=A-\sum_{i=1}^{k}\lambda_i\cdot X_i\cdot Y_i^{T}.
\]
Its remaining nonzero eigenvalues are
\[
\lambda_{k+1},\lambda_{k+2},\dots,\lambda_n.
\]
Thus, the deflation method makes it possible to transform the matrix step by step so that the eigenvalues that have already been found are replaced with zeros. After each transformation, the power method is applied to the resulting matrix to find the next eigenvalue in order of absolute value.
Practical Part: Eigenvalues of a Matrix Using the Deflation Method
Now let us see how the deflation method works for matrices of different orders. In each example, we will construct a matrix whose eigenvalue with the largest absolute value has been replaced with zero. We will then apply the power method to the transformed matrix with a tolerance of \( \varepsilon=0.1 \).
Example 1. Find the eigenvalue with the second-largest absolute value of the following matrix using the deflation method:
\[
A=
\begin{pmatrix}
4 & 1\\
1 & 2
\end{pmatrix}.
\]
For the original matrix, the eigenvalue with the largest absolute value and its corresponding normalized eigenvector are
\[
\lambda_1\approx4.414,
\qquad
X_1\approx
\begin{pmatrix}
0.924\\
0.383
\end{pmatrix}.
\]
Since the given matrix is symmetric, we construct \( A_1 \) using the formula
\[
A_1=A-\lambda_1\cdot X_1\cdot X_1^{T}.
\]
The transpose of the eigenvector is
\[
X_1^{T}\approx
\begin{pmatrix}
0.924 & 0.383
\end{pmatrix}.
\]
Multiply the column vector \( X_1 \) by the row vector \( X_1^{T} \):
\[
X_1\cdot X_1^{T}\approx
\begin{pmatrix}
0.924\\
0.383
\end{pmatrix}
\cdot
\begin{pmatrix}
0.924 & 0.383
\end{pmatrix}
=
\begin{pmatrix}
0.854 & 0.354\\
0.354 & 0.146
\end{pmatrix}.
\]
Multiply the resulting matrix by \( \lambda_1 \):
\[
\lambda_1\cdot X_1\cdot X_1^{T}\approx
4.414\cdot
\begin{pmatrix}
0.854 & 0.354\\
0.354 & 0.146
\end{pmatrix}
=
\begin{pmatrix}
3.768 & 1.561\\
1.561 & 0.646
\end{pmatrix}.
\]
Subtract the resulting spectral component from the original matrix:
\[
A_1\approx
\begin{pmatrix}
4 & 1\\
1 & 2
\end{pmatrix}
–
\begin{pmatrix}
3.768 & 1.561\\
1.561 & 0.646
\end{pmatrix}
=
\begin{pmatrix}
0.232 & -0.561\\
-0.561 & 1.354
\end{pmatrix}.
\]
Now apply the power method to the matrix \( A_1 \). Choose the initial normalized vector
\[
Y^{(0)}=
\frac{1}{\sqrt{2}}\cdot
\begin{pmatrix}
1\\
1
\end{pmatrix}
\approx
\begin{pmatrix}
0.707\\
0.707
\end{pmatrix}.
\]
At the first iteration, multiply the matrix \( A_1 \) by the vector \( Y^{(0)} \):
\[
Z^{(1)}=
A_1\cdot Y^{(0)}
\approx
\begin{pmatrix}
0.232 & -0.561\\
-0.561 & 1.354
\end{pmatrix}
\cdot
\begin{pmatrix}
0.707\\
0.707
\end{pmatrix}
=
\begin{pmatrix}
-0.232\\
0.561
\end{pmatrix}.
\]
Calculate the norm of the resulting vector:
\[
\left\lVert Z^{(1)}\right\rVert_2
\approx
\sqrt{(-0.232)^2+0.561^2}
\approx0.607.
\]
Normalize the vector:
\[
Y^{(1)}=
\frac{1}{0.607}\cdot
\begin{pmatrix}
-0.232\\
0.561
\end{pmatrix}
\approx
\begin{pmatrix}
-0.383\\
0.924
\end{pmatrix}.
\]
Find the first approximation of the eigenvalue with the second-largest absolute value using the Rayleigh quotient:
\[
\lambda_2^{(1)}=
\frac{\left(Y^{(1)}\right)^{T}\cdot A_1\cdot Y^{(1)}}
{\left(Y^{(1)}\right)^{T}\cdot Y^{(1)}}
\approx1.586.
\]
Perform the second iteration:
\[
Z^{(2)}=
A_1\cdot Y^{(1)}
\approx
\begin{pmatrix}
0.232 & -0.561\\
-0.561 & 1.354
\end{pmatrix}
\cdot
\begin{pmatrix}
-0.383\\
0.924
\end{pmatrix}
=
\begin{pmatrix}
-0.607\\
1.465
\end{pmatrix}.
\]
Calculate the norm:
\[
\left\lVert Z^{(2)}\right\rVert_2
\approx
\sqrt{(-0.607)^2+1.465^2}
\approx1.586.
\]
Normalize the vector:
\[
Y^{(2)}=
\frac{1}{1.586}\cdot
\begin{pmatrix}
-0.607\\
1.465
\end{pmatrix}
\approx
\begin{pmatrix}
-0.383\\
0.924
\end{pmatrix}.
\]
The new approximation is
\[
\lambda_2^{(2)}=
\frac{\left(Y^{(2)}\right)^{T}\cdot A_1\cdot Y^{(2)}}
{\left(Y^{(2)}\right)^{T}\cdot Y^{(2)}}
\approx1.586.
\]
Check the stopping condition:
\[
\left|\lambda_2^{(2)}-\lambda_2^{(1)}\right|
=
|1.586-1.586|
=
0
\leq0.1.
\]
The required tolerance has been achieved. Therefore,
\[
\lambda_2\approx1.586.
\]
Example 2. Find the eigenvalue with the second-largest absolute value of the following matrix using the deflation method:
\[
A=
\begin{pmatrix}
5 & 1 & 0\\
1 & 3 & 1\\
0 & 1 & 2
\end{pmatrix}.
\]
For the original matrix, the eigenvalue with the largest absolute value and its corresponding normalized eigenvector are
\[
\lambda_1\approx5.461,
\qquad
X_1\approx
\begin{pmatrix}
0.902\\
0.415\\
0.12
\end{pmatrix}.
\]
Since the given matrix is symmetric, construct the transformed matrix using the formula
\[
A_1=A-\lambda_1\cdot X_1\cdot X_1^{T}.
\]
The transpose of the eigenvector is
\[
X_1^{T}\approx
\begin{pmatrix}
0.902 & 0.415 & 0.12
\end{pmatrix}.
\]
Calculate the outer product:
\[
X_1\cdot X_1^{T}\approx
\begin{pmatrix}
0.902\\
0.415\\
0.12
\end{pmatrix}
\cdot
\begin{pmatrix}
0.902 & 0.415 & 0.12
\end{pmatrix}
=
\begin{pmatrix}
0.813 & 0.374 & 0.108\\
0.374 & 0.172 & 0.05\\
0.108 & 0.05 & 0.014
\end{pmatrix}.
\]
Multiply the resulting matrix by \( \lambda_1 \):
\[
\lambda_1\cdot X_1\cdot X_1^{T}\approx
5.461\cdot
\begin{pmatrix}
0.813 & 0.374 & 0.108\\
0.374 & 0.172 & 0.05\\
0.108 & 0.05 & 0.014
\end{pmatrix}
=
\begin{pmatrix}
4.44 & 2.045 & 0.591\\
2.045 & 0.942 & 0.272\\
0.591 & 0.272 & 0.079
\end{pmatrix}.
\]
Subtract the spectral component from the original matrix:
\[
A_1\approx
\begin{pmatrix}
5 & 1 & 0\\
1 & 3 & 1\\
0 & 1 & 2
\end{pmatrix}
–
\begin{pmatrix}
4.44 & 2.045 & 0.591\\
2.045 & 0.942 & 0.272\\
0.591 & 0.272 & 0.079
\end{pmatrix}
=
\begin{pmatrix}
0.56 & -1.045 & -0.591\\
-1.045 & 2.058 & 0.728\\
-0.591 & 0.728 & 1.921
\end{pmatrix}.
\]
Choose the initial normalized vector
\[
Y^{(0)}=
\frac{1}{\sqrt{3}}\cdot
\begin{pmatrix}
1\\
1\\
1
\end{pmatrix}
\approx
\begin{pmatrix}
0.577\\
0.577\\
0.577
\end{pmatrix}.
\]
At the first iteration, we obtain
\[
Z^{(1)}=
A_1\cdot Y^{(0)}
\approx
\begin{pmatrix}
0.56 & -1.045 & -0.591\\
-1.045 & 2.058 & 0.728\\
-0.591 & 0.728 & 1.921
\end{pmatrix}
\cdot
\begin{pmatrix}
0.577\\
0.577\\
0.577
\end{pmatrix}
=
\begin{pmatrix}
-0.621\\
1.005\\
1.188
\end{pmatrix}.
\]
Calculate the norm:
\[
\left\lVert Z^{(1)}\right\rVert_2
\approx
\sqrt{(-0.621)^2+1.005^2+1.188^2}
\approx1.676.
\]
Normalize the vector:
\[
Y^{(1)}=
\frac{1}{1.676}\cdot
\begin{pmatrix}
-0.621\\
1.005\\
1.188
\end{pmatrix}
\approx
\begin{pmatrix}
-0.371\\
0.6\\
0.709
\end{pmatrix}.
\]
The first approximation of the eigenvalue with the second-largest absolute value is
\[
\lambda_2^{(1)}=
\frac{\left(Y^{(1)}\right)^{T}\cdot A_1\cdot Y^{(1)}}
{\left(Y^{(1)}\right)^{T}\cdot Y^{(1)}}
\approx3.178.
\]
At the second iteration, we obtain
\[
Z^{(2)}=
A_1\cdot Y^{(1)}
\approx
\begin{pmatrix}
0.56 & -1.045 & -0.591\\
-1.045 & 2.058 & 0.728\\
-0.591 & 0.728 & 1.921
\end{pmatrix}
\cdot
\begin{pmatrix}
-0.371\\
0.6\\
0.709
\end{pmatrix}
=
\begin{pmatrix}
-1.253\\
2.138\\
2.018
\end{pmatrix}.
\]
Calculate the norm:
\[
\left\lVert Z^{(2)}\right\rVert_2
\approx
\sqrt{(-1.253)^2+2.138^2+2.018^2}
\approx3.196.
\]
Normalize the vector:
\[
Y^{(2)}=
\frac{1}{3.196}\cdot
\begin{pmatrix}
-1.253\\
2.138\\
2.018
\end{pmatrix}
\approx
\begin{pmatrix}
-0.392\\
0.669\\
0.631
\end{pmatrix}.
\]
The second approximation is
\[
\lambda_2^{(2)}=
\frac{\left(Y^{(2)}\right)^{T}\cdot A_1\cdot Y^{(2)}}
{\left(Y^{(2)}\right)^{T}\cdot Y^{(2)}}
\approx3.229.
\]
Check the stopping condition:
\[
\left|\lambda_2^{(2)}-\lambda_2^{(1)}\right|
=
|3.229-3.178|
=
0.051
\leq0.1.
\]
The required tolerance has been achieved. Therefore,
\[
\lambda_2\approx3.229.
\]
Example 3. Find the eigenvalue with the second-largest absolute value of the following matrix using the deflation method:
\[
A=
\begin{pmatrix}
6 & 1 & 0 & 0\\
1 & 4 & 1 & 0\\
0 & 1 & 3 & 1\\
0 & 0 & 1 & 2
\end{pmatrix}.
\]
For the original matrix, the eigenvalue with the largest absolute value and its corresponding normalized eigenvector are
\[
\lambda_1\approx6.464,
\qquad
X_1\approx
\begin{pmatrix}
0.899\\
0.417\\
0.129\\
0.029
\end{pmatrix}.
\]
Since the given matrix is symmetric, construct the transformed matrix using the formula
\[
A_1=A-\lambda_1\cdot X_1\cdot X_1^{T}.
\]
The transpose of the eigenvector is
\[
X_1^{T}\approx
\begin{pmatrix}
0.899 & 0.417 & 0.129 & 0.029
\end{pmatrix}.
\]
Calculate the outer product:
\[
X_1\cdot X_1^{T}\approx
\begin{pmatrix}
0.899\\
0.417\\
0.129\\
0.029
\end{pmatrix}
\cdot
\begin{pmatrix}
0.899 & 0.417 & 0.129 & 0.029
\end{pmatrix}
=
\begin{pmatrix}
0.809 & 0.375 & 0.116 & 0.026\\
0.375 & 0.174 & 0.054 & 0.012\\
0.116 & 0.054 & 0.017 & 0.004\\
0.026 & 0.012 & 0.004 & 0.001
\end{pmatrix}.
\]
Multiply the resulting matrix by \( \lambda_1 \):
\[
\lambda_1\cdot X_1\cdot X_1^{T}\approx
6.464\cdot
\begin{pmatrix}
0.809 & 0.375 & 0.116 & 0.026\\
0.375 & 0.174 & 0.054 & 0.012\\
0.116 & 0.054 & 0.017 & 0.004\\
0.026 & 0.012 & 0.004 & 0.001
\end{pmatrix}
=
\begin{pmatrix}
5.226 & 2.425 & 0.748 & 0.168\\
2.425 & 1.125 & 0.347 & 0.078\\
0.748 & 0.347 & 0.107 & 0.024\\
0.168 & 0.078 & 0.024 & 0.005
\end{pmatrix}.
\]
Subtract this spectral component from the original matrix:
\[
A_1\approx
\begin{pmatrix}
6 & 1 & 0 & 0\\
1 & 4 & 1 & 0\\
0 & 1 & 3 & 1\\
0 & 0 & 1 & 2
\end{pmatrix}
–
\begin{pmatrix}
5.226 & 2.425 & 0.748 & 0.168\\
2.425 & 1.125 & 0.347 & 0.078\\
0.748 & 0.347 & 0.107 & 0.024\\
0.168 & 0.078 & 0.024 & 0.005
\end{pmatrix}
=
\begin{pmatrix}
0.774 & -1.425 & -0.748 & -0.168\\
-1.425 & 2.875 & 0.653 & -0.078\\
-0.748 & 0.653 & 2.893 & 0.976\\
-0.168 & -0.078 & 0.976 & 1.995
\end{pmatrix}.
\]
Choose the initial normalized vector
\[
Y^{(0)}=
\frac{1}{2}\cdot
\begin{pmatrix}
1\\
1\\
1\\
1
\end{pmatrix}
=
\begin{pmatrix}
0.5\\
0.5\\
0.5\\
0.5
\end{pmatrix}.
\]
At the first iteration, we obtain
\[
Z^{(1)}=
A_1\cdot Y^{(0)}
\approx
\begin{pmatrix}
0.774 & -1.425 & -0.748 & -0.168\\
-1.425 & 2.875 & 0.653 & -0.078\\
-0.748 & 0.653 & 2.893 & 0.976\\
-0.168 & -0.078 & 0.976 & 1.995
\end{pmatrix}
\cdot
\begin{pmatrix}
0.5\\
0.5\\
0.5\\
0.5
\end{pmatrix}
=
\begin{pmatrix}
-0.784\\
1.013\\
1.887\\
1.363
\end{pmatrix}.
\]
Calculate the norm:
\[
\left\lVert Z^{(1)}\right\rVert_2
\approx
\sqrt{(-0.784)^2+1.013^2+1.887^2+1.363^2}
\approx2.656.
\]
Normalize the vector:
\[
Y^{(1)}=
\frac{1}{2.656}\cdot
\begin{pmatrix}
-0.784\\
1.013\\
1.887\\
1.363
\end{pmatrix}
\approx
\begin{pmatrix}
-0.295\\
0.381\\
0.71\\
0.513
\end{pmatrix}.
\]
The first approximation of the eigenvalue with the second-largest absolute value is
\[
\lambda_2^{(1)}=
\frac{\left(Y^{(1)}\right)^{T}\cdot A_1\cdot Y^{(1)}}
{\left(Y^{(1)}\right)^{T}\cdot Y^{(1)}}
\approx4.189.
\]
At the second iteration, calculate
\[
Z^{(2)}=
A_1\cdot Y^{(1)}
\approx
\begin{pmatrix}
0.774 & -1.425 & -0.748 & -0.168\\
-1.425 & 2.875 & 0.653 & -0.078\\
-0.748 & 0.653 & 2.893 & 0.976\\
-0.168 & -0.078 & 0.976 & 1.995
\end{pmatrix}
\cdot
\begin{pmatrix}
-0.295\\
0.381\\
0.71\\
0.513
\end{pmatrix}
=
\begin{pmatrix}
-1.389\\
1.94\\
3.025\\
1.736
\end{pmatrix}.
\]
Calculate the norm:
\[
\left\lVert Z^{(2)}\right\rVert_2
\approx
\sqrt{(-1.389)^2+1.94^2+3.025^2+1.736^2}
\approx4.226.
\]
Normalize the vector:
\[
Y^{(2)}=
\frac{1}{4.226}\cdot
\begin{pmatrix}
-1.389\\
1.94\\
3.025\\
1.736
\end{pmatrix}
\approx
\begin{pmatrix}
-0.329\\
0.459\\
0.716\\
0.411
\end{pmatrix}.
\]
The second approximation is
\[
\lambda_2^{(2)}=
\frac{\left(Y^{(2)}\right)^{T}\cdot A_1\cdot Y^{(2)}}
{\left(Y^{(2)}\right)^{T}\cdot Y^{(2)}}
\approx4.31.
\]
Check the stopping condition:
\[
\left|\lambda_2^{(2)}-\lambda_2^{(1)}\right|
=
|4.31-4.189|
=
0.121
>
0.1.
\]
The required tolerance has not yet been achieved, so perform the third iteration:
\[
Z^{(3)}=
A_1\cdot Y^{(2)}
\approx
\begin{pmatrix}
0.774 & -1.425 & -0.748 & -0.168\\
-1.425 & 2.875 & 0.653 & -0.078\\
-0.748 & 0.653 & 2.893 & 0.976\\
-0.168 & -0.078 & 0.976 & 1.995
\end{pmatrix}
\cdot
\begin{pmatrix}
-0.329\\
0.459\\
0.716\\
0.411
\end{pmatrix}
=
\begin{pmatrix}
-1.513\\
2.223\\
3.017\\
1.538
\end{pmatrix}.
\]
Calculate the norm:
\[
\left\lVert Z^{(3)}\right\rVert_2
\approx
\sqrt{(-1.513)^2+2.223^2+3.017^2+1.538^2}
\approx4.325.
\]
Normalize the vector:
\[
Y^{(3)}=
\frac{1}{4.325}\cdot
\begin{pmatrix}
-1.513\\
2.223\\
3.017\\
1.538
\end{pmatrix}
\approx
\begin{pmatrix}
-0.35\\
0.514\\
0.698\\
0.356
\end{pmatrix}.
\]
The third approximation is
\[
\lambda_2^{(3)}=
\frac{\left(Y^{(3)}\right)^{T}\cdot A_1\cdot Y^{(3)}}
{\left(Y^{(3)}\right)^{T}\cdot Y^{(3)}}
\approx4.359.
\]
Check the stopping condition:
\[
\left|\lambda_2^{(3)}-\lambda_2^{(2)}\right|
=
|4.359-4.31|
=
0.049
\leq0.1.
\]
The required tolerance has been achieved. Therefore,
\[
\lambda_2\approx4.359.
\]
What to Study Next: Other Methods for Computing the Spectrum
The deflation method is only one way to study the spectrum of a matrix. It is worth comparing it with other approaches and seeing how each method organizes the computation.
- Jacobi Rotation Method: How to Diagonalize a Symmetric Matrix — This article will explain how successive rotations eliminate off-diagonal elements and approximate the eigenvalues and eigenvectors.
- LU Decomposition Method: How to Approximate the Spectrum of a Matrix — This material will show how repeated LU decompositions transform a matrix and gradually bring its diagonal elements closer to the eigenvalues.
- Faddeev–LeVerrier Method: How to Construct the Characteristic Polynomial — This article will cover how to calculate the coefficients of the characteristic polynomial using the traces of matrix powers and then find its roots.
Eigenvalues of a Matrix: From Flowchart to Code
Interested in programming? Try implementing the algorithm shown in the flowchart using your favorite programming language. You can create a program that uses the deflation method to find the eigenvalue with the second-largest absolute value of a \( 2\times2 \) matrix. This implementation will help you better understand how to construct the transformed matrix, normalize vectors, check the tolerance, and organize iterative computations.
