Eigenvectors reveal directions that a matrix does not turn.
If a matrix has enough eigenvectors to form a basis, then the matrix becomes simple in that basis. Instead of mixing coordinates, it only scales them.
This is the idea of diagonalization.
A diagonal matrix is easy to understand:
\[ D= \begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}. \]
It acts by
\[ Dx= \begin{bmatrix} \lambda_1x_1\\ \lambda_2x_2\\ \vdots\\ \lambda_nx_n \end{bmatrix}. \]
So each coordinate evolves independently.
The central question of this chapter is:
When can we choose coordinates so that a matrix becomes diagonal?
This question is especially powerful for systems that evolve by repeated multiplication:
\[ x_{k+1}=Ax_k. \]
Such systems are called discrete dynamical systems.
After studying this chapter, you should be able to:
Let
\[ D= \begin{bmatrix} 3&0\\ 0&1 \end{bmatrix}. \]
Then
\[ D \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 3x\\ y \end{bmatrix}. \]
The first coordinate is multiplied by \(3\).
The second coordinate is multiplied by \(1\).
There is no mixing between coordinates.
Powers of diagonal matrices are also simple:
\[ D^k= \begin{bmatrix} 3^k&0\\ 0&1^k \end{bmatrix}. \]
More generally, if
\[ D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n), \]
then
\[ D^k=\operatorname{diag}(\lambda_1^k,\ldots,\lambda_n^k). \]
If
\[ D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n), \]
then
\[ D^k=\operatorname{diag}(\lambda_1^k,\ldots,\lambda_n^k). \]
This is the main reason diagonalization is useful.
Suppose \(A\) has eigenvectors
\[ v_1,\ldots,v_n \]
that form a basis of \(\mathbb{R}^n\), with corresponding eigenvalues
\[ \lambda_1,\ldots,\lambda_n. \]
Then every vector \(x\) can be written uniquely as
\[ x=c_1v_1+\cdots+c_nv_n. \]
Applying \(A\) gives
\[ Ax=c_1Av_1+\cdots+c_nAv_n. \]
Since
\[ Av_j=\lambda_jv_j, \]
we get
\[ Ax=c_1\lambda_1v_1+\cdots+c_n\lambda_nv_n. \]
So in the eigenvector basis, \(A\) simply scales each coordinate.
This is diagonalization.
Place the eigenvectors as columns of a matrix:
\[ P= \begin{bmatrix} v_1&v_2&\cdots&v_n \end{bmatrix}. \]
Place the eigenvalues on the diagonal:
\[ D= \begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}. \]
Since
\[ Av_j=\lambda_jv_j, \]
we have
\[ AP=PD. \]
If the eigenvectors form a basis, then \(P\) is invertible.
Multiplying by \(P^{-1}\) gives
\[ A=PDP^{-1}. \]
Equivalently,
\[ P^{-1}AP=D. \]
A square matrix \(A\) is diagonalizable if there is an invertible matrix \(P\) and a diagonal matrix \(D\) such that
\[ A=PDP^{-1}. \]
Equivalently,
\[ P^{-1}AP=D. \]
The columns of \(P\) are eigenvectors of \(A\).
The diagonal entries of \(D\) are the corresponding eigenvalues.
Let
\[ A= \begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}. \]
From the previous chapter, eigenvectors and eigenvalues are:
\[ v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad \lambda_1=3, \]
and
\[ v_2= \begin{bmatrix} 1\\ -1 \end{bmatrix}, \qquad \lambda_2=1. \]
Thus
\[ P= \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}, \qquad D= \begin{bmatrix} 3&0\\ 0&1 \end{bmatrix}. \]
Then
\[ A=PDP^{-1}. \]
This says that \(A\) becomes diagonal in the basis
\[ \left\{ \begin{bmatrix} 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -1 \end{bmatrix} \right\}. \]
import numpy as np
import matplotlib.pyplot as plt
A = np.array([[2, 1],
[1, 2]], dtype=float)
P = np.array([[1, 1],
[1, -1]], dtype=float)
D = np.array([[3, 0],
[0, 1]], dtype=float)
P @ D @ np.linalg.inv(P), A(array([[2., 1.],
[1., 2.]]),
array([[2., 1.],
[1., 2.]]))Check the coordinate version:
np.linalg.inv(P) @ A @ Parray([[3., 0.],
[0., 1.]])This should equal \(D\) up to numerical roundoff.
An \(n\times n\) matrix is diagonalizable if it has \(n\) linearly independent eigenvectors.
An \(n\times n\) matrix \(A\) is diagonalizable if and only if \(A\) has a basis of \(\mathbb{R}^n\) consisting of eigenvectors.
A useful sufficient condition is:
If an \(n\times n\) matrix has \(n\) distinct eigenvalues, then it is diagonalizable.
Why?
Eigenvectors corresponding to distinct eigenvalues are linearly independent.
However, a matrix can still be diagonalizable even if some eigenvalues repeat, as long as it has enough independent eigenvectors.
Consider
\[ A= \begin{bmatrix} 2&1\\ 0&2 \end{bmatrix}. \]
The only eigenvalue is
\[ \lambda=2 \]
with algebraic multiplicity \(2\).
Compute
\[ A-2I= \begin{bmatrix} 0&1\\ 0&0 \end{bmatrix}. \]
Solving
\[ (A-2I)v=0 \]
gives
\[ v= \begin{bmatrix} x\\ 0 \end{bmatrix}. \]
So the eigenspace is
\[ E_2=\operatorname{span} \left\{ \begin{bmatrix} 1\\ 0 \end{bmatrix} \right\}. \]
There is only one independent eigenvector.
Since a \(2\times 2\) matrix needs two independent eigenvectors to be diagonalizable, this matrix is not diagonalizable.
Recall:
For an eigenvalue \(\lambda\),
\[ 1\le \dim(E_\lambda)\le \text{algebraic multiplicity of }\lambda. \]
A matrix is diagonalizable if the total number of independent eigenvectors is \(n\).
Equivalently, for each eigenvalue, the geometric multiplicity must add up correctly so that all eigenspaces together provide \(n\) independent directions.
To diagonalize an \(n\times n\) matrix \(A\):
\[ \det(A-\lambda I)=0. \]
\[ E_\lambda=\operatorname{Null}(A-\lambda I). \]
\[ P=\begin{bmatrix}v_1&v_2&\cdots&v_n\end{bmatrix}. \]
\[ D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n) \]
in the same order as the columns of \(P\). 5. Then
\[ A=PDP^{-1}. \]
The ordering matters: the \(j\)th diagonal entry of \(D\) must correspond to the \(j\)th column of \(P\).
If
\[ A=PDP^{-1}, \]
then
\[ A^2=(PDP^{-1})(PDP^{-1})=PD^2P^{-1}. \]
Similarly,
\[ A^k=PD^kP^{-1}. \]
Since powers of diagonal matrices are easy,
\[ D^k=\operatorname{diag}(\lambda_1^k,\ldots,\lambda_n^k). \]
If
\[ A=PDP^{-1}, \]
then
\[ A^k=PD^kP^{-1}. \]
This formula is one of the main applications of diagonalization.
Let
\[ A= \begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}. \]
We have
\[ A=PDP^{-1}, \]
where
\[ P= \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}, \qquad D= \begin{bmatrix} 3&0\\ 0&1 \end{bmatrix}. \]
Therefore
\[ A^k = P \begin{bmatrix} 3^k&0\\ 0&1 \end{bmatrix} P^{-1}. \]
Since
\[ P^{-1} = \frac12 \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}, \]
we get
\[ A^k = \frac12 \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix} \begin{bmatrix} 3^k&0\\ 0&1 \end{bmatrix} \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}. \]
Multiplying gives
\[ A^k = \frac12 \begin{bmatrix} 3^k+1&3^k-1\\ 3^k-1&3^k+1 \end{bmatrix}. \]
def power_by_diagonalization(P, D, k):
return P @ np.linalg.matrix_power(D, k) @ np.linalg.inv(P)
for k in range(1, 5):
print("k =", k)
print(power_by_diagonalization(P, D, k))
print(np.linalg.matrix_power(A, k))
print()k = 1
[[2. 1.]
[1. 2.]]
[[2. 1.]
[1. 2.]]
k = 2
[[5. 4.]
[4. 5.]]
[[5. 4.]
[4. 5.]]
k = 3
[[14. 13.]
[13. 14.]]
[[14. 13.]
[13. 14.]]
k = 4
[[41. 40.]
[40. 41.]]
[[41. 40.]
[40. 41.]]
The two methods agree up to numerical roundoff.
A discrete dynamical system evolves step by step:
\[ x_{k+1}=Ax_k. \]
Starting from an initial vector \(x_0\), we get
\[ x_1=Ax_0, \]
\[ x_2=A^2x_0, \]
and in general,
\[ x_k=A^kx_0. \]
If \(A\) is diagonalizable,
\[ x_k=PD^kP^{-1}x_0. \]
This formula explains the long-term behavior of the system.
Suppose \(A\) has eigenvectors \(v_1,\ldots,v_n\) and eigenvalues \(\lambda_1,\ldots,\lambda_n\).
Write the initial vector as
\[ x_0=c_1v_1+\cdots+c_nv_n. \]
Then
\[ x_k=A^kx_0=c_1\lambda_1^kv_1+\cdots+c_n\lambda_n^kv_n. \]
Each eigenvector direction evolves independently.
If
\[ x_0=c_1v_1+\cdots+c_nv_n, \]
then
\[ x_k=A^kx_0=c_1\lambda_1^kv_1+\cdots+c_n\lambda_n^kv_n. \]
This is the main interpretation of diagonalization for dynamics.
For the system
\[ x_{k+1}=Ax_k, \]
eigenvalues determine growth or decay.
For an eigen-direction \(v\) with eigenvalue \(\lambda\),
\[ A^kv=\lambda^kv. \]
Therefore:
For discrete dynamics, \(|\lambda|\) controls growth or decay.
Let
\[ A= \begin{bmatrix} 0.6&0\\ 0&0.2 \end{bmatrix}. \]
Then
\[ x_{k+1}=Ax_k \]
has eigenvalues
\[ 0.6 \]
and
\[ 0.2. \]
Both have absolute value less than \(1\).
Therefore every trajectory tends to zero.
A_stable = np.array([[0.6, 0],
[0, 0.2]], dtype=float)
x = np.array([4, 3], dtype=float)
points = [x.copy()]
for k in range(20):
x = A_stable @ x
points.append(x.copy())
points = np.array(points)
plt.figure(figsize=(6, 6))
plt.plot(points[:, 0], points[:, 1], marker="o")
for k in [0, 1, 2, 5, 10, 20]:
plt.text(points[k, 0] + 0.03, points[k, 1] + 0.03, str(k))
plt.axhline(0, linewidth=0.8)
plt.axvline(0, linewidth=0.8)
plt.grid(True)
plt.axis("equal")
plt.title("Stable dynamics: eigenvalues have absolute value less than 1")
plt.show()
Let
\[ A= \begin{bmatrix} 1.2&0\\ 0&0.8 \end{bmatrix}. \]
The eigenvalue \(1.2\) causes growth in the first coordinate.
The eigenvalue \(0.8\) causes decay in the second coordinate.
Most initial vectors eventually align with the growing eigen-direction.
A_unstable = np.array([[1.2, 0],
[0, 0.8]], dtype=float)
x = np.array([1, 4], dtype=float)
points = [x.copy()]
for k in range(20):
x = A_unstable @ x
points.append(x.copy())
points = np.array(points)
plt.figure(figsize=(6, 6))
plt.plot(points[:, 0], points[:, 1], marker="o")
for k in [0, 1, 2, 5, 10, 20]:
plt.text(points[k, 0] + 0.03, points[k, 1] + 0.03, str(k))
plt.axhline(0, linewidth=0.8)
plt.axvline(0, linewidth=0.8)
plt.grid(True)
plt.axis("equal")
plt.title("Unstable dynamics: one eigenvalue has absolute value greater than 1")
plt.show()
Let
\[ A= \begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}. \]
The eigenvectors are
\[ v_1=\begin{bmatrix}1\\1\end{bmatrix}, \qquad v_2=\begin{bmatrix}1\\-1\end{bmatrix}, \]
with eigenvalues \(3\) and \(1\).
Let
\[ x_0= \begin{bmatrix} 2\\ 0 \end{bmatrix}. \]
Then
\[ x_0=v_1+v_2. \]
So
\[ x_k=3^kv_1+1^kv_2. \]
As \(k\) grows, the \(v_1\) direction dominates.
A = np.array([[2, 1],
[1, 2]], dtype=float)
x = np.array([2, 0], dtype=float)
points = [x.copy()]
for k in range(6):
x = A @ x
points.append(x.copy())
points = np.array(points)
plt.figure(figsize=(6, 6))
plt.plot(points[:, 0], points[:, 1], marker="o")
for k, p in enumerate(points):
plt.text(p[0] + 0.1, p[1] + 0.1, f"$x_{k}$")
plt.axhline(0, linewidth=0.8)
plt.axvline(0, linewidth=0.8)
plt.grid(True)
plt.axis("equal")
plt.title("Trajectory approaches the dominant eigen-direction")
plt.show()
An eigenvalue with largest absolute value is called a dominant eigenvalue.
If one eigenvalue has strictly largest absolute value and the initial vector has a component in its eigenvector direction, then repeated multiplication tends to align with the dominant eigenvector.
This idea is behind power iteration, a simple method for computing dominant eigenvectors.
It also appears in ranking algorithms, Markov chains, population models, and graph algorithms.
Power iteration repeatedly applies a matrix and normalizes:
\[ y_k=Ax_k, \]
\[ x_{k+1}=\frac{y_k}{\|y_k\|}. \]
If the conditions are favorable, \(x_k\) approaches a dominant eigenvector direction.
A = np.array([[2, 1],
[1, 2]], dtype=float)
x = np.array([1, 0], dtype=float)
history = []
for k in range(12):
y = A @ x
x = y / np.linalg.norm(y)
history.append(x.copy())
np.array(history)array([[0.89442719, 0.4472136 ],
[0.78086881, 0.62469505],
[0.73279349, 0.6804511 ],
[0.71578195, 0.69832385],
[0.71001067, 0.70419091],
[0.70807608, 0.70613615],
[0.70743003, 0.70678338],
[0.70721455, 0.706999 ],
[0.70714271, 0.70707086],
[0.70711876, 0.70709481],
[0.70711077, 0.70710279],
[0.70710811, 0.70710545]])history = np.array(history)
plt.figure(figsize=(6, 6))
for k, v in enumerate(history):
plt.arrow(0, 0, v[0], v[1], head_width=0.03, length_includes_head=True)
plt.text(v[0] + 0.03, v[1] + 0.03, str(k+1))
plt.axhline(0, linewidth=0.8)
plt.axvline(0, linewidth=0.8)
plt.grid(True)
plt.axis("equal")
plt.title("Power iteration approaches the dominant eigenvector")
plt.show()
A simple Markov chain can be written as
\[ x_{k+1}=Px_k, \]
where \(x_k\) is a probability vector and \(P\) is a transition matrix.
For example,
\[ P= \begin{bmatrix} 0.8&0.3\\ 0.2&0.7 \end{bmatrix}. \]
The columns of \(P\) add to \(1\), so if \(x_k\) is a probability vector, then \(x_{k+1}\) is also a probability vector.
A steady state is a vector \(\pi\) satisfying
\[ P\pi=\pi. \]
This is an eigenvector equation with eigenvalue \(1\).
Let
\[ P= \begin{bmatrix} 0.8&0.3\\ 0.2&0.7 \end{bmatrix}. \]
We want
\[ P\pi=\pi. \]
That is,
\[ (P-I)\pi=0. \]
Also, since \(\pi\) is a probability vector,
\[ \pi_1+\pi_2=1. \]
Solve:
\[ \begin{bmatrix} -0.2&0.3\\ 0.2&-0.3 \end{bmatrix} \begin{bmatrix} \pi_1\\ \pi_2 \end{bmatrix} =0. \]
The first equation is
\[ -0.2\pi_1+0.3\pi_2=0. \]
So
\[ 0.3\pi_2=0.2\pi_1. \]
Thus
\[ 3\pi_2=2\pi_1. \]
Together with
\[ \pi_1+\pi_2=1, \]
we get
\[ \pi_1=\frac35, \qquad \pi_2=\frac25. \]
So
\[ \pi= \begin{bmatrix} 3/5\\ 2/5 \end{bmatrix}. \]
P = np.array([[0.8, 0.3],
[0.2, 0.7]], dtype=float)
x = np.array([1.0, 0.0])
history = [x.copy()]
for k in range(20):
x = P @ x
history.append(x.copy())
history = np.array(history)
plt.figure(figsize=(7, 4))
plt.plot(history[:, 0], marker="o", label="state 1")
plt.plot(history[:, 1], marker="o", label="state 2")
plt.axhline(3/5, linestyle="--")
plt.axhline(2/5, linestyle="--")
plt.xlabel("step")
plt.ylabel("probability")
plt.grid(True)
plt.legend()
plt.title("Markov chain approaches steady state")
plt.show()
history[-1]
array([0.60000038, 0.39999962])The probabilities approach the eigenvector with eigenvalue \(1\).
Some real matrices have complex eigenvalues.
For example, a rotation matrix
\[ R_\theta= \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix} \]
has complex eigenvalues
\[ e^{i\theta} \]
and
\[ e^{-i\theta}. \]
Repeated multiplication rotates vectors.
If the eigenvalues have magnitude \(1\), motion may persist.
If the eigenvalues have magnitude less than \(1\), motion spirals inward.
If they have magnitude greater than \(1\), motion spirals outward.
This connects eigenvalues to oscillation and stability.
theta = np.pi / 8
scale = 0.9
A_spiral = scale * np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])
x = np.array([4.0, 0.0])
points = [x.copy()]
for k in range(60):
x = A_spiral @ x
points.append(x.copy())
points = np.array(points)
plt.figure(figsize=(6, 6))
plt.plot(points[:, 0], points[:, 1], marker="o", markersize=3)
plt.axhline(0, linewidth=0.8)
plt.axvline(0, linewidth=0.8)
plt.grid(True)
plt.axis("equal")
plt.title("Complex eigenvalues can create spiral dynamics")
plt.show()
Diagonalization is a special case of change of basis.
The matrix
\[ P^{-1}AP=D \]
means:
So diagonalization says:
In the right coordinate system, the transformation separates into independent one-dimensional actions.
This is why eigenvectors are powerful. They give the coordinate system that the matrix itself prefers.
Diagonalization is conceptually powerful, but not always the best numerical method.
Reasons:
For symmetric matrices, eigenvalue computations are especially well-behaved because eigenvectors can be chosen orthonormally. This leads to the spectral theorem, one of the central results of linear algebra.
A = np.array([[2, 1],
[1, 2]], dtype=float)
eigvals, P = np.linalg.eig(A)
D = np.diag(eigvals)
reconstructed = P @ D @ np.linalg.inv(P)
eigvals, P, D, reconstructed(array([3., 1.]),
array([[ 0.70710678, -0.70710678],
[ 0.70710678, 0.70710678]]),
array([[3., 0.],
[0., 1.]]),
array([[2., 1.],
[1., 2.]]))Measure reconstruction error:
np.linalg.norm(A - reconstructed)0.0For a non-diagonalizable matrix, NumPy may still return eigenvalues and some eigenvectors, but the eigenvector matrix may be singular or poorly conditioned.
B = np.array([[2, 1],
[0, 2]], dtype=float)
eigvals_B, P_B = np.linalg.eig(B)
eigvals_B, P_B, np.linalg.det(P_B)(array([2., 2.]),
array([[ 1.0000000e+00, -1.0000000e+00],
[ 0.0000000e+00, 4.4408921e-16]]),
4.4408920985006143e-16)The determinant of the eigenvector matrix is close to zero, reflecting that there are not enough independent eigenvectors.
Given an \(n\times n\) matrix \(A\):
If \(A\) has \(n\) independent eigenvectors:
\[ A=PDP^{-1}. \]
If
\[ A=PDP^{-1}, \]
then
\[ A^k=PD^kP^{-1}. \]
For
\[ x_{k+1}=Ax_k, \]
we have
\[ x_k=A^kx_0. \]
If \(A=PDP^{-1}\), then
\[ x_k=PD^kP^{-1}x_0. \]
Not every square matrix has enough eigenvectors.
For example,
\[ \begin{bmatrix} 2&1\\ 0&2 \end{bmatrix} \]
is not diagonalizable.
The diagonal entries of \(D\) must correspond to the columns of \(P\).
The eigenvectors must be linearly independent.
If \(P\) is singular, then \(A=PDP^{-1}\) is impossible.
The diagonalization formula is
\[ A=PDP^{-1}. \]
The coordinate-change form is
\[ P^{-1}AP=D. \]
Diagonalization is excellent for theory. For numerical computation, use stable algorithms designed for the task.
A matrix is diagonalizable if it has a basis of eigenvectors.
If
\[ P=\begin{bmatrix}v_1&\cdots&v_n\end{bmatrix} \]
and
\[ D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n), \]
with
\[ Av_j=\lambda_jv_j, \]
then
\[ A=PDP^{-1}. \]
Equivalently,
\[ P^{-1}AP=D. \]
If \(A=PDP^{-1}\), then
\[ A^k=PD^kP^{-1}. \]
For a discrete dynamical system
\[ x_{k+1}=Ax_k, \]
we have
\[ x_k=A^kx_0. \]
Eigenvalues determine growth, decay, oscillation, and stability.
Markov chain steady states are eigenvectors with eigenvalue \(1\).
Diagonalization reveals the coordinate system in which a transformation separates into independent directions.
For each matrix:
Find eigenvalues.
Find eigenvectors.
Determine whether the matrix is diagonalizable.
If diagonalizable, find \(P\) and \(D\).
\[ A= \begin{bmatrix} 2&1\\ 1&2 \end{bmatrix} \]
\[ A= \begin{bmatrix} 3&0\\ 0&2 \end{bmatrix} \]
\[ A= \begin{bmatrix} 1&2\\ 0&3 \end{bmatrix} \]
\[ A= \begin{bmatrix} 2&1\\ 0&2 \end{bmatrix} \]
Use diagonalization to compute \(A^k\).
\[ A= \begin{bmatrix} 2&1\\ 1&2 \end{bmatrix} \]
\[ A= \begin{bmatrix} 3&0\\ 0&2 \end{bmatrix} \]
\[ A= \begin{bmatrix} 1&2\\ 0&3 \end{bmatrix} \]
For each system, describe long-term behavior.
\[ x_{k+1}= \begin{bmatrix} 0.5&0\\ 0&0.25 \end{bmatrix} x_k. \]
\[ x_{k+1}= \begin{bmatrix} 1.2&0\\ 0&0.8 \end{bmatrix} x_k. \]
\[ x_{k+1}= \begin{bmatrix} 2&1\\ 1&2 \end{bmatrix} x_k. \]
\[ x_{k+1}= \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix} x_k. \]
Let
\[ A= \begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}. \]
Use eigenvectors
\[ v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ -1 \end{bmatrix}. \]
\[ x_0= \begin{bmatrix} 2\\ 0 \end{bmatrix} \]
as a linear combination of \(v_1\) and \(v_2\).
Let
\[ P= \begin{bmatrix} 0.8&0.3\\ 0.2&0.7 \end{bmatrix}. \]
\[ P\pi=\pi. \]
\[ x_0= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \]
compute several iterations and compare with the steady state.
np.linalg.matrix_power.Write a short paragraph answering each question.
Diagonalization explains how repeated linear processes behave. This matters in ranking algorithms, graph methods, Markov chains, recommendation systems, stability analysis, iterative algorithms, and spectral learning. In data science and AI, diagonalization and its relatives reveal dominant directions, long-term behavior, and hidden coordinate systems. The core idea is that a complicated transformation may become simple when expressed in the right basis.