Give an example of a 2 × 2 matrix where e₁ is an eigenvector, but e2 is not. Explain why your matrix has this property. (Hint: try an upper triangular matrix.)
Give an example of a 2 × 2 matrix where e₁ is an eigenvector, but e2 is not. Explain why your matrix has this property. (Hint: try an upper triangular matrix.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Example Problem: Eigenvectors of a 2x2 Matrix**
*Problem Statement:*
Give an example of a 2×2 matrix where **e₁** is an eigenvector, but **e₂** is not. Explain why your matrix has this property. (Hint: try an upper triangular matrix.)
---
*Explanation:*
To solve this problem, you will be looking for a 2x2 matrix form such that one of the standard basis vectors, **e₁**, is an eigenvector, while the other, **e₂**, is not. The hint suggests using an upper triangular matrix, which has the property that it can have less straightforward eigenvectors compared to diagonal matrices.
Here’s an example of such a matrix:
\[
A = \begin{pmatrix}
\lambda & 1 \\
0 & \mu
\end{pmatrix}
\]
Where:
- \(\lambda\) and \(\mu\) are constants.
- **e₁** = (1, 0) is indeed an eigenvector with eigenvalue \(\lambda\).
- **e₂** = (0, 1) is not an eigenvector unless a new condition is added, which is typically not the case in this type of question.
In this matrix:
- **e₁** is an eigenvector because:
\[
A \cdot \mathbf{e}_1 =
\begin{pmatrix}
\lambda & 1 \\
0 & \mu
\end{pmatrix}
\begin{pmatrix}
1 \\
0
\end{pmatrix}
=
\begin{pmatrix}
\lambda \\
0
\end{pmatrix}
=
\lambda \begin{pmatrix}
1 \\
0
\end{pmatrix}
\]
- **e₂** is not an eigenvector because the action of \(A\) on **e₂**:
\[
A \cdot \mathbf{e}_2 =
\begin{pmatrix}
\lambda & 1 \\
0 & \mu
\end{pmatrix}
\begin{pmatrix}
0 \\
1
\end{pmatrix}
=
\begin{pmatrix}
1 \\
\mu
\end{pmatrix}
\]
results in a vector that cannot be written as \(\mu \times \mathbf{e}_2\), unless specific conditions hold](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d5e88e6-b1af-4aea-9b08-2dadd85f5e2c%2F2eb6e671-ff04-4f32-b71c-e84eb9ff542c%2Fylnif7_processed.png&w=3840&q=75)
Transcribed Image Text:**Example Problem: Eigenvectors of a 2x2 Matrix**
*Problem Statement:*
Give an example of a 2×2 matrix where **e₁** is an eigenvector, but **e₂** is not. Explain why your matrix has this property. (Hint: try an upper triangular matrix.)
---
*Explanation:*
To solve this problem, you will be looking for a 2x2 matrix form such that one of the standard basis vectors, **e₁**, is an eigenvector, while the other, **e₂**, is not. The hint suggests using an upper triangular matrix, which has the property that it can have less straightforward eigenvectors compared to diagonal matrices.
Here’s an example of such a matrix:
\[
A = \begin{pmatrix}
\lambda & 1 \\
0 & \mu
\end{pmatrix}
\]
Where:
- \(\lambda\) and \(\mu\) are constants.
- **e₁** = (1, 0) is indeed an eigenvector with eigenvalue \(\lambda\).
- **e₂** = (0, 1) is not an eigenvector unless a new condition is added, which is typically not the case in this type of question.
In this matrix:
- **e₁** is an eigenvector because:
\[
A \cdot \mathbf{e}_1 =
\begin{pmatrix}
\lambda & 1 \\
0 & \mu
\end{pmatrix}
\begin{pmatrix}
1 \\
0
\end{pmatrix}
=
\begin{pmatrix}
\lambda \\
0
\end{pmatrix}
=
\lambda \begin{pmatrix}
1 \\
0
\end{pmatrix}
\]
- **e₂** is not an eigenvector because the action of \(A\) on **e₂**:
\[
A \cdot \mathbf{e}_2 =
\begin{pmatrix}
\lambda & 1 \\
0 & \mu
\end{pmatrix}
\begin{pmatrix}
0 \\
1
\end{pmatrix}
=
\begin{pmatrix}
1 \\
\mu
\end{pmatrix}
\]
results in a vector that cannot be written as \(\mu \times \mathbf{e}_2\), unless specific conditions hold
Expert Solution
Step 1
A non zero vector x is an eigen vector of a square matrix A if Ax = x where is a scalar
Then is called the eigen value of A and x is the eigen vector corresponding to eigen value
Step by step
Solved in 3 steps
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