31.7 - Examples
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Checking if two matrices are similar
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Today, we'll check if the two matrices A and B are similar. Can anyone tell me what it means for two matrices to be similar?
I think matrices A and B are similar if they represent the same linear transformation but in different bases.
Exactly! We can determine this by examining their eigenvalues and eigenvectors. Let's look at matrix A. It has an eigenvalue of 2 with algebraic multiplicity 2. What does that tell us?
It means the eigenvalue 2 is repeated.
Correct! But how many linearly independent eigenvectors does A have? If it has fewer than the algebraic multiplicity, what can we conclude?
It has only one eigenvector, so it's not diagonalizable.
That’s right! Therefore, matrices A and B cannot be similar since B is diagonalizable. Always remember: similarity relies not only on eigenvalues but also on the corresponding eigenvectors.
So A not being diagonalizable means it's not similar to B at all!
Exactly! Great summary. Now, let’s go to our second example about diagonalization.
Diagonalization of a Matrix
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For our second example, we have matrix A. Who can remind us what we need to do first in diagonalization?
We need to find the characteristic polynomial!
Correct! What's our characteristic polynomial here?
It's \( \text{det}(A - \lambda I) = (4 - \lambda)(3 - \lambda) \) which gives us eigenvalues 4 and 3.
Good job! Since we have distinct eigenvalues, what does this mean for diagonalization?
It means the matrix can be diagonalized!
Exactly! Now, how do we find the eigenvectors for each eigenvalue?
By solving the equations \( (A - 4I)x = 0 \) and \( (A - 3I)x = 0 \).
Correct! After finding our eigenvectors, we form our change-of-basis matrix P. Can anyone tell me what we do next?
We compute D using \( D = P^{-1}AP \).
Very well! This shows how A is similar to the diagonal matrix D, confirming its diagonalizability.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
The section presents two key examples explaining how to determine if two matrices are similar by analyzing their eigenvalues and eigenvectors. The first example demonstrates that two matrices are not similar due to lack of diagonalizability, while the second shows how to diagonalize a matrix by finding distinct eigenvalues and constructing the appropriate transformation matrix.
Detailed
Section 31.7: Examples
In this section, we explore two significant examples that highlight the concepts of matrix similarity and diagonalization.
Example 1: Checking Similarity
We analyze the matrices:
A = \( \begin{bmatrix} 2 & 1 \ 0 & 2 \end{bmatrix} \)
B = \( \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \)
We find that matrix A has the eigenvalue \( \lambda = 2 \) with an algebraic multiplicity of 2. However, upon calculating the eigenvectors corresponding to this eigenvalue, we observe that there is only one linearly independent eigenvector, which indicates that A is not diagonalizable. As a result, A is not similar to matrix B, which is diagonalizable and hence represents a different linear transformation.
Example 2: Diagonalization Using Similarity
Next, we consider matrix:
A = \( \begin{bmatrix} 4 & 1 \ 0 & 3 \end{bmatrix} \)
We start by deriving its characteristic equation:
\( \text{det}(A - \lambda I) = (4 - \lambda)(3 - \lambda) \)
The eigenvalues are distinct: \( \lambda_1 = 4 \) and \( \lambda_2 = 3 \). Since A has two distinct eigenvalues, it is diagonalizable. We proceed to find the eigenvectors, form the change-of-basis matrix P from these vectors, and compute the diagonal matrix D via diagonalization. Thus, we conclude that A is similar to the diagonal matrix D, confirming its diagonalizable nature.
These practical examples underscore the relationship between eigenvalues, eigenvectors, and the similarity of matrices, essential concepts in linear algebra, particularly in applications like solving systems of linear equations.
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Example 2: Diagonalization Using Similarity
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Chapter Content
Example 2: Diagonalization Using Similarity
Let
A = \( \begin{bmatrix} 4 & 1 \ 0 & 3 \end{bmatrix} \)
Find if A is diagonalizable.
Solution:
• Characteristic equation: \( \text{det}(A - \lambda I) = (4 - \lambda)(3 - \lambda) \)
• Eigenvalues: \( \lambda_1 = 4, \lambda_2 = 3 \)
• Since eigenvalues are distinct, A is diagonalizable.
• Find eigenvectors:
– For \( \lambda = 4:\) \( (A - 4I)x = 0 \)
– For \( \lambda = 3:\) \( (A - 3I)x = 0 \)
• Construct matrix P with eigenvectors, compute D = P^{-1}AP
Thus, A is similar to D, confirming diagonalization.
Detailed Explanation
In this example, we are determining if matrix A is diagonalizable. To do this, we first find the characteristic equation by taking the determinant of \( A - \lambda I \). The calculation results in the eigenvalues 4 and 3. Since these are distinct, we know that A can have a complete set of linearly independent eigenvectors, hence A is diagonalizable. Next, we find the eigenvectors for each eigenvalue. By forming a matrix P from these eigenvectors, we can compute the diagonal matrix D. This process of diagonalization confirms that A is indeed similar to a diagonal matrix D, reducing complexity in computations.
Examples & Analogies
Imagine a high school where students are organized by class based on their subjects. If each student (student being an eigenvector) can belong to their distinct classes (eigenvalues), we can arrange them neatly into separate groups without confusion (diagonalization). On the other hand, if there are too many mixed classes with overlapping subjects (like matrix A with insufficient distinct linearly independent eigenvectors), we cannot neatly group them by subjects (indicating non-diagonalizability). Diagonalization here is akin to creating a class schedule that clearly defines who belongs to which subject, making it easy to manage.
Key Concepts
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Matrix Similarity: Two matrices A and B are similar if they represent the same linear transformation under different bases.
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Diagonalization: A matrix A is diagonalizable if it can be expressed in the form A = PDP^{-1}, where D is diagonal.
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Eigenvalues and Eigenvectors: The eigenvalues and corresponding eigenvectors of a matrix are crucial for understanding its properties and performing diagonalization.
Examples & Applications
Example 1 demonstrates that matrices A and B are not similar as A is not diagonalizable.
Example 2 shows that matrix A is diagonalizable due to its distinct eigenvalues, allowing for the formation of a diagonal matrix D.
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Rhymes
Evals might seem like a quest, To find which vectors are the best! If they’re all distinct, you’ve got the key, To turn that matrix into D!
Stories
Once upon a time, matrices A and B were good friends. They shared the same eigenvalues but not the same eigenvectors. When the time came to change their bases, they could not be similar because A was stuck with just one vector, while B danced freely in diagonal form. They learned that to be similar, they needed the same kind of freedom—plenty of eigenvectors!
Memory Tools
For Diagonalization: 'E-V-C' means Eigenvalues first, then find Vectors, then Construct D.
Acronyms
D.O.P.E = Diagonalization Requires One Pair of Eigenvalues for Similarity.
Flash Cards
Glossary
- Eigenvalue
A scalar value that indicates how much a linear transformation stretches or compresses vectors associated with it.
- Diagonalizable
A matrix is diagonalizable if it can be expressed as a product of matrices such that one is a diagonal matrix.
- Changeofbasis Matrix
An invertible matrix that transforms vectors from one basis to another.
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