Do Similar Matrices Have the Same Eigenvalues?
The relationship between similar matrices and their eigenvalues is a fundamental concept in linear algebra that reveals deep connections between matrix transformations and their inherent properties. Two matrices A and B are said to be similar if there exists an invertible matrix P such that B = P⁻¹AP. In practice, this relationship preserves certain structural characteristics of the matrices, raising the question: **Do similar matrices have the same eigenvalues? ** The answer is yes, and this article explores why this is true, supported by mathematical proofs, examples, and practical implications It's one of those things that adds up. Nothing fancy..
Introduction to Similar Matrices and Eigenvalues
Before diving into the core question, let’s clarify the definitions:
- Similar Matrices: Two square matrices A and B are similar if there exists an invertible matrix P such that B = P⁻¹AP. The matrix P is called the similarity transformation matrix.
- Eigenvalues: For a square matrix A, a scalar λ is an eigenvalue if there exists a non-zero vector v (the eigenvector) such that Av = λv.
The eigenvalues of a matrix are solutions to the characteristic equation det(A - λI) = 0, where I is the identity matrix. The question of whether similar matrices share eigenvalues hinges on the relationship between their characteristic polynomials.
Proof: Similar Matrices Share the Same Eigenvalues
To prove that similar matrices A and B have the same eigenvalues, we start with their characteristic polynomials:
For matrix A: [ \det(A - \lambda I) = 0 ]
For matrix B (where B = P⁻¹AP): [ \det(B - \lambda I) = \det(P^{-1}AP - \lambda I) ]
Factor out P⁻¹ and P from the expression: [ \det(P^{-1}(A - \lambda I)P) = \det(P^{-1}) \cdot \det(A - \lambda I) \cdot \det(P) ]
Since det(P⁻¹) = 1/det(P), the determinant simplifies to: [ \frac{1}{\det(P)} \cdot \det(A - \lambda I) \cdot \det(P) = \det(A - \lambda I) ]
Thus, det(B - λI) = det(A - λI), meaning A and B have the same characteristic polynomial. Since eigenvalues are roots of the characteristic polynomial, similar matrices must have the same eigenvalues (including algebraic multiplicities).
Example: Verifying Eigenvalues Through Similarity
Let’s illustrate this with a concrete example Worth keeping that in mind..
Matrix A: [ A = \begin{bmatrix} 2 & 0 \ 0 & 3 \end{bmatrix} ]
Similarity transformation matrix P: [ P = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} ]
Compute B = P⁻¹AP:
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Find P⁻¹: [ P^{-1} = \begin{bmatrix} 1 & -1 \ 0 & 1 \end{bmatrix} ]
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Compute P⁻¹AP: [ P^{-1}AP = \begin{bmatrix} 1 & -1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \ 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} ]
First, multiply AP: [ AP = \begin{bmatrix} 2 & 0 \ 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \ 0 & 3 \end{bmatrix} ]
Then, multiply P⁻¹(AP): [ P^{-1}(AP) = \begin{bmatrix} 1 & -1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 2 \ 0 & 3 \end{bmatrix} = \begin{bmatrix} 2 & -1 \ 0 & 3 \end{bmatrix} ]
Thus, B = [[2, -1], [0, 3]].
Eigenvalues of A:
Eigenvalues of A
The characteristic polynomial of (A) is
[ \det(A-\lambda I)=\det!\begin{bmatrix} 2-\lambda & 0\[2pt] 0 & 3-\lambda \end{bmatrix} =(2-\lambda)(3-\lambda). ]
Setting this equal to zero gives the two eigenvalues
[ \lambda_1 = 2,\qquad \lambda_2 = 3, ]
each with algebraic multiplicity 1.
Eigenvalues of B
Recall that
[ B = P^{-1}AP = \begin{bmatrix} 2 & -1 \ 0 & 3 \end{bmatrix}. ]
Because (B) is upper‑triangular, its eigenvalues are simply the entries on the main diagonal:
[ \lambda_1 = 2,\qquad \lambda_2 = 3. ]
For completeness, we can verify this by computing the characteristic polynomial:
[ \det(B-\lambda I)=\det!\begin{bmatrix} 2-\lambda & -1\[2pt] 0 & 3-\lambda \end{bmatrix} =(2-\lambda)(3-\lambda), ]
which is exactly the same polynomial as for (A). Hence the eigenvalues (and their algebraic multiplicities) coincide.
What Similarity Preserves
The calculation above illustrates a fundamental fact:
- Similarity preserves the characteristic polynomial, and therefore all eigenvalues (including their algebraic multiplicities) remain unchanged.
- What does change under similarity are the eigenvectors and, in general, the basis in which the matrix is expressed. Two similar matrices may have completely different eigenvector sets, yet they describe the same linear transformation in different coordinate systems.
Conclusion
Similar matrices represent the same linear operator under different bases. So consequently, any intrinsic property of a linear operator—such as its eigenvalues and the way they appear in the characteristic polynomial—is invariant under similarity transformations. The example with (A=\begin{bmatrix}2&0\0&3\end{bmatrix}) and its similar partner (B=\begin{bmatrix}2&-1\0&3\end{bmatrix}) demonstrates that both matrices share the eigenvalues (2) and (3), confirming the general theorem that similar matrices always have identical eigenvalues (with matching algebraic multiplicities). This invariance is a cornerstone of matrix theory and underpins many applications in physics, engineering, and data science where change of basis is routinely employed.
