Adding a constant times the unit matrix and eigenvalues

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SUMMARY

The discussion centers on the relationship between eigenvalues of a matrix A and the addition of a constant times the identity matrix, denoted as eI. When solving the equation det |(A + eI) - λ_new I| = 0, it is established that λ_new = λ + e, where λ represents the original eigenvalues of matrix A. The eigenvectors remain unchanged, confirming that the addition of eI shifts the eigenvalues without altering the eigenvectors. This insight clarifies the impact of constant matrix additions on eigenvalue computations.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors in linear algebra
  • Familiarity with matrix determinants and their properties
  • Knowledge of identity matrices and their role in matrix operations
  • Basic proficiency in solving polynomial equations
NEXT STEPS
  • Study the implications of adding scalar multiples of identity matrices on eigenvalue problems
  • Explore the concept of matrix similarity and its relation to eigenvalues
  • Learn about the spectral theorem and its applications in linear algebra
  • Investigate numerical methods for computing eigenvalues using software tools like MATLAB or NumPy
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This discussion is beneficial for students and professionals in mathematics, particularly those studying linear algebra, as well as data scientists and engineers working with matrix computations in machine learning and data analysis.

julian
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To find the eigenvalues [itex]\lambda[/itex] of a matrix [itex]A[/itex] you solve the equation

[itex]det |A - \lambda I| = 0[/itex] eq(1)

but now what if you add [itex]e I[/itex] to the matrix A where e is a constant? Then you have to solve the equation,

[itex]det |(A + eI) - \lambda_{new} I| = 0[/itex] eq(2)

which is the same as solving

[itex]det |A - (\lambda_{new} - e) I| = 0[/itex] eq(3)

Doesn't comparison of eq(3) with eq(1) just imply [itex]\lambda_{new} = \lambda + e[/itex]?
 
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look at it this way [itex](A+eI)v=λ_{new}v[/itex] for an eigenvector v of A+eI.

Now suppose w is an eigenvector of A.

then [itex](A+eI)w=λw+ew=(λ+e)w[/itex]

so in fact even the eigenvectors coincide.
 
Thanks cus a PhD in maths didn't know what I was talking about...I was just checking I was right. Your way is more transparent.
 

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