Similar matrix and characteristic polynomial

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SUMMARY

Two similar matrices possess identical characteristic polynomials, while the reverse is not universally applicable; matrices with the same characteristic polynomial may not be similar. To demonstrate this, one can utilize matrices that differ in eigenvectors despite sharing eigenvalues. A practical example involves a 2x2 Jordan block and a 2x2 identity matrix, which both exhibit the same characteristic polynomial yet are not similar due to their differing eigenvector structures.

PREREQUISITES
  • Understanding of characteristic polynomials
  • Familiarity with eigenvalues and eigenvectors
  • Knowledge of matrix similarity and diagonalization
  • Basic concepts of linear transformations and change of basis
NEXT STEPS
  • Explore the properties of Jordan blocks in linear algebra
  • Learn about diagonalizable versus non-diagonalizable matrices
  • Study the implications of eigenvalues and eigenvectors in matrix theory
  • Investigate the process of change of basis in linear transformations
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Students and professionals in mathematics, particularly those studying linear algebra, matrix theory, or anyone interested in understanding the nuances of matrix similarity and characteristic polynomials.

Eric Nelson
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A true statement: Two similar matrices have the same characteristic polynomial.

The converse however is not true in general: two matrices with the same characteristic polynomial need not be similar.

HOw can I prove this?

Any help appreciated.
 
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it's enough to find two matrices which are not similar yet have the same char poly
 
Note: Pick an easy charisteristic polynomial to aim for.
 
can you suggest an easy characteristic polynomial, I'm stumped on this concept. thank you
 
You can try a difference of squares. For instance, x^2-1 Also, you can think of what you can do to a matrix that wouldn't change the characteristic polynomial.
 
If two matrices have the same characteristice polynomial, the obviously they have the same eigenvalues. What you want is that they have different eigenvectors. You can do that is one is diagonalizable but the other isn't. To give an example in 2 by 2 matrices, they must have only a single eigenvalue. On matrix would then have 2 \independent eigenvectors, the other only 1 eigenvalue.
 
to prove the part that is true, use change of basis. matrices A and B are similar if there exists an invertible Q such that B=Q^-1*A*Q. But all you're really doing is putting your linear transformation into another basis, so they'll have the same characteristic polynomial. You can check this by showing the change of basis, then taking the determinant of A and B.
 
Take a 2x2 jordan block and also a 2x2 identity matrix, and...
 

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