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Q23E

Expert-verifiedFound in: Page 336

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**23: Suppose matrix A is similar to B. What is the relationship between the characteristic polynomials of A and B? What does your answer tell you about the eigenvalues of A and B?**

Characteristic polynomials of A and B are the same, therefore their eigenvalues are also same.

In linear algebra, an eigenvector or characteristic vector of a linear transformation is a nonzero vector that changes at most by a scalar factor when that linear transformation is applied to it. The corresponding eigenvalue, often denoted by \lambda, is the factor by which the eigenvector is scaled.

If A and B are similar matrices then, ${T}^{-1}-AT=B$, for an invertible T. If det $\left(A-\lambda l\right)=0$, then

$0=de{t}^{-1}det\left(A-\lambda l\right)detT=det\left({T}^{-1}\left(A-\lambda l\right)T\right)=det\left({T}^{-1}AT-\lambda {T}^{-1}lT\right)=det\left({T}^{-1}AT-\lambda l\right)=det(B-\lambda l)$

This means that the characteristics polynomials of A and B are the same. From this also follows that their eigenvalues are exactly the same.

Hence, characteristic polynomials of A and B are the same, therefore their eigenvalues are also same.

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