The polynomial is $-\lambda^3+3\lambda -2$

which factorizes into ($\lambda-1$)($\lambda +1$)($\lambda -2$)

A matrix A has the above characteristic polynomial, and so its eigenvalues are 1, -1, and 2.

However, another matrix B, already in diagonal form, has the same characteristic polynomial, but with eigenvalues 1,1,-2, i.e., diagonal entries 1,1,-2.

Is this possible? Or have I gone wrong in my computations?

The problem statement does ask to show that the characteristic polynomials are the same but that the matrices A and B are not similar. So, perhaps I have found exactly what I needed, but it just seems weird...


  • 1
    $\begingroup$ If you get the same polynomial, it factors the same way. Thus, the eigenvalues are the same. There must be an issue with your computations. $\endgroup$ – cauchyproblem Dec 23 '15 at 1:15
  • 1
    $\begingroup$ The eigenvalues of a matrix are exactly the roots of the characteristic polynomial, so there must be a miscalculation somewhere. Howeover, it is true that two matrices can have the same characteristic polynomial without being similar. $\endgroup$ – C. Falcon Dec 23 '15 at 1:15

$-\lambda^3+3\lambda - 2 = -(\lambda-1)^2(\lambda+2) \neq -(\lambda-1)(\lambda+1)(\lambda-2)$.

  • $\begingroup$ Hi @Slade, I don't have a minus sign on the factorization...but is my factorization nonetheless still incorrect? I guessed the first root, and then used polynomial long division and the quadratic formula for the remaining two roots, thanks, $\endgroup$ – User001 Dec 23 '15 at 1:20
  • 1
    $\begingroup$ @User001 Plug in $\lambda=-1$ or $\lambda=2$. Do you get $0$? $\endgroup$ – Slade Dec 23 '15 at 1:22
  • $\begingroup$ no :-( ... I must have long-divided incorrectly ... thanks so much, @Slade, $\endgroup$ – User001 Dec 23 '15 at 1:24
  • Two matrices with the same characteristic polynomial necessarily have the same eigenvalues (the roots of the polynomial).
  • If an $n$-dimensional matrix has $n$ distinct eigenvalues, then it is diagonalizable. Consequently, all $n$-dimensional matrices with this set of $n$ distinct eigenvalues are similar.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.