Question

Prove that 2 similar matrices A and B have the same eigenvalues.

Answer 1

First of all by the definition of a similar matrix:
\[\text{A matrix $A$ is similar to $B$ if there exists $Q$ such that}\]\[A = Q^{-1}BQ\]

Answer 2

Consider the characteristic polynomials for A and B.

Answer 3

If the characteristic polynomials are the same the eigenvalues are the same.

Answer 4

how would i represent a general polynomial for one of these matrices?

Answer 5

c_0+c_1*a_1+...+c_n*a_n?

Answer 6

\[\det(A-\lambda I) \] is the characteristic polynomial for A

Answer 7

\begin{eqnarray*}
\det(A - \lambda I) &=& \det(Q^{-1}BQ - \lambda I) \\
&=& \det(Q^{-1}BQ - Q^{-1}(\lambda I)Q) \\
&=& \det(Q^{-1}(B - \lambda I)Q)\\
&=& \det(Q^{-1})\det(B - \lambda I)\det(Q) \\
&=& \det(Q^{-1})\det(Q)\det(B - \lambda I) \\
&=& \det(B-\lambda I)
\end{eqnarray*}

Answer 8

Therefore the characteristic polynomial of A is the same as the characteristic polynomial of B

Answer 9

Do you follow?