Show that if $A\sim B$, then $m_A(x)=m_B(x)$.

Are there any simple proofs of this result? I've seen something along the lines of, since $A = SBS^{-1}$, $p(SBS^{-1}) = Sp(B)S^{-1}$, but I'm not really sure what that means or how it's derived.

Additionally, if $A$ is diagonalizable and has eigenvalues $\lambda_1,...,\lambda_k$, how would we write the minimal polynomial?


1 Answer 1


Let's clarify that statement: $p(SBS^{-1}) = Sp(B)S^{-1}$.

Suppose that $p$ is a polynomial given by $p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0$. For a matrix $A$, we say that $p(A)$ is the matrix given by $$ p(A) = a_n A^n + a_{n-1}A^{n-1} + \cdots + a_1 A + a_0 I $$ Now, notice the following: $$ (SBS^{-1})^2 = SBS^{-1}SBS^{-1} = SB(S^{-1}S)BS^{-1} = S(B^2)S^{-1} $$ similarly, you can show that $$ (SBS^{-1})^3 = S(B^3)S^{-1} $$ in fact, you can inductively prove that $(SBS^{-1})^k = SB^kS^{-1}$ for every integer $k$. From there, we note that $$ p(SBS^{-1}) = \\ a_n (SBS^{-1})^n + a_{n-1}(SBS^{-1})^{n-1} + \cdots + a_1 (SBS^{-1}) + a_0 I =\\ a_n SB^nS^{-1} + a_{n-1}SB^{n-1}S^{-1} + \cdots + a_1 SBS^{-1} + a_0 SS^{-1} =\\ S(a_n B^n + a_{n-1}B^{n-1} + \cdots + a_1 B + a_0 I)S^{-1} = \\ Sp(B)S^{-1} $$ it follows that if $A = SBS^{-1}$, then for any polynomial $p$, $p(A)$ is only the zero matrix if $p(B)$ is also the zero matrix.


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