A matrix $A \in M_n(F)$ is invertible iff there exists a matrix like $B \in M_n(F)$ such that $AB=I_n$.

A scalar like $\lambda \in F$ is called an eigenvalue of $A$ iff $\lambda I_n-A$ is not invertible.

The characteristic polynomial of a square matrix is a polynomial which is invariant under matrix similarity and has the eigenvalues as roots. It has the determinant and the trace of the matrix as coefficients.

Question :

Assume that $A,B \in M_n(\mathbb C) $ and $f(x)=\det(xI_n-B)$ is the characteristic polynomial of $B$.
Prove that $f(A)$ is invertible iff $A,B$ have no common eigenvalues.

Note : Here, when we talk about $f(A)$, we mean something like this :
$f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0 \implies f(A)=a_nA^n+a_{n-1}A^{n-1}+\dots+a_1A+a_0I$

The problem :
I'm confused about the meaning of $f(A)$ being invertible and the characteristic polynomial of $A$ which is $det(xI_n-A)$. I don't know where to start and what to work on.


2 Answers 2


Let $\mu_i$ be eigenvalues of $A$. Since every matrix is similar to a upper triangular matrix, there is an invertible $P$ such that $$ P^{-1}AP=\pmatrix{\mu_1 \hspace{5 mm}*\\\ddots \\& \mu_{n-1} \hspace{5 mm}* \\& \hspace{10 mm}\mu_n }=J_A $$ where $J_A$ is an upper triangular matrix. Thus $$ f(A)=Pf(J_A)P^{-1}=P\pmatrix{f(\mu_1) \hspace{5 mm}*\\\ddots \\& f(\mu_{n-1} )\hspace{5 mm}* \\& \hspace{10 mm}f(\mu_n) }P^{-1} $$ So $f(\mu_i)$ are eigenvalues of $f(A)$. If a $\mu_k$ is also eigenvalue of $B$, then $f(\mu_k)=0$. Thus $f(A)$ has an eigenvalue of $0$ and is not invertible.

  • $\begingroup$ Thank you :) good explanation :) $\endgroup$ Oct 26, 2016 at 21:43
  • $\begingroup$ Here you go man :D $\endgroup$ Oct 26, 2016 at 22:53
  • $\begingroup$ @User1006 You should edit your post a little bit. $J_A$ does not have that expression you wrote. More precisely, the JCF is a block diagonal matrix, where each diagonal block (Jordan block) has one of the eigenvalues of $A$ on the diagonal and 1 on the first upperdiagonal. In fact, what you wrote is not even a possible JCF, since the last 2x2 block has 1 on the upper diagonal, but the diagonal entries are different. For instance, $\left[\begin{matrix}1 & 1\\0 &2\end{matrix}\right]$ is not in JCF... $\endgroup$
    – bartgol
    Oct 27, 2016 at 4:14
  • $\begingroup$ I'm sorry, but other people with little knowledge on JCF will stumble on this question and see your answer, and may be confused on why that is a JCF. I agree that the upper diagonal entries play little role here, but that does not mean we can write sloppy stuff. You can easily fix your $J_A$ by replacing the upper diagonal entries with binary numbers (0-1) or even with stars (as you did later). $\endgroup$
    – bartgol
    Oct 27, 2016 at 4:41
  • $\begingroup$ I would also point out that your $\mu_i$ may be repeated, otherwise one may be confused on why isn't $J_A$ a diagonal matrix. You can claim these are pedantic details. But as I said, people less prepared than you will look at your post months down the road and may get more questions than answers out of it. $\endgroup$
    – bartgol
    Oct 27, 2016 at 4:43

Hint: write $f$ factorized on its roots, that is,

$$ f(x)=(x-\lambda_1)^{m_1}\cdots(x-\lambda_k)^{m_k}, $$

where $m_i$ is the algebraic multiplicity of the eigenvalue $\lambda_i$ (of the matrix $B$). Clearly, $\sum m_i = n$, and $k\leq n$. Then

$$ f(A) = (A-\lambda_1I)^{m_1}\cdots(A-\lambda_kI)^{m_k} $$

From here, it should be easier to see why $f(A)$ is invertible iff $A$ and $B$ have no eigenvalue in common.

You may also need the fact that $A$ commutes with $p(A)$ for any polynomial $p$, and therefore you can write the characteristic polynomial in any order. For instance,

$$ f(A) = (A-\lambda_1I)^{m_1}\cdots(A-\lambda_kI)^{m_k} = (A-\lambda_kI)^{m_k}\cdots(A-\lambda_1I)^{m_1} $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .