# $f(A)$ is invertible iff $A,B$ have no common eigenvalues

A matrix like $A \in M_n(F)$ 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 : Hear, 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.

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,

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

Let $\mu_i$ be eigenvalues of $A$. Since every matrix is similar to a upper triangular matrix, there is 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 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 $f(\mu_k)=0$. So $f(A)$ has an eigenvalue of $0$ and thus not invertible.

• Thank you :) good explanation :) Oct 26 '16 at 21:43
• Here you go man :D Oct 26 '16 at 22:53
• @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... Oct 27 '16 at 4:14
• 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). Oct 27 '16 at 4:41
• 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. Oct 27 '16 at 4:43