The eigenvalue relation between $A$ and $A^{-1}$

Question

Let $A\in M_n(\mathbb{C})$ be invertible. What is the relationship between eigevalues of $A$ and $A^{-1}$? For example, if $A$ is $6\times 6$ and having eigenvalues $1,7,7,5,5,5$ (I mean $am(7)=2,~am(5)=3$). Then it is easy to prove that $A^{-1}$ has eigenvalues $1/1,~1/7,~1/5$. However, will the corresponding algebraic/geometric multiplicity the same? Why?

Answer 1

$A(x)=cx$ is equivalent to $A^{-1}A(x)=A^{-1}(cx)$ implies that $A^{-1}x={1\over c }x$. $(A-cI)^p=0$ is equivalent to $(A(I-cA^{-1})^p=0$. This is equivalent to $(I-cA^{-1})^p=0=$ which is equivalent to $(A^{-1}-{1\over c}I)^p=0$. Thus the multiplicity of the eigenvalue $c$ of $A$ is the multiplicity of the eigenvalue ${1\over c}$ of $A^{-1}$.

Answer 2

If $\lambda$ is an eigenvalue we have $A\vec v=\lambda \vec v$ for some eigenvectors $\vec v$. So, multiplying by $A^{-1}$, we have: $$ A^{-1}A\vec v=\lambda A^{-1} \vec v \iff \frac{1}{\lambda}\vec v= A^{-1}\vec v $$

This means that $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$ for the same set of eigenvectors.

Answer 3

The characteristic polynomial of $A^{-1}$ is $$ p_{A^{-1}}(X)= \det(A^{-1}-XI)= \frac{1}{\det A}\det(I-XA)= \frac{(-1)^nX^n}{\det A}\det(A-X^{-1}I)= \frac{(-1)^nX^n}{\det A}p_A(X^{-1}) $$ In other words, if $p_A(X)=a_0+a_1X+\dots+a_nX^n$, then $$ p_{A^{-1}}(X)=\frac{(-1)^n}{\det A}(a_n+a_{n-1}X+\dots+a_0X^n) $$ Now it's easy to conclude that if $\lambda$ has algebraic multiplicity $m$ as an eigenvalue of $A$, then $\lambda^{-1}$ has algebraic multiplicity $m$ as an eigenvalue of $A^{-1}$.

Also the geometric multiplicity is the same (easy exercise on eigenspaces).

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Note about the derivation. The computation above takes place in the field $F(X)$ of rational functions over the base field $F$. So the steps are just applying Binet's theorem and then writing $I-XA=-X(A-X^{-1}I)$, together with $\det(cA)=c^nA$ (for an $n\times n$ matrix $A$).

Answer 4

Yes, you are right about the multiplicity. Suppose that $\lambda$ is an eigenvalue with algebraic multiplicy $2$ (say). Then the Jordan form of the matrix will have two blocks of the type$$\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$$or one block of the type $(\lambda)$ and another one of the type$$\begin{pmatrix}\lambda&1&0\\0&\lambda&1\\0&0&1\end{pmatrix}.$$Suppose it's the former. When you invert the Jordan normal form, you'll invert these blocks, thereby getting$$\begin{pmatrix}\frac1\lambda&-\frac1{\lambda^2}\\0&\frac1\lambda\end{pmatrix}$$ and it follows from this that the algbraic multiplicity of $\frac1\lambda$ is $4$ and its geometric multiplicity is $2$. The same argument applies to that $1+3$ case and, in fact, it works in the general case too.

Answer 5

Yes, the multiplicities are also preserved.

According to Cayley Hamilton Theorem Matrix A satisfies its characteristic polynomial.

Now an eigenvalue of multiplicity $k$, results in a factor of $(A-\lambda )^k$ in the resulting polynomial.

Upon multiplication by $A^{-k}$ we get $(A^{-1} -\frac {1 }{\lambda})^k$ is a factor of the associated polynomial for $A^{-1}.$

Answer 6

Let $n\times n$ be the size of the matrix $A$ in question.

I will try to to compare the multiplicities.

1. The geometric multiplicity

The geometric multiplicity of a $a$ for a matrix $M$ is the dimension of the null space of $M-aI$

Let $\lambda$ be an eigenvalue of A, $$ \begin{split} v \in \ker(A-\lambda I)& \Leftrightarrow Av=\lambda v \\ & \Leftrightarrow A^{-1}v=\dfrac{1}{\lambda}v \\ & \Leftrightarrow v\in \ker(A^{-1}-\dfrac{1}{\lambda}I) \end{split} $$ Meaning the $A$ nullspace associated with $\lambda$ is the same as the $A^{-1}$ nullspace associated with $\dfrac{1}{\lambda}$.

Same space, same dimension, same geometric mutiplicities.

2. The algebraic multiplicity

The algebraic multiplicity of a $a$ for a matrix $M$ is the biggest integer $m$ such that $(X-a)^m$ divides $p_A$ (the characteristic polynomial of $A$ defined as $\det(XI-A)$ up to constant scaling).

For nonzero complex numbers $z$,

$\det(A) \times p_{A^{-1}}(z)=\det(A) \times \det(zI-A^{-1}) = \det(zA-I)$

$z^n\times p_A(\dfrac{1}{z})=\det(zI)\times \det(\dfrac{1}{z}I-A)=\det(zA-I)$

As such, $\det(A) \times p_{A^{-1}}(z)=z^n\times p_A(\dfrac{1}{z})$ for all nonzero complex numbers $z$.

We write $p_A=\displaystyle \prod_{\lambda \text{ root of $p_A$}}(X-\lambda)^{m_{\lambda}} $, $m_{\lambda}$ being the algebraic multiplicity of $\lambda$ with repect to $A$.

$$ \begin{split} z^n p_A(\dfrac{1}{z})& = \displaystyle ( \prod_{\lambda \text{ root of $p_A$}}z^{m_{\lambda}}) \times \displaystyle \prod_{\lambda \text{ root of $p_A$}}(\dfrac{1}{z}-\lambda)^{m_{\lambda}} \\ & = \displaystyle \prod_{\lambda \text{ root of $p_A$}}(-1)^{m_{\lambda}}(z-\dfrac{1}{\lambda})^{m_{\lambda}} \end{split} $$ Thus, $$p_{A^{-1}}(z)=\dfrac{1}{\det(A)}\displaystyle \prod_{\lambda \text{ root of $p_A$}}(-1)^{m_{\lambda}}(z-\dfrac{1}{\lambda})^{m_{\lambda}}$$

is true for all complex numbers but 0, but since this is a polynomial-like equality (with complex coefficients) then it's true everywhere. And,

$p_{A^{-1}} = \dfrac{1}{\det(A)}\displaystyle \prod_{\lambda \text{ root of $p_A$}}(-1)^{m_{\lambda}}(X-\dfrac{1}{\lambda})^{m_{\lambda}}$

From this, it's immediate that the algebraic multiplicity of some $\lambda$ with respect to $A$ is the same as the algebraic multiplicity of $\dfrac{1}{\lambda}$ with respect to $A^{-1}$.

Correct me if I made some mistake or typo.