6

Let $A \in \mathbb M_n(\mathbb C)$ be a normal matrix and $\lambda$ be an eigenvalue of $A.$ Let $\phi : C(\sigma(A)) \longrightarrow \mathbb M_n (\mathbb C)$ be the continuous functional calculus associated to $A.$ Consider the projection $P : = \phi \left (\mathbf {1}_{\{\lambda\}} \right ).$ Let the algebraic multiplicity of $\lambda$ be $m.$ Then show that $\text {rank}\ (P) = m.$

I am trying to figure out what $P$ is. Here $\phi = \mathscr {G}^{-1} \circ \xi,$ where $\mathscr {G} : A_1 \longrightarrow C \left (\widehat A_1 \right )$ is the Gelfand transform defined on $A_1 = \text {span}\ \{I,A\}$ (Here $\widehat {A_1}$ is the space of all non-zero multiplicative linear functionals on $A$ endowed with the induced weak*-topology) and $\xi : C(\sigma(A)) \longrightarrow C \left (\widehat {A_1} \right )$ is defined by $f \mapsto f \circ \mathscr {G} (A).$ Now what is $P\ $? By the definition of $\phi$ it turns out that $P = \mathscr {G}^{-1} \left (\mathbf {1}_{\{\lambda\}} \circ \mathscr {G} (A) \right ).$ Now what is $\mathbf {1}_{\{\lambda\}} \circ \mathscr {G} (A)\ $? In order to determine this let it act on some element $\psi \in \widehat {A_1}.$ Then we have $$\left ( \mathbf {1}_{\{\lambda\}} \circ \mathscr {G} (A) \right ) (\psi) = \mathbf {1}_{\{\lambda\}} (\psi (A)) = \begin{cases} 1 & \text {if}\ \psi (A) = \lambda \\ 0 & \text {if}\ \psi(A) \neq \lambda \end{cases}$$ Therefore it follows that $\mathbf {1}_{\{\lambda\}} \circ \mathscr {G} (A) = \mathbf {1}_{E},$ where $E = \left \{\varphi \in \widehat {A_1}\ \bigg |\ \varphi (A) = \lambda \right \}.$ So in order to find $P$ we only need to find what is $\mathscr {G}^{-1} (\mathbf {1}_{E}).$ What is that? Since $P \in A_1$ it turns out that $P = \sum\limits_{n \geq 0} c_n A^n,$ where $c_n \in \mathbb C.$ Hence $\mathbf {1}_{E} = \mathscr {G} (P) = \sum\limits_{n \geq 0} c_n \left (\mathscr {G} (A) \right )^n.$ Since $\text {ran} \left (\mathscr {G} (A) \right ) = \sigma (A)$ it turns out that that $\sum\limits_{n \geq 0} c_n z^n = 0$ for all $z \in \mathbb \sigma (A) \setminus \{\lambda\}$ and $\sum\limits_{n \geq 0} c_n \lambda^{n} = 1.$ Now let $c = \prod\limits_{\substack {\mu \in \sigma (A)} \\ \ \ \ {\mu \neq \lambda}} \frac {1} {\lambda - \mu}.$ Then it follows that $$P = c \prod\limits_{\substack {\mu \in \sigma (A)} \\ \ \ \ {\mu \neq \lambda}} \left (A - \mu I \right ).$$ In other words $P$ is the projection onto the eigenspace for the eigenvalue $\lambda$ of $A.$ Now what is $\text {rank}\ (P)\ $? Well since $A$ is diagonalizable the sum of the geometric multiplicities add up to $n$ and hence nullity of $P$ will be $n - \ell,$ where $\ell$ is the geometric multiplicity of $\lambda.$ So by the rank-nullity theorem $\text {rank}\ (P) = \ell,$ which coincides with the algebraic multiplicity of $\lambda$ since $A$ is diagonalizable. Am I right? Could anyone please check it once?

Thanks for your time.

ACB
  • 3,068
  • 1
    For what it's worth, here's a proof that I find much easier to understand. Show that if $v \in \Bbb C^n$ is an eigenvector of $A$ with $Av = \mu v$ and $f\in C(\sigma(A))$, then $$ \phi(f) v = f(\mu)v. $$ The desired result then follows as a consequence of the spectral theorem for normal matrices. – Ben Grossmann Oct 29 '21 at 09:58
  • @Ben Grossmann$:$ How do I conclude the result by the help spectral theorem for normal matrices having obtained the first equality? – ACB Oct 29 '21 at 16:35
  • By spectral theorem for normal matrices we can get hold of pairwise orthogonal projections $P_j$ with $\sum P_j = 1$ such that $A = \sum \lambda_j P_j.$ I think instead we can use rank-nullity theorem to conclude the result. – ACB Oct 29 '21 at 16:41
  • For example, we could proceed as follows. Consider an orthonormal eigenbasis $v_1,\dots,v_r,v_{r+1},\dots,v_n$ of $\Bbb C^n$ made of eigenvectors of $A$ such that $v_1,\dots,v_r$ are eigenvectors associated with $\lambda$. We have $$ P v_j = \begin{cases} v_j & 1 \leq j \leq r\ 0 & j > r.\end{cases} $$ Conclude that ${v_1,\dots,v_r}$ forms a basis for the image of $P$. – Ben Grossmann Oct 29 '21 at 16:42
  • @Ben Grossmann$:$ The result directly follows by rank-nullity theorem from your equality as nullity $(P) = n - m,$ where $m$ is the geometric multiplicity of $\lambda.$ – ACB Oct 29 '21 at 16:45
  • If you prefer to think about the nullity, then that is also fine. – Ben Grossmann Oct 29 '21 at 16:51

1 Answers1

2

Since you already know that $A$ is (orthogonally) diagonalizable, you may assume without loss of generality that $A$ is diagonal. We may write this as $$\tag1 A=\sum_{j=1}^r\lambda_jP_j, $$ where $\lambda_1,\ldots,\lambda_r$ are the eigenvalues, and $P_j$ the orthogonal projection onto the corresponding eigenspace. Because the projections $P_1,\ldots,P_r$ are pairwise orthogonal, it is trivial to check that $$\tag2 f(A)=\sum_{j=1}^rf(\lambda_j)\,P_j $$ for any polynomial $f$, and thus (by taking uniform limits) for any continuous function $f$ (which, since the spectrum is discrete, means any function).

By construction of the continuous functional calculus you have $\phi(f)=f(A)$ for all polynomials $f$, and thus $\phi(f)=f(A)$ for any function $f$, as in $(2)$.

So, for fixed, $\lambda_k$, you have $$ \phi(1_{\{\lambda_k\}})=\sum_{j=1}^r\phi_{\{\lambda_k\}}(\lambda_j)P_j=P_k. $$

From the expression $(1)$, we have $$ \det(A-\lambda I)=\det\Big(\sum_{j=1}^r(\lambda_j-\lambda)P_j\Big)=\prod_{j=1}^r(\lambda_j-\lambda)^{\operatorname{rank}{P_j}}. $$ So the algebraic multiplicity of $\lambda_k$ agrees with the geometric multiplicity of $\lambda_k$, which is the rank of $P_k$.

Martin Argerami
  • 217,281