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simultaneous commutativity of operators(Keith Conrad)
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In this note we work with linear operators on finite-dimensional complex vector spaces. Any such operator has an eigenvector, by the fundamental theorem of algebra. A linear operator is called diagonalizable if it has a basis of eigenvectors: there is a basis in which the matrix representation of the operator is a diagonal matrix. We are interested in conditions that make a finite set of linear operators simultaneously diagonalizable: there is a basis in which the matrix representation of each operator is diagonal.
Example 1. The matrices $A=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right)$ and $B=\left(\begin{array}{c}1&2 \\ 2&1\end{array}\right)$ acting on $\mathbf{C}^{2}$ are each diagonalizable, but they are not simultaneously diagonalizable: the eigenvectors of $A$ are scalar multiples of $\left(\begin{array}{l}i \\ 1\end{array}\right)$ and $\left(\begin{array}{c}-i \\ 1\end{array}\right)$ while the eigenvectors of $B$ are scalar multiples of $\left(\begin{array}{l}1 \\ 1\end{array}\right)$ and $\left(\begin{array}{c}-1 \\ 1\end{array}\right)$.
Example 2. The matrices$$
A=\left(\begin{array}{rrr}
7 & -10 & 5 \\
4 & -5 & 3 \\
-1 & 3 & 0
\end{array}\right) \text { and } B=\left(\begin{array}{rrr}
8 & -11 & 6 \\
5 & -6 & 4 \\
0 & 2 & 1
\end{array}\right)$$acting on $\mathbf{C}^{3}$ are simultaneously diagonalizable with common eigenbasis$$
\left(\begin{array}{l}
1 \\
1 \\
1
\end{array}\right),\left(\begin{array}{c}
-3 / 2+i / 2 \\
-1 / 2+i / 2 \\
1
\end{array}\right),\left(\begin{array}{c}
-3 / 2-i / 2 \\
-1 / 2-i / 2 \\
1
\end{array}\right) .
$$For linear operators to be simultaneously diagonalizable, they at least have to be individually diagonalizable, but more is needed (see Example 1). A further necessary condition is that the operators commute, since diagonal matrices commute. Check the matrices in Example 1 do not commute while those in Example 2 do commute. The following theorem shows that commuting linear operators already have something in common.
Theorem 3. If $A_{1}, \ldots, A_{r}$ are commuting linear operators on a finite-dimensional $\mathbf{C}$-vector space $V$ then they have a common eigenvector in $V$.
Proof. We induct on $r$, the result being clear if $r=1$ since we work over the complex numbers: every linear operator on a finite-dimensional $\bf C$-vector space has an eigenvector.
Now assume $r \geq 2$. Let the last operator $A_{r}$ have an eigenvalue $\lambda \in \mathbf{C}$ and let$$E_{\lambda}=\left\{v \in V: A_{r} v=\lambda v\right\}$$be the $\lambda$-eigenspace for $A_{r} .$ For $v \in E_{\lambda}, A_{r}\left(A_{i} v\right)=A_{i}\left(A_{r} v\right)=A_{i}(\lambda v)=\lambda\left(A_{i} v\right)$, so $A_{i} v \in E_{\lambda}$. Thus each $A_{i}$ restricts to a linear operator on the subspace $E_{\lambda}$.
The linear operators $\left.A_{1}\right|_{E_{\lambda}}, \ldots,\left.A_{r-1}\right|_{E_{\lambda}}$ commute since the $A_{i}$ 's commute as operators on the larger space $V$. There are $r-1$ of these operators, so by induction on $r$ (while quantifying over all finite-dimensional $\bf C$-vector spaces) these operators have a common eigenvector in $E_{\lambda}$. That vector is also an eigenvector of $A_{r}$ by the definition of $E_{\lambda}$. We're done.
A common eigenvector for $A_{1}, \ldots, A_{r}$ need not have the same eigenvalue for all the operators; the first eigenvector in Example 2 has eigenvalue 2 for $A$ and 3 for $B$.
Lemma 4. If $A: V \rightarrow V$ is a diagonalizable linear operator and $W$ is an $A$-stable subspace of $V$ then the restriction $\left.A\right|_{W}: W \rightarrow W$ is also diagonalizable.
Proof. We use a variation on the answer by Zorn. Letting $\lambda_{1}, \ldots, \lambda_{r}$ be the distinct eigenvalues of $A$ acting on $V$, we have $V=\bigoplus_{i=1}^{r} E_{\lambda_{i}}$, where $E_{\lambda_{i}}$ is the $\lambda_{i}$-eigenspace of $A$. For $w \in W$, write $w=v_{1}+\cdots+v_{r}$ where $v_{i} \in E_{\lambda_{i}}$. We'll prove each $v_{i}$ is in $W$. Then $W=\bigoplus_{i=1}^{r}\left(E_{\lambda_{i}} \cap W\right)$, so $\left.A\right|_{W}$ is diagonalizable.
Since $W$ is $A$-stable, $A^{k}(w) \in W$ for all $k \geq 0$. Also $A^{k} w=\lambda_{1}^{k} v_{1}+\cdots+\lambda_{r}^{k} v_{r}$. Taking $k=0,1, \ldots, r-1$, we have the following equation in $V^{r}$ :$$\left(\begin{array}{cccc}
1 & 1 & \cdots & 1 \\
\lambda_{1} & \lambda_{2} & \cdots & \lambda_{r} \\
\vdots & \vdots & \ddots & \vdots \\
\lambda_{1}^{r-1} & \lambda_{2}^{r-1} & \cdots & \lambda_{r}^{r-1}
\end{array}\right)\left(\begin{array}{c}
v_{1} \\
v_{2} \\
\vdots \\
v_{r}
\end{array}\right)=\left(\begin{array}{c}
w \\
A w \\
\vdots \\
A^{r-1} w
\end{array}\right) .
$$The vector on the right is in $W^{r}$ (a subspace of $V^{r}$ ) and the $r \times r$ matrix on the left is invertible (Vandermonde matrix with distinct $\lambda_{i}$ ). Therefore the vector on the left is in $W^{r}$, so each $v_{i}$ is in $W$.
Theorem 5. If $A_{1}, \ldots, A_{r}$ are commuting linear operators on $V$ and each $A_{i}$ is diagonalizable then they are simultaneously diagonalizable, i.e., there is a basis of $V$ consisting of simultaneous eigenvectors for the $A_{i}$.
Proof. We won't use Theorem 3, but the proof will be essentially the same type of argument as in the proof of Theorem 3; the stronger hypothesis (commutativity and individual diagonalizability) will lead to a stronger conclusion (a basis of simultaneous eigenvectors, not just one simultaneous eigenvector).
The result is clear if $r=1$, so assume $r \geq 2$. Since the last operator $A_{r}$ is diagonalizable on $V, V$ is the direct sum of the eigenspaces for $A_{r}$. Let $\lambda$ be an eigenvalue for $A_{r}$ and $E_{\lambda}$ be the $\lambda$-eigenspace of $A_{r}$ in $V$. As in the proof of Theorem 3, since each $A_{i}$ commutes with $A_{r}$ we have $A_{i}\left(E_{\lambda}\right) \subset E_{\lambda}$. Thus each $A_{i}$ restricts to a linear operator on the subspace $E_{\lambda}$ and the linear operators $\left.A_{1}\right|_{E_{\lambda}}, \ldots,\left.A_{r-1}\right|_{E_{\lambda}}$ commute since $A_{1}, \ldots, A_{r-1}$ commute as operators on $V$.
By Lemma 4 , the restrictions $\left.A_{1}\right|_{E_{\lambda}}, \ldots,\left.A_{r-1}\right|_{E_{\lambda}}$ are each diagonalizable on $E_{\lambda}$. Since the number of these operators is less than $r$, by induction on $r$ there is a basis for $E_{\lambda}$ consisting of simultaneous eigenvectorsThis choice of basis for $E_{\lambda}$ is not made by $A_{r}$, but by the other operators together. for $\left.A_{1}\right|_{E_{\lambda}}, \ldots,\left.A_{r-1}\right|_{E_{\lambda}}.$ The elements of this basis for $E_{\lambda}$ are eigenvectors for $A_r|_{E_{\lambda}}$ as well, since all nonzero vectors in $E_{\lambda}$ are eigenvectors for $A_{r}$. Thus $\left.A_{1}\right|_{E_{2}}, \ldots,\left.A_{r-1}\right|_{E_{1}},\left.A_{r}\right|_{E_{\lambda}}$ are all diagonalizable. The vector space $V$ is the direct sum of the eigenspaces $E_{\lambda}$ of $A_{r}$, so stringing together simultaneous eigenbases of $\left.A_{1}\right|_{E_{\lambda}}, \ldots,\left.A_{r-1}\right|_{E_{\lambda}},\left.A_{r}\right|_{E_{\lambda}}$ as $\lambda$ runs over the eigenvalues of $A_{r}$ gives a simultaneous eigenbasis of $V$ for all the $A_{1}, \ldots, A_{r}$.
Remark 6. Theorem 5 is not saying commuting operators diagonalize! It says commuting diagonalizable operators simultaneously diagonalize. For example, the matrices $\left(\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right)$ for all $a$ commute with each other, but none of them are diagonalizable when $a \neq 0$.
Because we are dealing with operators on finite-dimensional spaces, Theorem 5 extends to a possibly infinite number of commuting operators, as follows.
Corollary 7. Let $\left\{A_{i}\right\}$ be a set of commuting linear operators on a finite-dimensional $\bf C$-vector space $V$. If each $A_{i}$ is diagonalizable on $V$ then they are simultaneously diagonalizable.
Proof. Let $U$ be the subspace of $\operatorname{End}_{F}(V)$ spanned by the operators $A_{i}$ 's. Since $\operatorname{End}_{F}(V)$ is finite-dimensional, its subspace $U$ is finite-dimensional, so $U$ is spanned by a finite number of $A_{i}$ 's, say $A_{i_{1}}, \ldots, A_{i_{r}}$. By Theorem 5, there is a common eigenbasis of $V$ for $A_{i_{1}}, \ldots, A_{i_{r}}$. A common eigenbasis for linear operators is also an eigenbasis for any linear combination of the operators, so this common eigenbasis of $A_{i_{1}}, \ldots, A_{i_{r}}$ diagonalizes every element of $U$, and in particular diagonalizes each $A_{i}$.
Corollary 7 is important in number theory, where it implies the existence of eigenforms for Hecke operators. |
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Czhang271828
发表于 2021-6-18 11:27
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发表于 2021-7-18 21:42
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