如何创建一个链接/按钮，使我可以打开另一个窗口中的DataGrid在WPF？

// MainWindow.xaml.cs

using System.Windows;

namespace WpfApp
{
    public partial class MainWindow : Window
    {
        public MainWindow()
        {
            InitializeComponent();
        }

        private void OpenDataGridWindow(object sender, RoutedEventArgs e)
        {
            DataGridWindow dataGridWindow = new DataGridWindow();
            dataGridWindow.Show();
        }
    }
}

<!-- MainWindow.xaml -->

<Window x:Class="WpfApp.MainWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        Title="Main Window" Height="450" Width="800">
    <Grid>
        <Button Content="Open DataGrid Window" Click="OpenDataGridWindow" />
    </Grid>
</Window>

// DataGridWindow.xaml.cs

using System.Collections.Generic;
using System.Windows;

namespace WpfApp
{
    public partial class DataGridWindow : Window
    {
        public DataGridWindow()
        {
            InitializeComponent();

            // Simulated data source
            List<Person> people = new List<Person>
            {
                new Person { Name = "John", Age = 25 },
                new Person { Name = "Jane", Age = 30 },
                new Person { Name = "Bob", Age = 35 }
            };

            dataGrid.ItemsSource = people;
        }
    }

    public class Person
    {
        public string Name { get; set; }
        public int Age { get; set; }
    }
}

<!-- DataGridWindow.xaml -->

<Window x:Class="WpfApp.DataGridWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        Title="DataGrid Window" Height="450" Width="800">
    <Grid>
        <DataGrid x:Name="dataGrid" AutoGenerateColumns="True" />
    </Grid>
</Window>