求证矩阵$f A = f B + (f B - f I)f C (f B - f I)$特征值模长≤1

Setup and Key Known Properties #

First, let's recap the given information to ground our work:

• $\bf B$ is a symmetric idempotent matrix: $\bf B^T = B$ and $\bf B^2 = B$ (its eigenvalues can only be 0 or 1, and it's orthogonally diagonalizable).
• $\bf C$ is a symmetric matrix with all eigenvalues satisfying $|\lambda_C| ≤ 1$.
• The matrix $\bf A$ is defined as:
  $$\bf A = B + (B - I)C(B - I)$$
• We can leverage the commutation relation: $\bf AB = BA = B$.

Approach 1: Using Spectral Decomposition of $\bf B$ #

Since $\bf B$ is symmetric and idempotent, we can write its spectral decomposition as:
$$\bf B = Q\Lambda Q^T$$
where $Q$ is an orthogonal matrix ($Q^TQ = QQ^T = I$) and $\Lambda$ is a diagonal matrix with entries 0 or 1 (corresponding to $\bf B$'s eigenvalues).

Let's rewrite $\bf A$ using this decomposition:

1. Compute $\bf B - I = Q(\Lambda - I)Q^T$. Note that $\Lambda - I$ is a diagonal matrix with entries 0 (where $\Lambda$ has 1) and -1 (where $\Lambda$ has 0).
2. Substitute into the expression for $\bf A$:
   $$\bf A = Q\Lambda Q^T + Q(\Lambda - I)Q^T C Q(\Lambda - I)Q^T$$
3. Let $\bf C' = Q^T C Q$ (this is symmetric, like $\bf C$, and has the same eigenvalues, so $|\lambda_{C'}| ≤1$). Now $\bf A$ simplifies to:
   $$\bf A = Q\left[ \Lambda + (\Lambda - I)C'(\Lambda - I) \right] Q^T$$

Now look at the matrix inside the brackets, $\bf D = \Lambda + (\Lambda - I)C'(\Lambda - I)$. Split $\Lambda$ into block form (say $k$ ones and $n-k$ zeros):
$$\Lambda = \begin{pmatrix} I_k & 0 \ 0 & 0_{n-k} \end{pmatrix}, \quad \Lambda - I = \begin{pmatrix} 0_k & 0 \ 0 & -I_{n-k} \end{pmatrix}$$

Calculating the product $(\Lambda - I)C'(\Lambda - I)$ gives us a block matrix with zeros everywhere except the bottom-right block, which is $\bf C'{22}$ (the bottom-right submatrix of $\bf C'$). So:
$$\bf D = \begin{pmatrix} I_k & 0 \ 0 & C'{22} \end{pmatrix}$$

The eigenvalues of $\bf D$ are:

• $k$ copies of 1 (from the top-left block)
• The eigenvalues of $\bf C'{22}$, which all have magnitude ≤1 (since $\bf C'{22}$ is a symmetric submatrix of $\bf C'$, its eigenvalues lie within the range of $\bf C'$'s eigenvalues, per the Courant-Fischer theorem).

Since $\bf A$ is similar to $\bf D$ (via orthogonal similarity), they share the same eigenvalues. Thus, all eigenvalues of $\bf A$ have magnitude ≤1.

Approach 2: Using the Commutation Relation $\bf AB = B$ #

Let $\bf x$ be an eigenvector of $\bf A$ with eigenvalue $\lambda$, so $\bf Ax = \lambda x$. Multiply both sides by $\bf B$:
$$\bf BAx = \lambda Bx$$
But we know $\bf BA = B$, so this simplifies to:
$$\bf Bx = \lambda Bx \implies (\lambda - 1)\bf Bx = 0$$

This gives us two cases:

1. Case 1: $\lambda = 1$: This is trivially valid (magnitude 1 ≤1).
2. Case 2: $\bf Bx = 0$: Substitute this into the expression for $\bf Ax$:
   $$\bf Ax = [B + (B - I)C(B - I)]x = 0 + (-I)C(-I)x = Cx$$
   So $\lambda x = Cx$, meaning $\lambda$ is an eigenvalue of $\bf C$. By our given condition, $|\lambda| ≤1$.

In both cases, the eigenvalue's magnitude is at most 1.

内容的提问来源于stack exchange，提问作者mzp