如何求解协方差矩阵$oldsymbol{aBB'+R}$的特征值与特征向量？

Great question! Let's work through this systematically, starting with the column vector case you've already started on, then extend to the general low-rank $\boldsymbol{B}$ scenario.

First, let's restate the setup clearly:

• $\boldsymbol{R}$ is an $n\times n$ positive definite diagonal matrix (since we're dealing with a covariance matrix, it's safe to assume all diagonal entries $r_1, r_2, ..., r_n > 0$)
• $\boldsymbol{B}$ is an $n\times k$ low-rank matrix (rank $k$)
• $a$ is a positive real number

1. The Column Vector Case ($k=1$, $\boldsymbol{B} = \boldsymbol{b}$) #

You already started using the Matrix Determinant Lemma—let's finish that derivation to get full clarity:

The Matrix Determinant Lemma tells us:
$$\det(\boldsymbol{M} + \boldsymbol{u}\boldsymbol{v}^\top) = \det(\boldsymbol{M})(1 + \boldsymbol{v}^{\top\boldsymbol{M}}{-1}\boldsymbol{u})$$
Let $\boldsymbol{M} = \boldsymbol{R} - \lambda\boldsymbol{I}$, $\boldsymbol{u} = \sqrt{a}\boldsymbol{b}$, and $\boldsymbol{v}^\top = \sqrt{a}\boldsymbol{b}^\top$. Plugging these in gives:
$$\det(a\boldsymbol{b}\boldsymbol{b}^\top + \boldsymbol{R} - \lambda\boldsymbol{I}) = \det(\boldsymbol{R} - \lambda\boldsymbol{I}) \left(1 + a\boldsymbol{b}^\top(\boldsymbol{R} - \lambda\boldsymbol{I})^{-1}\boldsymbol{b}\right)$$

Since $\boldsymbol{R}$ is diagonal, $(\boldsymbol{R} - \lambda\boldsymbol{I})^{-1}$ is also diagonal with entries $\frac{1}{r_i - \lambda}$. So the quadratic form simplifies to:
$$\boldsymbol{b}^\top(\boldsymbol{R} - \lambda\boldsymbol{I})^{-1}\boldsymbol{b} = \sum_{i=1}^n \frac{b_i^2}{r_i - \lambda}$$

Our characteristic polynomial becomes:
$$\prod_{i=1}^n (r_i - \lambda) \left(1 + a\sum_{i=1}^n \frac{b_i^2}{r_i - \lambda}\right) = 0$$

Solving for Eigenvalues #

We split this into two cases:

• Case 1: $\prod_{i=1}^n (r_i - \lambda) = 0$: For any index $i$ where $b_i = 0$, $\lambda = r_i$ is an eigenvalue. The corresponding eigenvector is the standard basis vector $\boldsymbol{e}_i$, since $a\boldsymbol{b}\boldsymbol{b}^\top\boldsymbol{e}_i + \boldsymbol{R}\boldsymbol{e}_i = 0 + r_i\boldsymbol{e}_i$.
• Case 2: $1 + a\sum_{i=1}^n \frac{b_i^2}{r_i - \lambda} = 0$: Rearranging gives:
  $$\sum_{i=1}^n \frac{a b_i^2}{\lambda - r_i} = 1$$
  This equation has exactly one positive solution $\lambda^* > \max{r_i}$ (the left-hand side is strictly increasing in $\lambda$, going from $+\infty$ as $\lambda \to \max(r_i)^+$ to $0$ as $\lambda \to +\infty$). The corresponding eigenvector is $(\boldsymbol{R} - \lambda^{*\boldsymbol{I})}{-1}\boldsymbol{b}$—you can verify this by plugging it back into the eigenvalue equation!

2. Generalizing to Low-Rank $\boldsymbol{B}$ (Rank $k$) #

When $\boldsymbol{B}$ has rank $k$, we can use Singular Value Decomposition (SVD) to simplify the problem:

Let $\boldsymbol{B} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^\top$, where:

• $\boldsymbol{U}$ is an $n\times k$ orthogonal matrix (columns form an orthonormal basis for $\boldsymbol{B}$'s column space)
• $\boldsymbol{\Sigma}$ is a $k\times k$ diagonal matrix with positive singular values $\sigma_1, ..., \sigma_k$
• $\boldsymbol{V}$ is a $k\times k$ orthogonal matrix

Now, $a\boldsymbol{BB}^\top = a\boldsymbol{U}\boldsymbol{\Sigma}^{2\boldsymbol{U}}\top$. Let $\boldsymbol{C} = \boldsymbol{U}^\top\boldsymbol{R}\boldsymbol{U}$—this is a $k\times k$ positive definite matrix (since $\boldsymbol{R}$ is positive definite and $\boldsymbol{U}$ is orthogonal).

Eigenvalue Classification #

We can split the eigenvalues of $\boldsymbol{M} = a\boldsymbol{BB}^\top + \boldsymbol{R}$ into two disjoint sets:

• Eigenvalues from the orthogonal complement of $\boldsymbol{B}$'s column space: For any vector $\boldsymbol{x}$ that's orthogonal to all columns of $\boldsymbol{U}$ (i.e., $\boldsymbol{U}^\top\boldsymbol{x} = 0$), $\boldsymbol{BB}^\top\boldsymbol{x} = 0$, so $\boldsymbol{M}\boldsymbol{x} = \boldsymbol{R}\boldsymbol{x}$. This means any diagonal entry $r_i$ of $\boldsymbol{R}$ where the standard basis vector $\boldsymbol{e}_i$ isn't in $\boldsymbol{B}$'s column space remains an eigenvalue of $\boldsymbol{M}$, with eigenvector $\boldsymbol{e}_i$.
• Eigenvalues from $\boldsymbol{B}$'s column space: Restricted to $\boldsymbol{B}$'s column space, $\boldsymbol{M}$ acts as $\boldsymbol{U}(a\boldsymbol{\Sigma}^2 + \boldsymbol{C})\boldsymbol{U}^\top$. So the eigenvalues here are exactly the eigenvalues of the $k\times k$ matrix $\boldsymbol{D} = a\boldsymbol{\Sigma}^2 + \boldsymbol{C}$. The corresponding eigenvectors are $\boldsymbol{U}\boldsymbol{y}$, where $\boldsymbol{y}$ is an eigenvector of $\boldsymbol{D}$.

Alternatively, we can use the rank-$k$ extension of the Matrix Determinant Lemma:
$$\det(\boldsymbol{R} - \lambda\boldsymbol{I} + a\boldsymbol{BB}^\top) = \det(\boldsymbol{R} - \lambda\boldsymbol{I}) \det(\boldsymbol{I}_k + a\boldsymbol{B}^\top(\boldsymbol{R} - \lambda\boldsymbol{I})^{-1}\boldsymbol{B})$$
This holds when $\lambda$ isn't an eigenvalue of $\boldsymbol{R}$. The roots of the characteristic polynomial are:

• The eigenvalues $r_i$ of $\boldsymbol{R}$ (as described above)
• The $k$ solutions to $\det(\boldsymbol{I}_k + a\boldsymbol{B}^\top(\boldsymbol{R} - \lambda\boldsymbol{I})^{-1}\boldsymbol{B}) = 0$, which will all be greater than the smallest eigenvalue of $\boldsymbol{R}$.

3. Quick Recap #

• For $\boldsymbol{B}$ as a column vector: Eigenvalues are the $r_i$ where $b_i=0$, plus one additional eigenvalue from the scalar equation.
• For low-rank $\boldsymbol{B}$: Eigenvalues are the $r_i$ not associated with $\boldsymbol{B}$'s column space, plus $k$ eigenvalues from the $k\times k$ matrix $\boldsymbol{D} = a\boldsymbol{\Sigma}^2 + \boldsymbol{U}^\top\boldsymbol{R}\boldsymbol{U}$.
• Eigenvectors can be constructed by splitting vectors into the space spanned by $\boldsymbol{B}$ and its orthogonal complement.

内容的提问来源于stack exchange，提问作者Taylor