矩阵行列式证明中(tI+sB)⁻¹A对称化的技术咨询

Let's walk through this clearly—since you're working through a proof involving symmetric matrices and determinant manipulations, this symmetrization trick is a common move in linear algebra, so it's good to unpack the details.

1. How to Symmetrize $(tI+sB)^{-1}A$ #

First, let's ground ourselves in the properties we can leverage here:

• $I$ (identity matrix) and $B$ are symmetric, so their linear combination $tI + sB$ is also symmetric (real scalars preserve symmetry in matrix combinations).
• The inverse of a symmetric invertible matrix is symmetric: $[(tI+sB)^{-1}]T = (tI+sB)^{-1}$.
• $A$ is symmetric, so $A^T = A$.

The standard method to symmetrize any square matrix $M$ is to take the average of the matrix and its transpose—this guarantees the result is symmetric. Applying this to $M = (tI+sB)^{-1}A$:

1. Calculate the transpose of $M$:
   $$M^T = \left[(tI+sB)^{-1}A\right]T = A^T \cdot \left[(tI+sB)^{-1}\right]T = A \cdot (tI+sB)^{-1}$$
   (We used the transpose rule $(AB)^T = B^T A^T$, plus the symmetry of $A$ and $(tI+sB)^{-1}$.)

2. Average $M$ and its transpose to get the symmetricized matrix:
   $$\frac{1}{2}\left[(tI+sB)^{-1}A + A(tI+sB)^{-1}\right]$$

You can verify this is symmetric by taking its transpose—it will match the original expression exactly.

2. Impact on the Determinant $\det\left(I + (tI+sB)^{-1}Ax\right)$ #

Now, the big question: how does this symmetrization affect the determinant you care about? Let's break this down:

First, rewrite your determinant to isolate the matrix $M = (tI+sB)^{-1}A$ (since $x \in \mathbb{R}$):
$$\det\left(I + (tI+sB)^{-1}Ax\right) = \det\left(I + x \cdot (tI+sB)^{-1}A\right) = \det(I + xM)$$

Key Points: #

• $\det(I + xM) = \det(I + xM^T)$: The determinant of a matrix equals the determinant of its transpose. So:
  $$\det(I + xM) = \det\left((I + xM)^T\right) = \det(I + xM^T)$$
  For our case, this means $\det\left(I + x(tI+sB)^{-1}A\right) = \det\left(I + xA(tI+sB)^{-1}\right)$—the order of multiplication doesn't change the determinant here.

• Symmetrization does not preserve the original determinant: In general, $\det\left(I + x \cdot \frac{M + M^T}{2}\right) \neq \det(I + xM)$. For example, take $M = \begin{pmatrix}0 & 1 \ 0 & 0\end{pmatrix}$: $\det(I + xM) = 1$, but the determinant of $I + x$ times the symmetricized $M$ is $1 - \frac{x^2}{4}$, which is clearly different.

• Why does the proof suggest symmetrization then?: The goal is almost certainly to convert $M$ into a symmetric matrix so that you can use properties unique to symmetric matrices—like having real eigenvalues, being orthogonally diagonalizable, or satisfying spectral theorems. Even though the determinant of $I + xM$ isn't the same as the determinant involving the symmetricized matrix, $M$ and its symmetricized version share key properties (e.g., their traces are equal, since $\text{tr}(M^T) = \text{tr}(M)$). Additionally, $M$ is similar to $M^T$ (since $M^T = (tI+sB)M(tI+sB)^{-1}$), meaning they have identical eigenvalues—so products like $\det(I + xM)$ (which is the product of $(1 + x\lambda_i)$ for each eigenvalue $\lambda_i$ of $M$) are preserved when swapping $M$ with $M^T$.

内容的提问来源于stack exchange，提问作者ajay pawar