矩阵C=B+ghᵀ的可逆性证明及逆矩阵推导问询

Hey there! No worries at all—this is a classic linear algebra result called the Sherman-Morrison formula, and I’ll walk you through it step by step, making sure we verify every matrix dimension rule along the way (since that’s often the tricky part when you’re getting back into algebra!).

First, let’s recap the given details and confirm all dimensions are valid for the operations we’ll use:

• $B$ is an invertible $n \times n$ matrix, so its inverse $B^{-1}$ is also $n \times n$.
• $g$ and $h$ are $n \times 1$ column vectors, which means $h^T$ (the transpose of $h$) is a $1 \times n$ row vector.
• The product $gh^T$ is an $n \times n$ matrix (since $n \times 1$ multiplied by $1 \times n$ gives $n \times n$), so adding it to $B$ (another $n \times n$ matrix) is totally valid—$C = B + gh^T$ is indeed an $n \times n$ matrix.

To prove $C$ is invertible and that $D = B^{-1} - \frac{B^{-1}ghTB^{-1}}{1 + h^TB{-1}g}$ is its inverse, we just need to show that multiplying $C$ and $D$ gives the $n \times n$ identity matrix $I$ (by definition of matrix inverses). Let’s do that calculation:

Step 1: Expand the product $C \times D$ #

Using the distributive property of matrix multiplication (just like the hint mentions: $A(B+C)D = ABD + ACD$), we expand:
$$
C \times D = \left(B + gh^{T\right)\left(B}{-1} - \frac{B^{-1}ghTB^{-1}}{1 + h^TB{-1}g}\right)
$$
This splits into four terms:
$$
= B \cdot B^{-1} - B \cdot \frac{B^{-1}ghTB^{-1}}{1 + h^TB{-1}g} + gh^T \cdot B^{-1} - gh^T \cdot \frac{B^{-1}ghTB^{-1}}{1 + h^TB{-1}g}
$$

Step 2: Simplify each term one by one #

Let’s go through each term and simplify, checking dimensions as we go:

1. First term: $B \cdot B^{-1} = I$ (the $n \times n$ identity matrix—this is the definition of a matrix inverse).
2. Second term: $B \cdot B^{-1}ghTB^{-1} = (B \cdot B^{-1})ghTB^{-1} = I \cdot gh^TB{-1} = gh^TB{-1}$. Here, we used matrix associativity, and all dimensions check out: $B \cdot B^{-1}$ is $n \times n$, multiplied by $gh^TB{-1}$ (also $n \times n$) gives $n \times n$.
3. Third term: $gh^T \cdot B^{-1}$ is just $gh^TB{-1}$ (an $n \times n$ matrix, since $gh^T$ is $n \times n$ and $B^{-1}$ is $n \times n$).
4. Fourth term: This is the tricky one! Notice that $h^TB{-1}g$ is a scalar (a single number): $h^T$ is $1 \times n$, $B^{-1}$ is $n \times n$, $g$ is $n \times 1$—multiplying these gives a $1 \times 1$ matrix, which is just a scalar. We can factor this scalar out:
   $$
   gh^T \cdot B^{-1}ghTB^{-1} = g(h^TB{-1}g)h^TB{-1} = (h^TB{-1}g)gh^TB{-1}
   $$
   Since scalars commute with matrices, we can move $h^TB{-1}g$ to the front without changing the product.

Step 3: Combine terms and simplify #

Let’s substitute the simplified terms back into the expanded product, and let’s use $S = 1 + h^TB{-1}g$ (the denominator, which is non-zero per the problem statement, so division is valid):
$$
C \times D = I - \frac{gh^TB{-1}}{S} + gh^TB{-1} - \frac{(h^TB{-1}g)gh^TB{-1}}{S}
$$
Now combine the terms with $gh^TB{-1}$:
$$
= I + gh^TB{-1}\left(1 - \frac{1}{S}\right) - \frac{(h^TB{-1}g)gh^TB{-1}}{S}
$$
Since $1 - \frac{1}{S} = \frac{S - 1}{S} = \frac{h^TB{-1}g}{S}$, substitute that in:
$$
= I + gh^TB{-1} \cdot \frac{h^TB{-1}g}{S} - \frac{(h^TB{-1}g)gh^TB{-1}}{S}
$$
The last two terms are identical (remember, scalars commute with matrices), so they cancel each other out:
$$
= I + \frac{(h^TB{-1}g)gh^TB{-1} - (h^TB{-1}g)gh^TB{-1}}{S} = I + 0 = I
$$

Conclusion #

We’ve shown that $C \times D = I$, which means $C$ is invertible, and its inverse is exactly the $D$ given in the problem. All matrix products and operations were valid because we checked dimensions at every step—this is key to avoiding mistakes with matrix algebra!

Hope that clears things up for you. It’s a great problem to refresh your linear algebra skills—kudos to your daughter for giving you a fun challenge!

备注：内容来源于stack exchange，提问作者Robert Beattie