Let A ∈ℂ^n ×n be invertible and let 𝐮, 𝐯 ∈ℂ^n. Show that the matrix A+𝐮 𝐯^T is invertible if and

step 1 Hi, so in this occasion we need to prove this statement here. So let's suppose that a is a squar

Let A ∈ℂ^n ×n be invertible and let 𝐮, 𝐯 ∈ℂ^n. Show that the...

Let $A \in \mathbb{C}^{n \times n}$ be invertible and let $\mathbf{u}, \mathbf{v} \in \mathbb{C}^n$. Show that the matrix $A+\mathbf{u v}^T$ is invertible if and only if $1+\mathbf{v}^T A^{-1} \mathbf{u} \neq 0$.

Key Concepts

Sherman-Morrison Formula

The Sherman-Morrison formula is a specific result in numerical linear algebra that provides an explicit expression for the inverse of a matrix after a rank-one update. It shows that if an invertible matrix A is updated by a rank-one matrix uv^T, then the inverse of the updated matrix can be expressed in terms of A, u, and v, given that a certain scalar condition holds.

Condition for Invertibility via Determinant Adjustment

The key condition that arises from applying the Sherman-Morrison formula is that the matrix A + uv^T is invertible if and only if the scalar 1 + v^T A^(-1) u is not equal to zero. This condition essentially adjusts for the impact of the rank-one update on the determinant of the original matrix, ensuring that the modified determinant remains non-zero.

Invertible Matrix

An invertible matrix is one that has a unique inverse, meaning that when multiplied by its inverse, it yields the identity matrix. This property is equivalent to the determinant of the matrix being non-zero and implies that the matrix represents a bijective linear transformation.

Rank-One Update

A rank-one update involves adding a matrix of rank one, typically formed as an outer product of two vectors, to an existing matrix. In many applications, this slight alteration allows for efficient computation of the inverse or determinant modification without recomputing from scratch.

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