Let A ∈ℂ^p ×p, 𝐮 ∈ℂ^p, 𝐯 ∈ℂ^p and assume that A is invertible. Show that [ A -𝐮 𝐯^H

Let A ∈ℂ^p ×p, 𝐮 ∈ℂ^p, 𝐯 ∈ℂ^p and assume that A is...

Let $A \in \mathbb{C}^{p \times p}, \mathbf{u} \in \mathbb{C}^p, \mathbf{v} \in \mathbb{C}^p$ and assume that $A$ is invertible. Show that $\left[\begin{array}{cc}A & -\mathbf{u} \\ \mathbf{v}^H & 1\end{array}\right]$ is invertible $\Longleftrightarrow A+\mathbf{u v}^H$ is invertible $\Longleftrightarrow 1+\mathbf{v}^H A^{-1} \mathbf{u} \neq 0$ and that if these conditions are met, then $$ \left(I_p+\mathbf{u} \mathbf{v}^H A^{-1}\right)^{-1} \mathbf{u}=\mathbf{u}\left(1+\mathbf{v}^H A^{-1} \mathbf{u}\right)^{-1} . $$

Let $A \in \mathbb{C}^{p \times p}, \mathbf{u} \in \mathbb{C}^p, \mathbf{v} \in \mathbb{C}^p$ and assume that $A$ is invertible. Show that
$\left[\begin{array}{cc}A & -\mathbf{u} \\ \mathbf{v}^H & 1\end{array}\right]$ is invertible $\Longleftrightarrow A+\mathbf{u v}^H$ is invertible $\Longleftrightarrow 1+\mathbf{v}^H A^{-1} \mathbf{u} \neq 0$ and that if these conditions are met, then
$$
\left(I_p+\mathbf{u} \mathbf{v}^H A^{-1}\right)^{-1} \mathbf{u}=\mathbf{u}\left(1+\mathbf{v}^H A^{-1} \mathbf{u}\right)^{-1} .
$$

Key Concepts

Rank-One Perturbation

Rank-one perturbation describes the modification of a matrix by adding or subtracting a matrix of rank one, typically represented as the outer product of two vectors. This concept underlies many results in numerical linear algebra, including the Sherman–Morrison formula, and is used to analyse how slight changes affect the invertibility and spectrum of the matrix.

Block Matrix Inversion

Block matrix inversion techniques are used to invert matrices that are partitioned into submatrices. These techniques often rely on the Schur complement and provide conditions under which the entire matrix is invertible based on the invertibility of its submatrices or associated expressions. This approach is central in establishing equivalences relating to the invertibility of the original block matrix and its updates.

Sherman–Morrison Formula

The Sherman–Morrison formula is a specific case of a rank-one update to an invertible matrix. It provides an explicit formula for the inverse of a matrix that has been perturbed by the outer product of two vectors. This result is particularly useful when dealing with matrices expressed as the sum of an invertible matrix and a rank-one matrix, and it allows for efficient computations and comparisons of invertibility conditions.

Schur Complement

The Schur complement is a fundamental concept in matrix analysis used to study the invertibility and factorization of block matrices. When a block matrix is partitioned and one of the blocks is invertible, the invertibility of the entire matrix can be determined by the invertibility of the Schur complement. This tool connects the properties of the submatrices and offers a method to explicitly compute inverses of block matrices.

Transcript

00:05 For chapter 1, section 6, exercise 48, we have m -inverted equals i, and we're going to prove that.

00:29 So we have m -inverted equals a -1 -u -w inverted times v, a -inverted plus a -inverted -inverted plus a -inverted u times w.

00:50 V a inverted times you close parentheses inverted this be a inverted so from here we get i plus u w minus v a inverted u inverted v a inverted minus u a inverted minus okay i'll write this below u w inverted times v a inverted times u open parentheses w minus v a inverted times u inverted times v times a inverted all right so we're going to call this asterisk and this we're going to call a star so first looking at our asterisk, we have va inverted u, w minus va inverted u equals va inverted u minus w plus w.

02:42 And we have w minus va inverted u inverted u inverted so this is going to equal v a inverted u minus w w minus va inverted u and invert this plus w w minus va inverted u inverted u and so what this is going to equals negative identity plus w minus va inverted all right, now for the star.

03:51 This is uw inverted, w -w -w -1 -v -a -inverted, u -inverted minus identity -inverted, times v -a -inverted.

04:08 So this is going to equal u -w -minus v -a -inverted, u -inverted, v -a -inverted, minus u -w -inverted, inverted, v -a -inverted.

04:29 So now we can say that m inverted has i well hold on let's go back to the top or it's already written.

05:07 Okay, so we see i'm going to highlight it this portion here and the other portion is going to be subtracted so it's going to be minus that.

05:48 So what we get here, when we separate this into its components and blue are both of these work.

06:33 They're being subtracted.

06:47 So we can conclude that that m and an inverted does equal its identity.

07:05 And so to get that into four, that implies that a equals i, who equals u, v.

07:22 Equals v transpose and w equals i this is for one for two we have to have a equals a u equals u equals v transpose and w equals zero for three a has equal i sub m u equals u v equals v and w equals i sub m so now i can say on position on position one we will have a b c inverted equal to a minus b d inverted c inverted this this will be negative d inverted c inverted this will be negative d inverted c a minus negative b, d, inverted, c, inverted, negative a minus b, d inverted, times c inverted, b, d inverted...

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