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Questions tagged [generalized-inverse]

A generalized inverse of a matrix $A$ is any matrix $A^{-}$ satisfying $AA^{-}A = A$. When $A$ is nonsingular, $A^{-}$ is unique and $A^{-} = A^{-1}$; otherwise, there are infinitely many solutions to $A^{-}$. Generalized inverses arise in linear models for statistics, for when the design matrix of a linear model is not invertible and the ordinary least squares estimate of the parameter vector is not unique.

A generalized inverse of a matrix $A$ is any matrix $A^{-}$ satisfying $AA^{-}A = A$. When $A$ is nonsingular, $A^{-}$ is unique and $A^{-} = A^{-1}$; otherwise, there are infinitely many solutions to $A^{-}$. Generalized inverses arise in linear models for statistics, for when the design matrix of a linear model is not invertible and the ordinary least squares estimate of the parameter vector is not unique. Further information can be found here.

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Trace inequality for a product of p.s.d. matrices and their pseudo inverse.

Let $A, B_i$ be positive semidefinite real matrices. Let $\dagger$ stand for the Moore-Penrose generalized inverse. I managed to prove that if $\operatorname{Ran}B_1\subseteq\operatorname{Ker}B_2$ then $$\operatorname{trace}\left((A + B_1 +…
Manuel
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Regular semigroups -- intuition

I'm trying to develop some intuition around the definition of the pseudoinverse in a regular semigroup. Let the semigroup be $S$ with its associative operation written by juxtaposition. The pseudoinverse of an element $a \in S$ is an element $x \in…
Jamie Ballingall
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$A \in \mathbb{C}^{m\times n}$,$A=FG^*$ and $r(A)=r(F)=r(G)$. Prove $A^\dagger = G(F^*AG)^{-1}F^*$ and $A^\dagger = (G^\dagger)^*F^\dagger$

Let $A^\dagger$ be a Moore-Penrose inverse of a matrix $A$. If $A \in \mathbb{C}^{m\times n}$ and $A=FG^*$, for some $F,G$ and $r(A)=r(F)=r(G)$, prove that $$A^\dagger = G(F^*AG)^{-1}F^*$$ and $$A^\dagger = (G^\dagger)^*F^\dagger.$$ I need to show…
mathbbandstuff
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Calculating generalized inverse of singular matrix using singular value decomposition (SVD)

I became confused about how singular value decomposition can be used to find generalized inverse of singular matrix. Specifically, I am dealing with the matrix $G=\begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & \sqrt{2} & \sqrt{2} & 0 \\…
Senna
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Convergence of generalized inverses

I copy-paste the post from Math Overflow - maybe someone can give me a tip. During the reading about Fisher–Tippett–Gnedenko theorem (it can be found easily on wiki - I don't have enough reputation to post more links), I've got stuck, trying to…
user2280549
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Prove: if $x_0$ is a function of $Ax=b$, then there exist some generalized inverse matrix of A, G, s.t. $x_0 = Gb$

Suppose $x_0$ is a solution of $Ax=b$, where $b\neq0$. How to prove that $x_0 = Gb$, where $G$ is a generalized inverse matrix of A? This is the Lemma 9.3 of Linear Algebra and Matrix. Here is proof, but I just cannot get it. The last line says…
mxdxzxyjzx
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How to understand quasi-inverse of a function f∘g∘f = f?

Recently I was studying the quasi-inverse. Before I studied the quasi-inverse, I revisited the inverse and the left-right inverse. inverse function: Let $f : X → Y$, $g : Y → X$ is inverse of $f$, if only if, $f∘g = id_{Y}$ and $g∘f = id_{X}$. It is…
chansey
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A question on positive semi definite matrices

Suppose $A,B$ are symmetric, positive semi-definite matrices of same order such that $A \preceq B \preceq \kappa A$. How to prove that this is equivalent to $\frac{1}{\kappa}A^+ \preceq B^+ \preceq A^+$? Here $A^+$ is the generalized inverse of $A$.
Sudipta Roy
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Prove that $A^2=A\iff \Sigma K=I_r$

Let $A$ be a square complex matrix and let $A=U\Sigma V^*$ be a singular value decomposition. Then $A$ can be written as $$A=U\begin{bmatrix} \Sigma K & \Sigma L\\ 0 & 0 \end{bmatrix} U^*$$ where $V^* =\begin{bmatrix} K &…
David
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If $f : U \rightarrow X$ is an isometric inclusion of Banach spaces, does $f' : X' \rightarrow U'$ have a bounded generalized inverse?

Let $X$ be a Banach space and let $U$ be a closed Banach subspace. The inclusion mapping $$ f : U \rightarrow X $$ induces a dual mapping $$ f' : X' \rightarrow U' $$ I am wondering about the existence of a bounded operator $$ g : U' \rightarrow X'…
shuhalo
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A similarity-like transfromation with left-invertible matrix.

Suppose $X\in \mathbb R^{m\times n}$ $(m>n)$ is a left-invertible matrix, and its left-inverse is $X^+=(X^\top X)^{-1}X^\top$ (so that $X^+X=I$). Now I have a matrix $A\in \mathbb R^{n\times n}$. Define $B\triangleq XAX^+$. By computing some…
fibon
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Questions about Moore–Penrose inverse

I have some questions about Moore–Penrose inverse. Let $A, P\in \mathbb{R}^{d\times d}$. Suppose $A$ is positive definite and $P$ is a projection matrix with $P^2=P, P^\top=P$. I try to prove that $A^{-1}-(PAP)^{-}$ is semi-positive definite. Here…
beginner
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How to prove that any matrices have their own generalized inverse.

Let $A$ be a matrix with a form $(m.n)$, and $X$ be a matrix with a form of $(n,m)$. If $AXA = A$, $X$ is called a generalized inverse of $A$. How can we prove that any matrices have their own generalized inverse? \begin{eqnarray} \\…
ohisamadaigaku
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Invertibility modulo the intersection of ideals in $C^*$-algebras

Crossposted to mathoverflow due to low attention. Let $\mathcal{A}$ be a $C^*$-algebra and $A \in \mathcal{A}$. I am interested in the invertibility of $A$ modulo certain (two-sided) ideals $\mathcal{J} \subseteq \mathcal{A}$, i.e. the existence of…
Klaus
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Generalized inverse matrix rules

I would like to ask how free you are when you are calculating generalized inverses? I know, that it is said that there are infinite many of them, but we always usually choose submatrix such that elements are picked in square shape so it is easy to…
Mirjan Pecenko
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