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conformable

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conformable
conformable, a. (kənˈfɔːməb(ə)l) [f. conform v. + -able: perh. after agree-able, the suffix having here a like force: cf. also comfortable, amicable, etc. It. has conformabile and conformevole in Florio. Formerly also written confirmable, by confusion with that word q.v. Cf. confirm, conform.] 1. Ac... Oxford English Dictionary
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Conformable matrix
In this case, we say that and are conformable for multiplication (in that sequence). Only a square matrix is conformable for matrix inversion. wikipedia.org
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non-conformable
† non-conˈformable, a. Obs. [non- 3. See conformable 3 c.] Nonconforming.1647 Clarendon Hist. Reb. iii. 257 The non-conformable party of the kingdom. 1672 Baxter Bagshaw's Scand. iii. 32 In 1640 there were not found near half so many Non conformable Ministers as are Counties in England. 1691 Wood At... Oxford English Dictionary
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Conformation
standards for its breed Conformation show, a dog show in which dogs are judged according to how well they conform to the established breed type See also Conformable wikipedia.org
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when is the annihilator matrix conformable
It states that if the regression we are concerned with is:where and are and matrices respectively and where and are conformable, then the estimate
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Parasequence
The succession is supposed to be relatively conformable in the sense that breaks in deposition within the parasequence are much shorter than the time of Properties Since parasequences are relatively conformable, so Walther's law applies within a parasequence. wikipedia.org
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Conformable matrices for multiplication proof I need help resolving this exercise, any indication would be of great help to me. If anyone knows which book they belong to, I appreciate the information. Let A and B be ...
Hint: The equalities are just equivalent to showing that $$(I+BA)B=B(I+AB)$$ $$I=(I+A)-A$$ provided those matrix inverse are well defined. Edit to be more specific for the first equation: Since $(I+BA)B=B(I+AB)$, we can multiply post-multiply $(I+AB)^{-1}$ on both sides. $$(I+BA)B(I+AB)^{-1}=B$$ Now...
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Lambda distribution
The lambda distribution is either of two probability distributions used in statistics: Tukey's lambda distribution is a shape-conformable distribution wikipedia.org
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An elementary proof in matrix algebra **If $A$ and $B$ are two conformable matrices,and $AB = A$ and $BA = B$ then prove that A and B are idempotent matrices.** This may be trivial but I am not sure how to proceed on...
$$Ax = ABx = A(BA)x = (AB)Ax = AAx$$
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Tonoloway Formation
Lower contact with the Wills Creek is probably conformable. Upper contact is conformable and undulatory, occurring at the base of the "calico" limestone of the Keyser Formation. wikipedia.org
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Same rank implies same column space True or False? If $X$ and $Y$ are conformable matrices for the product $XY$ and if the rank of $XY$ equals the rank of $X$, then the span of the columns of $XY$ equals the span of t...
**True** : Note that a column of $XY$ is a linear combination of columns of $X$. Hence $$ {\rm col}\ (XY)\subset {\rm col}\ (X)$$ Since rank are same, they are same.
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Does AB = 0 mean that BZA = 0? Simple linear algebra question, but I can't offer a rigorous proof. Given $A$ and $B$ non-square matrices and knowing $ AB = 0$ does this imply that $BZA = 0$, where $Z$ is an arbi...
Try $A=\left(\begin{smallmatrix}0&1\\\0&0\end{smallmatrix}\right)$, $B=\left(\begin{smallmatrix}1&0\\\0&0\end{smallmatrix}\right)$, $Z=I$.
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Forward-looking difference equation: closed-form solution to $\sum_{j=0}^{\infty}A^jB\Lambda^j$? Assume $$Z_t = A\mathbb{E}_tZ_{t+1} + B\nu_t$$ and $$\nu_t = \Lambda \nu_{t-1} + e_t$$ Where $A, B$, and $\Lambda$ are ...
Note that $\Lambda$ can be written as $$\Lambda=diag\\{\lambda_1,\cdots,\lambda_n\\}$$ and therefore $$\Lambda^j=diag\\{\lambda_1^j,\cdots,\lambda_n^j\\}$$ or equivalently $$\Lambda^j=\sum_{i=1}^n{\lambda_i^j e_ie_i^T}$$ where $e_i$ is the $i$-th column of the identity matrix. Then > $$\sum_{j=1}^{\...
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Inverse of $FR(X)R'F'$ Suppose I have three matrices $F$, $R$ and $X$ , where $F$ and $X$ are non-singular and $R$ is full row rank and whose respective dimensions are `N×N`, `N×K` and `K×K`. I have to solve \begin{...
Before answering your question, few comments: a) It is not possible to invert a matrix that has not an equal number of rows and columns; so it is not possible to invert $R$ which not a square matrix. b) Even if the matrix has an equal number of rows and columns it is not always possible to compute t...
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How to show $\text{rank}(p)=\text{trace}(p)$ for every projector $p$ defined on $\mathbb{R}^n$? I know this is an old question and there are several answers for this using eigenvalues and matrix factorization but they...
Consider a rank-factorization $p = cf$ where $f:\Bbb R^n \to \Bbb R^r$ and $c:\Bbb R^r \to \Bbb R^n$, where $r = \operatorname{rank}(p)$, $c$ is injective, and $f$ is surjective. We note that $$ p^2 = p \implies cfcf = cf. $$ That is, we have $c(fc)f = c(\operatorname{id}_{\Bbb R^r})f$. Because $f$ ...
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