d-

SIAM J. MUMER. ANAL. Vol. 23, No. 5, October 1986 (C) 1986 Society for Industrial andApplied Mathemaics 008 STABLE ATrRACTING SETS IN DYNAMICALSYSTEMS AND IN THEIR ONE-STEP DISCRETIZATIONS*

FINDING RANGE, INTERQUARTILE RANGE, …

FINDING STANDARD DEVIATION, RANGE AND INTERQUARTILE RANGE. Step 1. When you are done entering data, press STAT and go to CALC. Step 2. Highlight the first option, 1-Var Stats, and press ENTER.

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UNIVERSITY OF NEW MEXICO

take University of New Mexico courses while simultaneously attending high school or during the summer between junior and senior years. This is a part-time status and should not be confused with Early Admission. Please refer to page 22 of the

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DETERMINANTS BY ROW AND COLUMN EXPANSION

2 TERRY A. LORING 2.1. Expand on the second row (twice). 0 1 4 32 3 1 4 40 6 0 2 32 3 1 6 60 = −3 X0 1 4 32 X3 X1 X4 X4X0 X6 0 2 32

Multivariate Regression (Chapter 10)

Multivariate regression As in the univariate, multiple regression case, you can whether subsets of the x variables have coe cients of 0. In this case, there is a matrix in the null hypothesis, H 0: B d = 0. The E and H matrices are given by E = Y0Y Bb0X0Y H = bB0X0Y Bb0 rX 0 rY And the test statistics are given as before.

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Chapter 4. Gauss-Markov Model - University of New Mexico

Chapter 4, page 4 Result 4.1. b b The BLUE of estimable is uncorrelated with all unbiased estimators of--TT^ zero. Proof: a y First, characterize unbiased estimators of zero as c such that T E( c ) c 0 for all , œ œay aXb bTT or c 0 and , or ( ). Computing the covariance between and weœœ−Xa 0 a X b ayTT TTa -^ have

DEGREE SEQUENCE degree sequence - University of New …

DEGREE SEQUENCE 2 Example 0.1. Up to isomorphism, find all simple graphs with degree sequence (1,1,1,1,2,2,4). Solution: The degree 4 vertex must be adjacent to 0, 1 or 2 of the vertices of degree 2, so we

Lebesgue Measure and The Cantor Set - University of New …

1 Overview De ne a measure. De ne when a set has measure zero. Find the measure of [0;1], I and Q. Construct the Cantor set. Find the measure of the Cantor set.

[1] Eigenvectors and Eigenvalues - MIT Mathematics

[4] Computing Eigenvectors [5] Computing Eigenvalues [1] Eigenvectors and Eigenvalues Example from Di erential Equations Consider the system of rst order, linear ODEs. dy 1 dt = 5y 1 + 2y 2 dy 2 dt = 2y 1 + 5y 2 We can write this using the companion matrix form: y0 1 y0 2 = 5 2 2 5 y 1 y 2 : Note that this matrix is symmetric. Using notation ...

  Eigenvalue, Eigenvectors

1 Eigenvalues and Eigenvectors - Calvin University

1 = (1,0,3) and u 2 = (1,1,3). It is a fact that all other eigenvectors associated with λ 2 = −2 are in the span of these two; that is, all others can be written as linear combinations c 1u 1 +c 2u 2 using an appropriate choices of the constants c 1 and c 2. Example: Find the eigenvalues and associated eigenvectors of the matrix A = −1 2 0 ...

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Eigenvalues and Eigenvectors §5.2 Diagonalization

0 3 1 0 0 3 1 A: Show that A is not diagonalizable. Solution: Use Theorem 5.2.2 and show that A does not have 3 linearly independent eigenvectors. I To nd the eigenvalues, we solve det( I A) = 1 1 1 0 + 3 1 0 0 + 3 = ( 1)( +3)2 = 0: So, = 1; 3 are the only eigenvalues of A: Satya Mandal, KU Eigenvalues and Eigenvectors x5.2 Diagonalization

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Eigenvalues and Eigenvectors - MIT Mathematics

P is symmetric, so its eigenvectors .1;1/ and .1; 1/ are perpendicular. The only eigenvalues of a projection matrix are 0 and 1. The eigenvectors for D 0 (which means Px D 0x/ fill up the nullspace. The eigenvectors for D 1 (which means Px D x/ fill up the column space. The nullspace is projected to zero. The column space projects onto itself.

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Eigenvalues and Eigenvectors - MIT Mathematics

P is symmetric, so its eigenvectors .1;1/ and .1; 1/ are perpendicular. The only eigenvalues of a projection matrix are 0 and 1. The eigenvectors for D 0 (which means Px D 0x/ fill up the nullspace. The eigenvectors for D 1 (which means Px D x/ fill up the column space. The nullspace is projected to zero. The column space projects onto itself.

  Eigenvalue, Eigenvalues and eigenvectors, Eigenvectors

Eigenvalues and Eigenvectors

If all 3eigenvalues are distinct →-−%≠0 Hence, /1"=0, i.e., the eigenvectors are orthogonal (linearly independent), and consequently the matrix !is diagonalizable. Note that a diagonalizable matrix !does not guarantee 3distinct eigenvalues.

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Eigenvalues, Eigenvectors, and Diagonalization

Find all of the eigenvalues and eigenvectors of A= 1 1 0 1 : The characteristic polynomial is ( 1)2, so we have a single eigenvalue = 1 with algebraic multiplicity 2. The matrix A I= 0 1 0 0 has a one-dimensional null space spanned by the vector (1;0). Thus, the geometric multiplicity of this eigenvalue is 1.

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Eigenvalues and eigenvectors of rotation matrices

λ2 − 1 = 0, (12) which yields the eigenvalues, λ = ±1. The interpretation of this result is immediate. The matrix R(θ) when operating on a vector ~v represents a reflection of that vector through a line of reflection that passes through the origin. In the case of λ = 1 we have R(θ)~v = ~v, which means that ~v is a

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Eigenvalues, eigenvectors, and eigenspaces of linear ...

form, so we can read o the three eigenvalues: 1 = 1, 2 = 3, and 3 = 2. (It doesn’t matter the order you name them.) Thus, the spectrum of this matrix is the set f1;2;3g. Let’s nd the 1-eigenspace. We need to solve Ax = 1x. That’s the same as solving (A 1I)x = 0. The matrix A 1Iis 2 4 0 0 0 3 2 0 3 2 1 3 5 which row reduces to 2 4 1 0 1 6 ...

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