Orthogonal matrices, SVD, low rank
... To develop fast, stable methods for matrix computation, it will be crucial that we understand different types of structures that matrices can have. This includes both “basis-free” properties, such as orthogonality, singularity, or self-adjointness; and properties such as the location of zero element ...
... To develop fast, stable methods for matrix computation, it will be crucial that we understand different types of structures that matrices can have. This includes both “basis-free” properties, such as orthogonality, singularity, or self-adjointness; and properties such as the location of zero element ...
Multivariate Analysis (Slides 2)
... Eigenvectors of Covariance Matrices • Fact: An m × m covariance matrix Σ has m orthonormal eigenvectors. • Proof Omitted, but uses Spectral Decomposition (or similar theorem) of a symmetric matrix. • This result will be used when developing principal components analysis. ...
... Eigenvectors of Covariance Matrices • Fact: An m × m covariance matrix Σ has m orthonormal eigenvectors. • Proof Omitted, but uses Spectral Decomposition (or similar theorem) of a symmetric matrix. • This result will be used when developing principal components analysis. ...
Reformulated as: either all Mx = b are solvable, or Mx = 0 has
... Any vector space over R of dimension n is isomorphic to Rn . Any vector space over C of dimension n is isomorphic to Cn . Indeed, let U be a vector space of dimension n over F (where F is R or C). Let u1 , . . . , un be a basis for U . Define T : U ! F n by T (uj ) = ej for j = 1, . . . n and then e ...
... Any vector space over R of dimension n is isomorphic to Rn . Any vector space over C of dimension n is isomorphic to Cn . Indeed, let U be a vector space of dimension n over F (where F is R or C). Let u1 , . . . , un be a basis for U . Define T : U ! F n by T (uj ) = ej for j = 1, . . . n and then e ...
Proofs Homework Set 5
... (a) Show that if A ∼ B (that is, if they are row equivalent), then EA = B for some matrix E which is a product of elementary matrices. Proof. By definition, if A ∼ B, there is some sequence of elementary row operations which, when performed on A, produce B. Further, multiplying on the left by the co ...
... (a) Show that if A ∼ B (that is, if they are row equivalent), then EA = B for some matrix E which is a product of elementary matrices. Proof. By definition, if A ∼ B, there is some sequence of elementary row operations which, when performed on A, produce B. Further, multiplying on the left by the co ...
Solve xT*A*x +b*x+c=0
... xTAy , yTAx, etc are bilinear forms and xTAx, yTAy, etc are quadratic forms. The bilinear forms are cross products of x and y, etc: eg xTAy = x1y1a11 + x1y2a21 + x2y1a12 + x2y2a22 yTAx = x1y1a11 + x1y2a12 + x2y1a21 + x2y2a22; The quadratic forms are: xTAx = a11x12 + (a12+a21)x1x2 + a22x22 (provided ...
... xTAy , yTAx, etc are bilinear forms and xTAx, yTAy, etc are quadratic forms. The bilinear forms are cross products of x and y, etc: eg xTAy = x1y1a11 + x1y2a21 + x2y1a12 + x2y2a22 yTAx = x1y1a11 + x1y2a12 + x2y1a21 + x2y2a22; The quadratic forms are: xTAx = a11x12 + (a12+a21)x1x2 + a22x22 (provided ...
Always attach the data frame
... #The transpose of the p x q matrix A =(aij) is a q x p matrix obtained by #interchanging the rows and columns of A. It is written as A' = (aji) #The command in R to do a transpose is t(A). #Note that (AB)' = B'A' (or in R notation, t(A%*%B) will equal t(B)%*%t(A) #A square matrix A is called symmetr ...
... #The transpose of the p x q matrix A =(aij) is a q x p matrix obtained by #interchanging the rows and columns of A. It is written as A' = (aji) #The command in R to do a transpose is t(A). #Note that (AB)' = B'A' (or in R notation, t(A%*%B) will equal t(B)%*%t(A) #A square matrix A is called symmetr ...
![[Review published in SIAM Review, Vol. 56, Issue 1, pp. 189–191.]](http://s1.studyres.com/store/data/017041202_1-65d6c90f40ee4c0571100dd75b61a7f5-300x300.png)
Matrices
... i. Step through each value of A multiplying it by each value in the column of B, and sum them all up to get the value for C: N ...
... i. Step through each value of A multiplying it by each value in the column of B, and sum them all up to get the value for C: N ...
Semidefinite and Second Order Cone Programming Seminar Fall 2001 Lecture 10
... m1 (αx) = 0, we conclude that m2 (t) is identical to the minimum polynomial of x, that is m1 (t) ≡ m2 (t). Therefore, αd aj (x) = αd−j aj (αx) which means that aj (αx) = αj aj (x). Thus we have proven Lemma 2 The coefficients aj (x) of the minimum polynomials of x are homogeneous polynomials of degr ...
... m1 (αx) = 0, we conclude that m2 (t) is identical to the minimum polynomial of x, that is m1 (t) ≡ m2 (t). Therefore, αd aj (x) = αd−j aj (αx) which means that aj (αx) = αj aj (x). Thus we have proven Lemma 2 The coefficients aj (x) of the minimum polynomials of x are homogeneous polynomials of degr ...
aa9pdf
... there are uncountably many λ ∈ C such that the element a − λ is not a zero divisor but, at the same time, it is not invertible. (Thus, as opposed to the case of finite dimensional algebras, ‘most’ of noninvertible elements of A are not zero-divisors!) 2. Let A be a central simple k-algebra. Prove th ...
... there are uncountably many λ ∈ C such that the element a − λ is not a zero divisor but, at the same time, it is not invertible. (Thus, as opposed to the case of finite dimensional algebras, ‘most’ of noninvertible elements of A are not zero-divisors!) 2. Let A be a central simple k-algebra. Prove th ...
June 2014
... a) Define what it means for a matrix G to be a generator matrix, and a matrix H to be a parity check matrix, for a linear code C over a finite field F. b) Let G, H be full rank matrices over F, where G is k × n and H is (n − k) × n. Prove that there exists a code C with generator matrix G and parity ...
... a) Define what it means for a matrix G to be a generator matrix, and a matrix H to be a parity check matrix, for a linear code C over a finite field F. b) Let G, H be full rank matrices over F, where G is k × n and H is (n − k) × n. Prove that there exists a code C with generator matrix G and parity ...
Jordan normal form
In linear algebra, a Jordan normal form (often called Jordan canonical form)of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called a Jordan matrix, representing the operator with respect to some basis. Such matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal (on the superdiagonal), and with identical diagonal entries to the left and below them. If the vector space is over a field K, then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix lie in K, or equivalently if the characteristic polynomial of the operator splits into linear factors over K. This condition is always satisfied if K is the field of complex numbers. The diagonal entries of the normal form are the eigenvalues of the operator, with the number of times each one occurs being given by its algebraic multiplicity.If the operator is originally given by a square matrix M, then its Jordan normal form is also called the Jordan normal form of M. Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given M is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed; it is conventional to group blocks for the same eigenvalue together, but no ordering is imposed among the eigenvalues, nor among the blocks for a given eigenvalue, although the latter could for instance be ordered by weakly decreasing size.The Jordan–Chevalley decomposition is particularly simple with respect to a basis for which the operator takes its Jordan normal form. The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.The Jordan normal form is named after Camille Jordan.