Mutually Orthogonal Latin Squares and Finite Fields
... from the deck. Can you arrange these cards into a 4 × 4 array, so that in each column and row, no two cards share the same suit or same face value? This question should feel similar to the problem of constructing a Latin square: we have an array, and we want to fill it with symbols that are not repe ...
... from the deck. Can you arrange these cards into a 4 × 4 array, so that in each column and row, no two cards share the same suit or same face value? This question should feel similar to the problem of constructing a Latin square: we have an array, and we want to fill it with symbols that are not repe ...
POLYNOMIALS IN ASYMPTOTICALLY FREE RANDOM MATRICES
... Recent work of Belinschi, Mai and Speicher [2] resulted in a general algorithm to calculate the distribution of any selfadjoint polynomial in free variables. Since many classes of independent random matrices become asymptotically free if the size of the matrices goes to infinity, this algorithm appl ...
... Recent work of Belinschi, Mai and Speicher [2] resulted in a general algorithm to calculate the distribution of any selfadjoint polynomial in free variables. Since many classes of independent random matrices become asymptotically free if the size of the matrices goes to infinity, this algorithm appl ...
How Much Does a Matrix of Rank k Weigh?
... basis of the row space of A, each row of A is a linear combination of the rows of R. That means there is a unique m × k matrix C such that A = C R. Note that the rank of C must also be k. Using the factorization A = C R we can express the set of rank k matrices as the Cartesian product of the set of ...
... basis of the row space of A, each row of A is a linear combination of the rows of R. That means there is a unique m × k matrix C such that A = C R. Note that the rank of C must also be k. Using the factorization A = C R we can express the set of rank k matrices as the Cartesian product of the set of ...
Proposition 2 - University of Bristol
... For shapes other than stars, there do not seem to be any inequalities that are as notable as Propositions 1 and 2. It is possible to write down the inequalities that result from Sylvester’s criterion, but it is generally not easy to rearrange them into a meaningful form. The conditional independence ...
... For shapes other than stars, there do not seem to be any inequalities that are as notable as Propositions 1 and 2. It is possible to write down the inequalities that result from Sylvester’s criterion, but it is generally not easy to rearrange them into a meaningful form. The conditional independence ...
11 Linear dependence and independence
... 3. In the definition, we require that not all of the scalars c1 , . . . , cn are 0. The reason for this is that otherwise, any set of vectors would be linearly dependent. 4. If a set of vectors is linearly dependent, then one of them can be written as a linear combination of the others: (We just do ...
... 3. In the definition, we require that not all of the scalars c1 , . . . , cn are 0. The reason for this is that otherwise, any set of vectors would be linearly dependent. 4. If a set of vectors is linearly dependent, then one of them can be written as a linear combination of the others: (We just do ...
Optimal strategies in the average consensus problem
... Bcde,ghi Cab,jk (notice the shift of the first indices by two places). We will denote this matrix by B ⊙C. This definition applies to any two square matrices whose dimensions are powers of n. In general, we can write (B ⊙ C)p,q = (B ⊗ C)σt (p),q , where σ operates a cyclic permutation by one place t ...
... Bcde,ghi Cab,jk (notice the shift of the first indices by two places). We will denote this matrix by B ⊙C. This definition applies to any two square matrices whose dimensions are powers of n. In general, we can write (B ⊙ C)p,q = (B ⊗ C)σt (p),q , where σ operates a cyclic permutation by one place t ...
TGchapter2USAL
... MATLAB does not compute the inverse matrix; instead it solves the linear system directly). When used with a non-square matrix, the backslash operator solves the appropriate system in the least-squares sense; see help slash for details. Of course, as with the other arithmetic operators, the matrices ...
... MATLAB does not compute the inverse matrix; instead it solves the linear system directly). When used with a non-square matrix, the backslash operator solves the appropriate system in the least-squares sense; see help slash for details. Of course, as with the other arithmetic operators, the matrices ...
Eigenvectors and Eigenvalues
... λ is an eigenvalue of A iff (A – λI)x= 0 has non-trivial solutions A – λI is not invertible IMT all false The set {xεRn: (A – λI)x= 0} is the nullspace of (A – λI)x= 0, A a subspace of Rn The set of all solutions is called the eigenspace of A corresponding to λ ...
... λ is an eigenvalue of A iff (A – λI)x= 0 has non-trivial solutions A – λI is not invertible IMT all false The set {xεRn: (A – λI)x= 0} is the nullspace of (A – λI)x= 0, A a subspace of Rn The set of all solutions is called the eigenspace of A corresponding to λ ...
1 Inner product spaces
... Theorem: For any linear map T : V → V there exists unique linear map T ∗ : V → V such that for all u, v ∈ V , hT (u), vi = hu, T ∗ (v)i. T ∗ is the adjoint operator of T . Properties: (a) (T ∗ )∗ = T , (b) (αT )∗ = αT ∗ , (c) (T1 + T2 )∗ = T1∗ + T2∗ , (d) (T1 T2 )∗ = T2∗ T1∗ , (e) if e1 , . . . , e ...
... Theorem: For any linear map T : V → V there exists unique linear map T ∗ : V → V such that for all u, v ∈ V , hT (u), vi = hu, T ∗ (v)i. T ∗ is the adjoint operator of T . Properties: (a) (T ∗ )∗ = T , (b) (αT )∗ = αT ∗ , (c) (T1 + T2 )∗ = T1∗ + T2∗ , (d) (T1 T2 )∗ = T2∗ T1∗ , (e) if e1 , . . . , e ...
![[1] Eigenvectors and Eigenvalues](http://s1.studyres.com/store/data/014314239_1-836177f5554eb265785889c7c5fb970f-300x300.png)
[1] Eigenvectors and Eigenvalues
... initial conditions y1 (0) and y2 (0). It makes sense to multiply by this parameter because when we have an eigenvector, we actually have an entire line of eigenvectors. And this line of eigenvectors gives us a line of solutions. This is what we’re looking for. Note that this is the general solution ...
... initial conditions y1 (0) and y2 (0). It makes sense to multiply by this parameter because when we have an eigenvector, we actually have an entire line of eigenvectors. And this line of eigenvectors gives us a line of solutions. This is what we’re looking for. Note that this is the general solution ...
Lecture 9: 3.2 Norm of a Vector
... A point P in 2-space now has both (x, y) coordinates and (x ...
... A point P in 2-space now has both (x, y) coordinates and (x ...
Real-Time Endmember Extraction on Multicore Processors
... Exploiting that s n and that only a few columns of M̄T are required, these computations can be performed in approximately 2n(ns − n2 /3) + 2ns(p − 1) ≈ 2sn2 flops. 3) Computation of Determinants: The determinant of a nonsingular matrix V is usually obtained from the factorization P V = LU (where P ...
... Exploiting that s n and that only a few columns of M̄T are required, these computations can be performed in approximately 2n(ns − n2 /3) + 2ns(p − 1) ≈ 2sn2 flops. 3) Computation of Determinants: The determinant of a nonsingular matrix V is usually obtained from the factorization P V = LU (where P ...