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Any matrix can be decomposed as = for some isometries , and diagonal nonnegative real matrix . The pseudoinverse can then be written as A + = V D + U ∗ {\displaystyle A^{+}=VD^{+}U^{*}} , where D + {\displaystyle D^{+}} is the pseudoinverse of D {\displaystyle D} and can be obtained by transposing the matrix and replacing the nonzero values ...
The adjugate of a diagonal matrix is again diagonal. Where all matrices are square, A matrix is diagonal if and only if it is triangular and normal. A matrix is diagonal if and only if it is both upper-and lower-triangular. A diagonal matrix is symmetric. The identity matrix I n and zero matrix are diagonal. A 1×1 matrix is always diagonal.
An anti-diagonal matrix is invertible if and only if the entries on the diagonal from the lower left corner to the upper right corner are nonzero. The inverse of any invertible anti-diagonal matrix is also anti-diagonal, as can be seen from the paragraph above. The determinant of an anti-diagonal matrix has absolute value given by the product ...
Although an explicit inverse is not necessary to estimate the vector of unknowns, it is the easiest way to estimate their accuracy, found in the diagonal of a matrix inverse (the posterior covariance matrix of the vector of unknowns). However, faster algorithms to compute only the diagonal entries of a matrix inverse are known in many cases. [19]
In general, the inverse of a tridiagonal matrix is a semiseparable matrix and vice versa. [11] The inverse of a symmetric tridiagonal matrix can be written as a single-pair matrix (a.k.a. generator-representable semiseparable matrix) of the form [12] [13]
Let A be a square n × n matrix with n linearly independent eigenvectors q i (where i = 1, ..., n).Then A can be factored as = where Q is the square n × n matrix whose i th column is the eigenvector q i of A, and Λ is the diagonal matrix whose diagonal elements are the corresponding eigenvalues, Λ ii = λ i.
The fundamental fact about diagonalizable maps and matrices is expressed by the following: An matrix over a field is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to , which is the case if and only if there exists a basis of consisting of eigenvectors of .
Find the matrix V of eigenvectors of A (each column of V is an eigenvector of A). Find the inverse V −1 of V. Let ′ =. Then A ′ will be a diagonal matrix whose diagonal elements are eigenvalues of A. Replace each diagonal element of A ′ by its (natural) logarithm in order to obtain ′. Then