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This is all we need to get. The steps are: (1) Start with (A + B)x = y (A + B) x = y. (2) Then Ax = y − Bx A x = y − B x, so x =A−1y −A−1Bx x = A − 1 y − A − 1 B x. (3) Multiply x x in step (2) by B B to get. Bx = BA−1y − BA−1Bx B x = B A − 1 y − B A − 1 B x. which is equivalent to.
Suppose that we have two invertible matrices, A and B. Then it holds: (AB)−1 =B−1A−1, and, in general: (∏k=0N Ak)−1 =∏k=0N A−1 N−k. For the sake of simplicity, let's assume ∏N−1 k=0 Ai = A and AN = B. You can easily verify that both A and B are invertible. Now you are looking for a matrix C such that C ⋅ (AB) = I.
If we use matrices we have A A as the input-output matrix, I I as an identity matrix and d d as final demand. In order to find the final input x x we may solve the Leontief Inverse: x = (I − A)−1 ⋅ d x = (I − A) − 1 ⋅ d. So here's my question: Is there a simple rationale behind this inverse?
Trace of an Inverse Matrix. I want to know if there is a way to simplify, or a closed form solution of tr(Σ−1) t r (Σ − 1) where Σ Σ is a symmetric positive definite matrix. Let A A be symmetric positive definite matrix hence ∃ ∃ a diagonal matrix D D whose diagonal entries are nonzero and A = PDP−1 A = P D P − 1 so A−1 = PD ...
By definition matrix C is a (Moore–Penrose) pseudo-inverse of A if. ACA = A. CAC = C. (AC)T = AC. (CA)T = CA. Assume that A has linearly independent columns, so that the matrix ATA is invertible. Then. (1) (2) AT =(ACA)T = AT(AC)T AT =AT(AC)T =ATAC C = (ATA)−1AT .
Lets have invertible matrix A, so you can write following equation (definition of inverse matrix): AA − 1 = I. Lets transpose both sides of equation. (using IT = I , (XY)T = YTXT) (AA − 1)T = IT. (A − 1)TAT = I. From the last equation we can say (based on the definition of inverse matrix) that AT is inverse of (A − 1)T.
My sticking points so far concern the logic of certain steps in matrix algebra: i.e. why do we take the inverse of a matrix? Would be useful (and extremely generous and informative) if someone could present a numerical example. The closest explanation I could find on-line is the 'inverse is useful for decrypting a coded message'.
All the proofs here use algebraic manipulations. But I think it may be more illuminating to think of a symmetric matrix as representing an operator consisting of a rotation, an anisotropic scaling and a rotation back. This is provided by the Spectral theorem, which says that any symmetric matrix is diagonalizable by an orthogonal matrix. With ...
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Here is the proof. Since B is the inverse matrix, then (cA)B = I, c(AB) = I, AB = 1 cI, finally we multiply both sides with A−1 on the left, A−1AB = A−1 1 cI, we get IB = 1 cA−1I = 1 cA−1. Multiplying a matrix A by a constant c is the same as scaling every row of a matrix by c. You can then consider c to be the n × n matrix cId, (so ...