If we have a system of linear equations A x = b , then in case A is invertible it easy

Leah Pope

Leah Pope

Answered question

2022-06-15

If we have a system of linear equations A x = b, then in case A is invertible it easy to say that the solution is x = A 1 b. In all other cases, even if A is invertible, the solution if it exist will be in the form x = A + b + ( I A + A ) w, where A + is called pseudoinvers of A, and w is a vector of free parameters. My question here is how to compute A + for any matrix A? what is the way for doing that? It is easy to compute A + if the column are linearly independent (so that m n, A + = ( A T A ) 1 A, and also if the rows are linearly independent (so that m n), A + = A T ( A A T ) 1 , but I dont know how to compute A + if A is sequare non invertible matrix or if the columns or rows are not linearly independents. The above does not works. If anyone have any idea about computing pseudoinverse or if there is easier method for finding the solution x for system of linear equations A x = b, please help me in all cases, and please keep in mind that I want to find the general form for the solution if it exist not just for a particular b.

Answer & Explanation

Hadley Cunningham

Hadley Cunningham

Beginner2022-06-16Added 20 answers

Let A = U Σ V T be an SVD of A, i.e., U and V are orthogonal and Σ = diag ( σ 1 , σ 2 , , σ r , 0 , , 0 ) is real diagonal with σ k > 0 for all k = 1 , , r. Then
A + = V Σ + U T ,
with
Σ + = diag ( σ 1 1 , σ 2 1 , , σ r 1 , 0 , , 0 ) .

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