Can I avoid inverting a matrix?

General Usage
,
1

I have the following code:

  M = s*I-SS.A
  G = SS.C*inv(M)*SS.B + SS.D

It calculates the transfer function of a linear system in state space form.

Can I avoid inverting the matrix M? I heard that this can result in bad accuracy
and should be avoided.

2

I am not sure to what extend it improves the accuracy but generally it is preferred to use an appropriate matrix factorization of M and the use one of the division operators to apply the inverse. See docs of factorize for some more info on what factorization to use. It looks like your matrix might be positive definite so you should use cholesky! to factorize.

M = s*I-SS.A
Mfact = cholesky!(M) # lu! or other suitable other factorization
G = SS.C*(Mfact(M)\SS.B) + SS.D
3

Just use M\SS.B

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4

It’s less a question of accuracy (the loss of accuracy is usually minor), and more a question of efficiency. Inverting a matrix to compute A^-1 * b is simply a lot of pointless extra effort compared to directly solving the system via A \ b (which forms a factorization of A, but does not form the inverse).

For example, with A = randn(1000,1000) and b = randn(1000), inverting the matrix is slower by a factor of nearly 3:

julia> @btime inv($A) * $b;
  14.184 ms (6 allocations: 8.13 MiB)

julia> @btime $A \ $b;
  5.108 ms (4 allocations: 7.64 MiB)

Whereas if you are computing C * A^-1 * B where C and B are also 1000x1000 matrices, the slowdown is proportionately less:

julia> @btime $C * inv($A) * $B;
  31.570 ms (9 allocations: 23.38 MiB)

julia> @btime $C * ($A \ $B);
  22.695 ms (7 allocations: 22.90 MiB)

(It is much worse when A is sparse, because inverting the matrix throws away all the sparsity whereas factorizing it does not.)

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5

ControlSystemsBase.freqresp uses the Hessenberg factorization to efficiently compute C(i\omega I - A)^{-1}B + D for multiple different \omega, is there a reason you can’t just call freqresp(SS, w)? Or if you have a complex s not on the imaginary axis, evalfr(SS, s)

6 Likes