When computing the inverse for some square matrix A in MATLAB, using
Ai = inv(A)
% should be the same as:
Ai = A^-1
MATLAB usually notifies me that this is not the most efficient way of inverting. So what's more efficient? If I have an equation system, using the /,\ operators probably is. But sometimes I need the inverse for other computations.
What's the most efficient way to invert?
I would recommend to use svd (unless you are really absolute sure that your matrix is not ill-conditioned). Then, based on singular values you make your decisions on further actions to take. This may sound like a 'overkill' approach, but in long run it will pay back.
Now if your matrix A is actually invertible, then the pseudo inverse of A coincidence with inv(A), however if you are close to 'singularity' you'll easily make appropriate decision how to proceed to actually make the pseudo inverse. Naturally these decisions will depend on your application.
Added a straightforward example:
> A= randn(3, 2); A= [A A(:, 1)+ A(:, 2)]
A =
-1.520342 -0.239380 -1.759722
0.022604 0.381374 0.403978
0.852420 1.521925 2.374346
> inv(A)
warning: inverse: matrix singular to machine precision, rcond = 0
ans =
Inf Inf Inf
Inf Inf Inf
Inf Inf Inf
> [U, S, V]= svd(A)
U =
-0.59828 -0.79038 0.13178
0.13271 -0.25993 -0.95646
0.79022 -0.55474 0.26040
S =
Diagonal Matrix
3.6555e+000 0 0
0 1.0452e+000 0
0 0 1.4645e-016
V =
0.433921 0.691650 0.577350
0.382026 -0.721611 0.577350
0.815947 -0.029962 -0.577350
> s= diag(S); k= sum(s> 1e-9) % simple thresholding based decision
k = 2
> Ainv= (U(:, 1: k)* diag(1./ s(1: k))* V(:, 1: k)')'
Ainv =
-0.594055 -0.156258 -0.273302
0.483170 0.193333 0.465592
-0.110885 0.037074 0.192290
> A* Ainv
ans =
0.982633 0.126045 -0.034317
0.126045 0.085177 0.249068
-0.034317 0.249068 0.932189
> A* pinv(A)
ans =
0.982633 0.126045 -0.034317
0.126045 0.085177 0.249068
-0.034317 0.249068 0.932189