Compute the logarithm of a matrix

Question

There have already been challenges about computing the exponential of a matrix , as well as computing the natural logarithm of a number. This challenge is about finding the (natural) logarithm of matrix.

You task is to write a program of function that takes an invertible \$n \times n\$ matrix \$A\$ as input and returns the matrix logarithm of that matrix. The matrix logarithm of a matrix \$ A\$ is defined (similar to the real logarithm) to be a matrix \$L\$ with \$ exp(L) = A\$.

Like the complex logarithm the matrix logarithm is not unique, you can choose to return any of the possible results for a given matrix.

Examples (rounded to five significant digits):

log( [[ 1,0],[0, 1]] ) = [[0,0], [0,0]]
log( [[ 1,2],[3, 4]] ) = [[-0.3504 + 2.3911i, 0.9294 - 1.0938i], [1.3940 - 1.6406i, 1.04359 + 0.75047i]]
log( [[-1,0],[0,-1]] ) = [[0,pi],[-pi,0]] // exact
log( [[-1,0],[0,-1]] ) = [[0,-pi],[pi,0]] // also exact
log( [[-1,0],[0,-1]] ) = [[pi*i,0],[0,pi*i]] // also exact

log( [[-1,0,0],[0,1,0],[0,0,2]] ) = [[3.1416i, 0, 0], [0, 0, 0], [0, 0, 0.69315]]
log( [[1,2,3],[4,5,4],[3,2,1]] ) = [[0.6032 + 1.5708i, 0.71969, -0.0900 - 1.5708i],[1.4394, 0.87307, 1.4394],[-0.0900 - 1.5708i, 0.71969, 0.6032 + 1.5708i]]

If you want to try out more examples use the function digits 5 matrix logarithm followed by a matrix in Wolfram Alpha

Rules:

You can Input/Output matrices as nested lists
You can Input/Output complex numbers as pairs of real numbers
You can assume the logarithm of the input matrix exists
Your result should be accurate up to at least 5 significant (decimal) digits
You only have to handle matrices of sizes \$2\times2\$ and \$3\times3\$
You program may return different results when called multiple times on the same input as long as all of them are correct
Please add builtin answers (including libraries) to the community wiki instead of posting them separately
This is code-golf the shortest solution (per language) wins

SuperStormer · Accepted Answer · 2023-07-30 13:06:22Z

3

Python + scipy, 30 bytes

from scipy.linalg import*
logm

answered Jul 30, 2023 at 13:06

SuperStormer

2,01311 silver badges21 bronze badges

1

\$\begingroup\$ Should this count as builtin? \$\endgroup\$

l4m2
– l4m2

2023-07-30 14:28:24 +00:00
Commented Jul 30, 2023 at 14:28
1

\$\begingroup\$ @l4m2 Yes. But it was posted before I added the rule against built-in solutions. \$\endgroup\$

bsoelch
– bsoelch

2023-07-30 14:34:22 +00:00
Commented Jul 30, 2023 at 14:34

Add a comment |

Dominic van Essen · Accepted Answer · 2023-07-31 14:33:51Z

R, 61 bytes

function(m,e=eigen(m))e$ve%*%diag(log(e$va+0i))%*%solve(e$ve)

Try it online!

Uses this method from Wikipedia for finding the logarithm of a diagonalizable (and so invertible) matrix.

R, 105 bytes*

(*)Only for matrices M for which norm(M-I)<1

function(m,i=diag(nrow(m)),R=Reduce)R(`+`,lapply(1:1e3,function(n)-(-1)^n*R(`%*%`,rep(list(m-i),n),i)/n))

Try it online!

I didn't initially spot the 'invertible' in the challenge specification, so tried to use this other method from Wikipedia which works for non-diagonalizable matrices M, but with the restriction that norm(M-I) must be <1.
In many cases, this can be extended to matrices with larger-valued elements, using L == log( M %*% X) == log( M ) + log( X ): in these cases, we divide the matrix M by a sufficiently-large power-of-2 l, and add l times the log of 2x the identity matrix to the result.
Nevertheless, this is not fully-general, and fails for matrices M that cannot be scaled to satisfy norm(M-I)<1 (for instance, the matrix [[1,2],[3,4]]).

R, 158 bytes*

(*)Only for matrices M for which norm(M*l-I)<1 for some value l

function(m,i=diag(nrow(m)),l=log2(norm(m)*2)%/%1,f=function(m,R=Reduce)R(`+`,lapply(1:1e3,function(n)-(-1)^n*R(`%*%`,rep(list(m-i),n),i)/n)))f(m/2^l)+l*f(i*2)

Try it online!