Why Matlab do not respect the exponential matrix property \exp^{YXY^​{-1}}=Y\ex​p^XY^{-1}?

Given two square matrices X, Y, with Y invertible, it holds that

,

but when coding it seems not.

For example, given the two matrices

>> X = [1 1 1; 1 1 1; 1 1 1];  
>> Y = [1 1 1; 0 1 1; 0 0 1];

Y is invertible as it is full rank.

Defining its inverse as

>> IY = inv(Y)

IY =

     1    -1     0
     0     1    -1
     0     0     1

it is numerically correct.

When checking for the above property a get

>> exp(Y*X*IY)

ans =

   20.0855    1.0000    1.0000
    7.3891    1.0000    1.0000
    2.7183    1.0000    1.0000

>> Y*exp(X)*IY

ans =

    8.1548         0         0
    5.4366         0         0
    2.7183         0         0

Clearly the property does not hold.

What am I missing?

Answer 1

[Paul]

The identity holds for the matrix exponential, which is implemented with expm(), not exp() (which is the element-wise exponential function)

X = [1 1 1; 1 1 1; 1 1 1];  
Y = [1 1 1; 0 1 1; 0 0 1];

% numerically
expm(Y*X/Y) - Y*expm(X)/Y

ans = 3×3

1.0e+-13 * 0.5684 0.0711 -0.1066 0.3375 0.0355 -0.0711 0.1599 0.0178 -0.0266

% symbolically
expm(sym(Y)*sym(X)*inv(sym(Y))) - sym(Y)*expm(sym(X))*inv(sym(Y))

ans = 

[Alberto Cavarzeran]

Yes! Thank you very much Paul

Answer 2

[Walter Roberson]

Let X and Y be scalar matrices.

syms X Y
lhs = exp(Y * X * inv(Y))

lhs = 

rhs = Y * exp(X * inv(Y))

rhs = 

eqn = lhs == rhs

eqn = 

solve(eqn, X)

ans = 

So your proposed equation only holds if X has that particular relationship to Y.

I would suggest to you that your relationship is incorrect.

rhs2 = exp(X * inv(Y))^Y

rhs2 = 

eqn2 = lhs == rhs2

eqn2 = 

assume(X, 'real')
assume(Y, 'real')
simplify(eqn2)

ans = 

symtrue

[Alberto Cavarzeran]

Actually in my identity I have

rhs = Y * exp(X) * inv(Y)

and not

rhs = Y * exp(X * inv(Y))

so in the scalar case lhs = rhs is immediately verified.

[Walter Roberson]

Are you sure it is an identity?

syms X [3 3] real
syms Y [3 3] real
lhs = exp(Y * X * inv(Y))

lhs = 

rhs = Y * exp(X) * inv(Y)

rhs = 

eqn = lhs - rhs

eqn = 

Seqn = simplify(eqn)

Seqn = 

x = [1 1 1; 1 1 1; 1 1 1];  
y = [1 1 1; 0 1 1; 0 0 1];

subs(Seqn, [X,Y], [x, y])

ans =