Is this matrix exponentiation method correct?

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The matrix exponentiation method presented contains multiple errors. The assumption that the exponential of a sum of matrices equals the product of their exponentials is incorrect unless the matrices commute. Additionally, the claim that scaling a matrix by a scalar can be expressed as the exponential of the matrix raised to that scalar is also flawed. The discussion highlights the need for a correct understanding of matrix exponentiation properties. Overall, the method proposed does not yield valid results due to these misconceptions.
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Is correct my step by step below?

<br /> \begin{aligned}<br /> \exp \left (\begin{bmatrix} a_{11} &amp; a_{12} \\ a_{21} &amp; a_{22} \\ \end{bmatrix} \right ) &amp;= \exp \left ( a_{11} \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 0 \\ \end{bmatrix} + a_{22} \begin{bmatrix} 0 &amp; 0 \\ 0 &amp; 1 \\ \end{bmatrix} + a_{12} \begin{bmatrix} 0 &amp; 1 \\ 0 &amp; 0 \\ \end{bmatrix} + a_{21} \begin{bmatrix} 0 &amp; 0 \\ 1 &amp; 0 \\ \end{bmatrix} \right ) \\<br /> &amp; = \exp\left ( \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 0 \\ \end{bmatrix} \right )^{a_{11}} \exp\left ( \begin{bmatrix} 0 &amp; 0 \\ 0 &amp; 1 \\ \end{bmatrix} \right )^{a_{22}} \exp\left ( \begin{bmatrix} 0 &amp; 1 \\ 0 &amp; 0 \\ \end{bmatrix} \right )^{a_{12}} \exp\left ( \begin{bmatrix} 0 &amp; 0 \\ 1 &amp; 0 \\ \end{bmatrix} \right )^{a_{21}} \\<br /> &amp;= \begin{bmatrix} 1 &amp; 1 \\ 1 &amp; 0 \end{bmatrix}^{a_{11}} \begin{bmatrix} 0 &amp; 1 \\ 1 &amp; 1 \end{bmatrix}^{a_{22}} \begin{bmatrix} 1 &amp; 1 \\ 0 &amp; 1 \end{bmatrix}^{a_{12}} \begin{bmatrix} 1 &amp; 0 \\ 1 &amp; 1 \end{bmatrix}^{a_{21}}<br /> \end{aligned}
 
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Hi Jhenrique! :smile:

Hint:

i] what is exp##\begin{bmatrix} a & 0 \\ 0 & a \\ \end{bmatrix}## ? what is exp##\begin{bmatrix} 0 & a \\ a & 0 \\ \end{bmatrix}## ?

ii] do your final matrices commute? :wink:
 
Jhenrique, I fixed your LaTeX so it doesn't spill across the screen.

The answer to your question is no. You made multiple errors. Your second line erroneously assumes ##\exp(A+B) = \exp(A)\exp(B)## and also erroneously assumes ##\exp(sA)=\exp(A)^s##.
 
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