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I want to ask why

"If $A$ is skew-symmetric ($A^T=-A$) then $e^{At}$ is an orthogonal matrix".

Here is my solution step:

$e^{At}*(e^{At})^T =e^{At}*e^{-At}=e^{At-At}=e^0$

I think the matrix $e^0$ should be a matrix full of $1$, instead of identity matrix.

Jean Marie
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Ruizhe Pang
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  • See https://math.stackexchange.com/questions/4431306/when-a-is-skew-symmetric-q-ea-t-is-orthogonal-i-e-qt-q-i ($A$ is just $Q$ in this case) – FishDrowned Jan 13 '24 at 07:58

3 Answers3

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Your interpretation of $e^{O}$ is wrong. $e^{O}=I+O/1!+O^{2}/2!+\cdots=I$.

geetha290krm
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$e^0$ is the identity matrix since any matrix to the power of $0$ is the identity matrix.

This is because for any matrix $A$ we want $A^{0}A = A$ (just as $\forall a \in \mathbb{R}$, $a^{0} \cdot a = a$) thus $A^{0} = A^{-1}A = I$.

So $e^0 = 0^0 + 0^1 + \frac{0^2}{2} + \frac{0^3}{6} + \ldots = 0^0 = I$.

Note: $0$ is just a matrix full of zeros.

FishDrowned
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If the exp() function is applied element-wise then $e^0$ is indeed $J,\,$ the all-ones matrix.

However, the convention for matrix functions is to apply them matrix-wise, i.e. as a power series in $A.\;$ And it is only in the matrix-wise sense that the exponential of a skew symmetric matrix is orthogonal $$\eqalign{ \def\qiq{\quad\implies\quad} A &= -A^T \\ E &= \exp(A) \qiq E\cdot E^T = I \\ }$$ In the element-wise sense, the exponential of a skew symmetric matrix is a matrix whose off-diagonal terms are reciprocals of one another $$ F = \exp_{\varepsilon}(A) \qiq F\odot F^T = J $$ where $\odot$ denotes the elementwise/Hadamard product.

greg
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