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Let $\alpha:(-1,1)\to O(n,\mathbb R)$ be a smooth map such that $\alpha(0)=I$, the identity matrix. Then which of the following is true?

(a) $\alpha'(0)$ is symmetric.

(b) $\alpha'(0)$ is skew-symmetric.

(c) $\alpha'(0)$ is singular.

(d) $\alpha'(0)$ is non-singular.

$\alpha'(0)=\displaystyle \lim_{t\to 0}\frac{\alpha(t)-\alpha(0)}{t}=\lim_{t\to 0}\frac{A_t-I}{t}$ where $\alpha(t)=A_t\in O(n,\mathbb R).$

What can we say from here about $\alpha'(0)?$

Here $O(n,\mathbb R)$ denotes the set of all $n\times n$ real orthogonal matrices.

Any help is appreciated. Thank you.

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Mini_me
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1 Answers1

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An orthogonal matrix satisfies $A^T \cdot A = I$. If $\alpha(t) = A(t)$ is a smooth path of orthogonal matrices, we have

$$ \alpha(t)^T \cdot \alpha(t) = I. $$

Differentiating the identity using the product rule, we get

$$ (\alpha(t)^T)' \cdot \alpha(t) + \alpha(t)^T \cdot \alpha'(t) = (\alpha'(t))^T \cdot \alpha(t) + \alpha(t)^T \cdot \alpha'(t) = 0. $$

Plugging in $t = 0$, we get

$$ \alpha'(0)^T + \alpha'(0) = 0 $$

so $\alpha'(0)$ is skew-symmetric. It might or might not be symmetric (if it is symmetric, it must be the zero matrix), it might or might not be singular.

levap
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