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Let $X$ be an orthogonal matrix, i.e. $XX^T=I$

Is there any special properties of the matrix product $XX$?

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The only thing "special" about $X^2$ is that it belongs to the special orthogonal group. Other than that, nothing special can be said about it.

As $X$ is real orthogonal, $X^2$ is also orthogonal and $\det(X^2)=1$, i.e. $Y=X^2$ belongs to the special orthogonal group. Conversely, if $Y\in SO(n)$, then $Y=X^2$ for some real orthogonal matrix $X$.

In fact, as $Y$ is a real normal matrix, it is orthogonally equivalent to its real Jordan form, which is the direct sum of $I_p,\,-I_q$ and some $r$ rotation matrices of size $2$, where $p+q+2r=n$. Since $\det Y=1$, $q$ must be an even number and in turn $-I_q$ is the direct sum of $q/2$ rotation matrices for angle $\pi$. In other words, $Y=Q\left(I_p\oplus R(\theta_1)\oplus\cdots\oplus R(\theta_{(n-p)/2})\right)Q^T$ for some orthogonal matrix $Q$ and some angles $\theta,\ldots,\theta_{(n-p)/2}\in\mathbb{R}$, where $R(\theta)$ denotes the rotation matrix of size $2$ corresponding to the angle $\theta$. Therefore $Y=X^2$ for $X=Q\left(I_p\oplus R\left(\frac{\theta_1}2\right)\oplus\cdots\oplus R\left(\frac{\theta_{(n-p)/2}}2\right)\right)Q^T$.

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As the Orthogonal group is a group unter matrix multiplication the square will be orthogonal too. Its Inverse will be $X^T X^T$. When $X$ is symmetric $X^2$ will be the identiy. Of which properties did you think?

As the matrix will be orthogonal too it does have all properites of orthogonal matrices, and in special all eigenvalues will be 1 in the real case.

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