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"In a finite dimensional vector space a identity operator (which is positive) can have infinitely many square roots". I see only $2 ^ {C(n,2)}$ when dimension is $n (>1)$ by swapping two basis elements. Where am I wrong ?

Don Larynx
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user103623
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  • In the field of quaternions, the equation $x^2=1$ has a non countable number of solutions. Since quaternions have a representation with real matrices of order $4$ with $1$ represented by the identity matrix, you got many more examples of square roots of the identity matrix. – Taladris Nov 01 '13 at 07:34

3 Answers3

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Let $X: \mathbb{R} \to \mathbb{R}^{2 \times 2}$,

$$X(t) := \begin{bmatrix} 1 & t \\ 0 & -1 \end{bmatrix}.$$

Then

$$(X(t))^2 := \begin{bmatrix} 1 & t \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$

for all $t \in \mathbb{R}$. I leave it to you to generalize this example to higher dimension spaces.

Note that these are merely all triangular matrices $X$ such that $X^2 = I$. There are many other, nontriangular ones, with the same property.

Vedran Šego
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To elaborate on your example, $Y=\pmatrix{0&1\\ 1&0}$ is a square root of $I$. Therefore, $X=PYP^{-1}$ is also a square root of $I$ for every invertible $P$. One can obtain infinitely many such $X$s by using different $P$s. And basically, all square roots of $I$ are obtained in this manner, i.e. they are similar to a diagonal matrix whose diagonal entries (or eigenvalues) are $\pm1$.

user1551
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    For the sake of completeness, one should show that $P \mapsto X$ has an infinite image. For example, if $Y = \pm I$, this is obviously not the case, since $X = \pm I$ for any nonsingular $P$. – Vedran Šego Nov 01 '13 at 14:08
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First of all, a positive operator and an identity operator are much different.

Second of all, the square root of an identity is an identity, and so forth.

Don Larynx
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