Find three $2\times 2$ complex matrices $h_1, h_2, h_3$ satisfying $h_ih_j+h_jh_i = 2\delta_{ij}I.$ 
  
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*Find three $2\times 2$ complex matrices $h_1, h_2, h_3$ satisfying    $$h_ih_j+h_jh_i = 2\delta_{ij}I.$$ 
  
*Use induction to find three    matrices of size $2^n\times 2^n$ with    this property
  

Ideas: All the matrices are square roots of $I$, so they are invertible with eigenvalues only $\pm 1$. They must have one of each, or else they commute with everything, and that doesn't work. So they are reflections across some $1d$ subspace along some other $1d$ subspace
For the second part, we can just build block matrices out of matrices of the form like the first part.
 A: You may solve the problem by brute force. Let the three matrices be $X,Y,Z$. Then we have, in particular, $Y^2=I$ and $XY=-YX$. The first equation means $Y=Y^{-1}$ and hence the second equation means $X$ is similar to $-X$. Since $X^2=I$, it follows that $X$ is similar to $X_0=\operatorname{diag}(1,-1)$ and by a change of basis, we may assume that $X=X_0$. Having $X$ fixed, one can infer from the other given conditions that $Y=\pmatrix{0&y\\ \frac1y&0}$ and $Z=\pmatrix{0&\pm iy\\ \frac1{\pm iy}&0}$. Apply the similar transform $M\mapsto\pmatrix{\frac1{\sqrt{y}}&0\\ 0&\sqrt{y}}M\pmatrix{\sqrt{y}&0\\ 0&\frac1{\sqrt{y}}}$ to $X,Y$ and $Z$, we may assume further that $y=1$.
A: So basically, you are asked for matrix representations for an algebra with three generators, lets call them $\mathfrak i$, $\mathfrak j$ and $\mathfrak k$ that are anti-commuting and have square $1$. By abuse of notation, lets use the same letter for the matrix representations of the generators.
Any representation vector space $V$ is decomposed into $V_+=ker(\mathfrak i-1)=im(1+\mathfrak i)$ and $V_-=ker(\mathfrak i+1)=im(1-\mathfrak i)$. 
If $v\in V_-$, then $v=(1-\mathfrak i)w$ for some $w\in V$. Then $jv=(1+\mathfrak i)\mathfrak jw\in V_+$, so that the action of $\mathfrak j$ exchanges $V_+$ and $V_-$.
The action of $\mathfrak k$ excanges the subspaces as well, so that the action of $(\mathfrak{jk})$ stays inside the subspaces. Since $(\mathfrak{jk})^2=\mathfrak{jkjk}=-\mathfrak{jjkk}=-1$, $\mathfrak jk$ acts as a complex unit on $V_+$ and $V_-$. The only compatible way is to have it act as $+i$ on $V_+$ and $-i$ on $V_-$ or vice versa, so that $\mathfrak{jk}=i\mathfrak i$.
