 Solution

For a one-dimensional character, write $\chi (i)=ϵ$ and $\chi (j)=\eta$. Since $\chi (-1)=1$, the relations $i^2=j^2=-1$ require $ϵ,\eta \in \{\pm 1\}$. Also $ij=k$ gives $\chi (k)=ϵ\eta$. The four sign choices therefore supply all four linear characters.

For the remaining two-dimensional character ${\chi }_2$, its value at the identity is two. Column orthogonality between the identity and $-1$ gives $4+2{\chi }_2(-1)=0$, so ${\chi }_2(-1)=-2$. Between the identity and each of the other classes, the four linear-character values sum to zero, forcing the remaining character values to vanish. Thus the character table isHere column orthogonality is the standard finite-group character theorem ${\sum }_{\rho }{\chi }_{\rho }(e{)}^{\ast }{\chi }_{\rho }(g)=0$ for $g\ne e$. The rows also have unit norm and are mutually orthogonal in the character inner product with the displayed class-size weights. This completes the character table of the quaternion group.