= Solution
\b[Representations.] A complex <group representation> is a <group homomorphism> $\rho:G\to\operatorname{GL}(W)$ to the invertible linear maps on a vector space. It is an <irreducible representation> when $W\ne0$ and the only invariant subspaces are $0$ and $W$.
\b[Group table.] The supplied matrices satisfy $A=-I$, $C=-B$, and $A^2=B^2=C^2=I$. Their <multiplication table> is
$$
\boxed{\begin{array}{c|rrrr}
&I&A&B&C\\\hline
I&I&A&B&C\\
A&A&I&C&B\\
B&B&C&I&A\\
C&C&B&A&I
\end{array}.}
$$
The table proves closure and gives identity $I$ and an inverse for every element; associativity is inherited from <matrix multiplication>. The table is symmetric, so the <group> is abelian. It is the <Klein four-group> $V\cong C_2\times C_2$.
\b[Dimensions and conjugacy.] The <sum of squares of irreducible representation dimensions> is $|V|=4$. A two-dimensional irreducible would alone contribute four, leaving no room for the ever-present one-dimensional <trivial representation>. Hence the given two-dimensional representation is reducible. Since the <group> is abelian, every <conjugacy class> is a singleton. There are therefore four inequivalent complex <irreducible representations>, each of dimension one:
$$
\boxed{\text{classes }\{I\},\{A\},\{B\},\{C\};\qquad\text{irreducible dimensions }1,1,1,1.}
$$
Taking the <trace> of the given matrices gives their <character of a representation>:
$$
\boxed{\chi_{\rm given}(I,A,B,C)=(2,-2,0,0).}
$$
\b[Irreducible characters.] In a one-dimensional representation, the values on $A$ and $B$ must each be $\pm1$, since both elements square to $I$. Their values determine the value on $C=AB$. The four choices give the complete <character table>:
$$
\boxed{\begin{array}{c|rrrr}
&I&A&B&C\\\hline
\chi_1&1&1&1&1\\
\chi_2&1&1&-1&-1\\
\chi_3&1&-1&1&-1\\
\chi_4&1&-1&-1&1
\end{array}.}
$$
Each row is a character because it respects the generating relations, and the rows have unit norm and pairwise zero <character inner product>. The multiplicities of these rows in $(2,-2,0,0)$ are respectively $0,0,1,1$, by <character orthogonality>. Thus
$$
\boxed{\rho_{\rm given}\cong\rho_3\oplus\rho_4.}
$$
\b[Diagonal form.] The common invariant lines are spanned by $(1,1)^T$ and $(1,-1)^T$. Use their normalized vectors as columns of
$$
U=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad U^{-1}=U^T=U.
$$
Then simultaneous <diagonalization> gives
$$
\boxed{U^TIU=\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad U^TAU=\begin{pmatrix}-1&0\\0&-1\end{pmatrix},\quad U^TBU=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad U^TCU=\begin{pmatrix}-1&0\\0&1\end{pmatrix}.}
$$
Reading the first and second diagonal entries across the four matrices recovers $\chi_3$ and $\chi_4$, respectively, verifying the decomposition explicitly.
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