= Solution
\b[Definitions and labels.] A complex <group representation> on a vector space $V$ is a <group homomorphism> $\rho:G\to\operatorname{GL}(V)$, so $\rho(gh)=\rho(g)\rho(h)$ and $\rho(e)=I$. It is an <irreducible representation> if $V\ne0$ and its only invariant subspaces are $0$ and $V$.
Read permutation matrices by their action on column basis vectors. The labels in the original PDF are
$$
I=e,\quad A=(123),\quad B=(132),\quad C=(13),\quad D=(12),\quad E=(23).
$$
In particular, the original PDF gives the last row of $C$ as $(1,0,0)$; the provided TeX transcription's $(0,0,1)$ is incorrect and would not give an invertible permutation matrix.
\b[Multiplication and conjugacy.] Matrix products act on columns with the right factor first. For example $A^2=B$, $AD=C$, and $DA=E$. The complete <multiplication table>, with row factor multiplied by column factor, is
$$
\boxed{\begin{array}{c|rrrrrr}
&I&A&B&C&D&E\\\hline
I&I&A&B&C&D&E\\
A&A&B&I&E&C&D\\
B&B&I&A&D&E&C\\
C&C&D&E&I&A&B\\
D&D&E&C&B&I&A\\
E&E&C&D&A&B&I
\end{array}.}
$$
Conjugating a <permutation> relabels its cycles. Thus the <conjugacy classes> are
$$
\boxed{\{I\},\qquad\{A,B\},\qquad\{C,D,E\},}
$$
of sizes one, two and three. These are respectively the identity, the three-cycles and the transpositions in the <symmetric group on three letters>.
\b[Dimensions.] The <sum of squares of irreducible representation dimensions> equals the <group order>, which here is six. A three-dimensional <irreducible representation> would already contribute nine, so the displayed representation is reducible. There are three inequivalent complex <irreducible representations>, one for each <conjugacy class>, including the <trivial representation> of dimension one. If their dimensions are $1,d_2,d_3$, then $d_2^2+d_3^2=5$, whose positive integer solution is $\{d_2,d_3\}=\{1,2\}$. Hence
$$
\boxed{\text{irreducible dimensions: }1,1,2.}
$$
One can also see reducibility directly: every permutation matrix fixes $(1,1,1)^T$.
\b[Characters.] The <character of a representation> is the <trace> of its representing matrix, constant on each <conjugacy class>. The given <permutation representation> has character equal to the number of fixed basis vectors:
$$
\boxed{\chi_{\rm perm}=(3,0,1)}
$$
in class order $(I;A,B;C,D,E)$. A <trivial representation> has character $(1,1,1)$. For the second one-dimensional character, a transposition has value $\pm1$ because its square is the identity. The two three-cycles are conjugate and inverse, so their common value $a$ obeys both $a^2=1$ and $a^3=1$, forcing $a=1$. <Character orthogonality> with the trivial character then requires $1+2+3b=0$, giving $b=-1$. This determines the <sign representation> without constructing its matrices.
Write the two-dimensional character as $(2,x,y)$. <Character orthogonality> with the first two rows gives
$$
2+2x+3y=0,\qquad2+2x-3y=0,
$$
so $x=-1$, $y=0$. Its squared norm is $(4+2)/6=1$, as required. The complete <character table> is therefore
$$
\boxed{\begin{array}{c|rrr}
\text{class}&I&A,B&C,D,E\\\hline
\text{class size}&1&2&3\\\hline
\chi_{\rm triv}&1&1&1\\
\chi_{\rm sign}&1&1&-1\\
\chi_{\rm std}&2&-1&0
\end{array}.}
$$
The last row belongs to the <standard representation of the symmetric group>.
To decompose the given representation, use the <character inner product>
$$
\langle\chi,\psi\rangle=\frac16\left[\chi(I)\overline{\psi(I)}+2\chi(A)\overline{\psi(A)}+3\chi(C)\overline{\psi(C)}\right].
$$
Its multiplicities are
$$
\langle\chi_{\rm perm},\chi_{\rm triv}\rangle=\frac{3+3}{6}=1,\qquad
\langle\chi_{\rm perm},\chi_{\rm sign}\rangle=\frac{3-3}{6}=0,\qquad
\langle\chi_{\rm perm},\chi_{\rm std}\rangle=\frac6{6}=1.
$$
Thus
$$
\boxed{\rho_{\rm perm}\cong\mathbf1_{\rm triv}\oplus\rho_{\rm std},\qquad3=1+2.}
$$
Concretely the invariant line is spanned by $(1,1,1)^T$ and its invariant complement is $\{(x,y,z):x+y+z=0\}$. Their characters are exactly the first and third rows, agreeing with the decomposition obtained entirely from <character orthogonality>.
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