= Solution
Use $|m,m'\rangle$ as shorthand for the <tensor product> of two total-<isospin>-one states with third components $m,m'$. A total-<isospin>-zero state must have total third component zero, so write $a|1,-1\rangle+b|0,0\rangle+c|-1,1\rangle$. Applying the total <raising operator> gives
$$
I_+\bigl(a|1,-1\rangle+b|0,0\rangle+c|-1,1\rangle\bigr)
=\sqrt2\bigl[(a+b)|1,0\rangle+(b+c)|0,1\rangle\bigr].
$$
An <isospin> singlet must be annihilated by this operator; hence $b=-a$ and $c=a$. Normalization gives the <two-isovector singlet>
$$
\boxed{|I=0,M=0\rangle=\frac{|1,-1\rangle-|0,0\rangle+|-1,1\rangle}{\sqrt3}.}
$$
It is also annihilated by the total <lowering operator>. A raising or lowering action changes the third-component label by one; the unchanged output label in the PDF's preliminary ladder formula is a typographical omission, not the action used here.
Exchange commutes with the total <isospin> operators. The displayed singlet is symmetric. The $I=2$ multiplet has symmetric highest state $|1,1\rangle$; applying the symmetric total <lowering operator> preserves that exchange symmetry. The symmetric square of a three-dimensional space has dimension six, exactly $1+5$, while its antisymmetric square has dimension three. The remaining $I=1$ multiplet is therefore antisymmetric. Thus the <Clebsch-Gordan decomposition> is
$$
\mathbf3\otimes\mathbf3=\mathbf1_S\oplus\mathbf3_A\oplus\mathbf5_S.
$$
<Pions> have zero intrinsic <spin> and form an <isospin> triplet. <Bose-Einstein statistics> requires the complete two-pion state to be symmetric under exchange. Its relative orbital state changes by $(-1)^\ell$, so odd <orbital angular momentum> requires antisymmetric <isospin>: \b[an odd-$\ell$ two-pion state has $I=1$]. This statement treats the charge components as states of the same <isospin> multiplet, rather than ignoring exchange when their charges differ.
For the decay, let $|\chi\rangle=H_I|K^0\rangle$. The stated commutator with $I_3$ gives $I_3|\chi\rangle=0$. More decisively, the commuting <raising operator> gives
$$
I_+^2|\chi\rangle=H_I I_+^2|K^0\rangle=0.
$$
The kaon doublet cannot be raised twice. An $I=2,M=0$ component would survive two raisings, so it is excluded. Meanwhile a spinless <kaon> decaying into two spinless <pions> has $\ell=0$ by conservation of total <angular momentum>, independently of whether parity is conserved by the decay. The exchange condition excludes $I=1$. This proves the <highest-weight weak-isospin selection in kaon decay>: \b[the two-pion final state has $I=0$].
In a consistent pion phase convention, the singlet's charge components have the structure
$$
|0,0\rangle=\frac{|\pi^+\pi^-\rangle+|\pi^-\pi^+\rangle-|\pi^0\pi^0\rangle}{\sqrt3}.
$$
The normalized symmetric charged channel has coefficient $\sqrt{2/3}$ and the neutral channel coefficient $-1/\sqrt3$. Thus, in the <isospin> limit with equal pion masses and matching two-body phase space,
$$
\boxed{\frac{\Gamma(K^0\to\pi^+\pi^-)}{\Gamma(K^0\to\pi^0\pi^0)}=2.}
$$
Equivalently, ordered charged and neutral amplitudes have equal magnitudes, but neutral identical particles contribute the phase-space factor $1/2!$. This is the same counting expressed in another basis, and must not be included a second time after using the normalized charge-channel coefficients. Small pion-mass and electromagnetic effects can modify the ideal ratio.
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