I tried asking this on stackexchange but was unsuccessful.
On page 150 of section 4.5.3 of Peter Petersen's Riemannian Geometry it is noted that, given an orthonormal basis $X,iX,Y,iY$ for $T_p\mathbb{C}P^2$, the following basis diagonalizes the curvature operator $\mathfrak{R}:\Lambda^2T_p\mathbb{C}P^2 \to \Lambda^2T_p\mathbb{C}P^2 $:
\begin{align*} &X \wedge iX \pm Y \wedge iY\\ & X \wedge Y \pm iX \wedge iY \\ & X \wedge iY \pm Y \wedge iX \end{align*}
with eigenvalues lying in $[0,6]$. I have attempted the calculations using the O'Neill formula but get stuck. For example assume $\mathfrak{R}(X \wedge iX + Y \wedge iY)=c \left(X \wedge iX + Y \wedge iY\right)$. Then:
\begin{align*} g(\mathfrak{R}(X \wedge iX + Y \wedge iY),X \wedge iX + Y \wedge iY)&=cg(X \wedge iX + Y \wedge iY,X \wedge iX + Y \wedge iY)\\ &=2c \end{align*}
and by the definition of $\mathfrak{R}$:
\begin{align*} &g(\mathfrak{R}(X \wedge iX + Y \wedge iY),X \wedge iX + Y \wedge iY) \\ &=R(X,iX,iX,X)+R(Y,iY,iY,Y)+2R(X,iX,iY,Y) \\ &=\sec(X,iX)+\sec(Y,iY)+2R(X,iX,iY,Y) \\ &=8+2R(X,iX,iY,Y) \end{align*}
so:
\begin{align*}2c &= 8+2R(X,iX,iY,Y) \\ c&=4+R(X,iX,iY,Y) \end{align*}
At this point I do not know how to proceed. I know of a formula for expanding $R(X,iX,iY,Y)$ in terms of sectional curvatures but it is quite complicated. Alternatively, going back to the O'Neill formula we have:
\begin{align*}R(X,iX,iY,Y)=&\overline{R}(\overline{X},\overline{iX},\overline{iY},\overline{Y})+\frac14 \overline{g} ([\overline{iX},\overline{Y} ],[\overline{X},\overline{iY}])-\frac14 \overline{g}([\overline{X},\overline{Y}],[\overline{iX},\overline{iY}]) \\ &+ \frac12 \overline{g} ([ \overline{X},\overline{iX}],[\overline{iY},\overline{Y}] )\end{align*}
where $\overline{R}$ denotes the curvature tensor on $S^5$, $\overline{g}$ denotes the metric on $S^5$, and $\overline{V}$ denotes a horizontal lift. Since $X,iX,Y,iY$ are orthonormal and their lifts are also orthonormal I suppose $\overline{R}(\overline{X},\overline{iX},\overline{iY},\overline{Y})=0$ so we are left with
\begin{align*}R(X,iX,iY,Y)=&\frac14 \overline{g} ([\overline{iX},\overline{Y} ],[\overline{X},\overline{iY}])-\frac14 \overline{g}([\overline{X},\overline{Y}],[\overline{iX},\overline{iY}]) + \frac12 \overline{g} ([ \overline{X},\overline{iX}],[\overline{iY},\overline{Y}] )\end{align*}
Have I missed something that should make this easier to work out?
edit: I followed through with the first idea (expanding out $R(X,iX,iY,Y)$ in terms of sectional curvatures) and got that the eigenvalues are $0,0,2,2,2,6$. Given that $\mathbb{C}P(2)$ is Einstein with Einstein constant $6$ and Scalar curvature $12$, I am tempted to believe that the eigenvalues are correct (as their sum does equal $12$).