[proofplan]
We argue by contradiction. If $TS^{2n}$ splits as $E \oplus E^\perp$ for some proper subbundle $E$ of rank $k$ with $0 < k < 2n$, then the multiplicativity of the Euler class gives $e(TS^{2n}) = e(E) \smile e(E^\perp)$. Since $e(E) \in H^k(S^{2n}; \mathbb{Z})$ and $e(E^\perp) \in H^{2n-k}(S^{2n}; \mathbb{Z})$, and both groups vanish for $0 < k < 2n$, the product is zero. This contradicts the fact that $e(TS^{2n}) \neq 0$, which follows from the identity $e(TS^{2n})[S^{2n}] = \chi(S^{2n}) = 2$.
[/proofplan]
[step:Verify that $e(TS^{2n}) \neq 0$]
The Euler class of the tangent bundle satisfies $e(TM)[M] = \chi(M)$ for any compact oriented manifold $M$. For $M = S^{2n}$, the [Euler Characteristic from the Diagonal](/theorems/2297) (or direct computation from the [Homology of Spheres](/theorems/1945)) gives
\begin{align*}
\chi(S^{2n}) = 1 + (-1)^{2n} = 1 + 1 = 2.
\end{align*}
Therefore $e(TS^{2n})[S^{2n}] = 2 \neq 0$, which implies $e(TS^{2n}) \neq 0 \in H^{2n}(S^{2n}; \mathbb{Z})$.
[/step]
[step:Assume a splitting exists and derive a contradiction from the cohomology of $S^{2n}$]
Suppose for contradiction that $TS^{2n} \cong E \oplus E^\perp$ for some subbundle $E$ of rank $k$ with $0 < k < 2n$, where $E^\perp$ denotes the orthogonal complement of rank $2n - k$ (which exists because $S^{2n}$ is compact and an inner product on $TS^{2n}$ exists by the [Inner Products on Vector Bundles](/theorems/2277) theorem). The multiplicativity of the Euler class under direct sums gives
\begin{align*}
e(TS^{2n}) = e(E) \smile e(E^\perp).
\end{align*}
The Euler class $e(E)$ lies in $H^k(S^{2n}; \mathbb{Z})$, and $e(E^\perp)$ lies in $H^{2n-k}(S^{2n}; \mathbb{Z})$.
By the [Homology of Spheres](/theorems/1945), the cohomology of $S^{2n}$ is
\begin{align*}
H^j(S^{2n}; \mathbb{Z}) \cong \begin{cases} \mathbb{Z} & j = 0 \text{ or } j = 2n, \\ 0 & \text{otherwise.} \end{cases}
\end{align*}
Since $0 < k < 2n$, both $H^k(S^{2n}; \mathbb{Z}) = 0$ and $H^{2n-k}(S^{2n}; \mathbb{Z}) = 0$. Therefore $e(E) = 0$ and $e(E^\perp) = 0$, giving
\begin{align*}
e(TS^{2n}) = 0 \smile 0 = 0.
\end{align*}
This contradicts $e(TS^{2n}) \neq 0$, so no such splitting can exist.
[guided]
The argument exploits a striking mismatch between what the Euler class of $TS^{2n}$ must be (non-zero, since $\chi(S^{2n}) = 2$) and what a splitting would force it to be (zero, since intermediate cohomology of $S^{2n}$ vanishes).
Why does the Euler class multiply under direct sums? The Thom class of a direct sum $E \oplus F$ is the external product of the individual Thom classes, and pulling back to the base space converts the external product to a cup product of the Euler classes. This is a general property of the [Thom Isomorphism](/theorems/2283) and the Euler class construction.
The key geometric content is the cohomological "gap" in $S^{2n}$: there is no non-trivial cohomology between degrees $0$ and $2n$. This means any subbundle of intermediate rank would have a trivial Euler class, which would force the Euler class of the ambient bundle to vanish. Since $\chi(S^{2n}) = 2 \neq 0$, this is impossible.
Note that this argument does not apply to $S^{2n+1}$, where $\chi(S^{2n+1}) = 0$ by the [Euler Characteristic Vanishes in Odd Dimensions](/theorems/2292) theorem. Indeed, $TS^{2n+1}$ does admit proper subbundles — for instance, odd-dimensional spheres have nowhere-zero vector fields by the [Hairy Ball Theorem](/theorems/2248), giving a rank-$1$ subbundle.
[/guided]
[/step]
