 Solution

For real orthonormal column vectors, $u^{T}u=v^{T}v=1$ and $u^{T}v=v^{T}u=0$. Expand using associativity of matrix multiplication:HenceMore precisely, it is the orthogonal projection onto the span of $u,v$: it fixes both vectors and annihilates every vector orthogonal to both.

The transpose convention presumes real vectors, or the explicitly stated bilinear identities above. For complex vectors orthonormal in the Hermitian inner product, the correct orthogonal projection is $u{u}^{†}+v{v}^{†}$. The printed transpose formula need not work in that setting: $u=(i,0{)}^T$ and $v=(0,1{)}^T$ are Hermitian-orthonormal, but $u{u}^T+v{v}^T=\mathrm{diag}(-1,1)$ has square $I_2$, which is different from itself.