 Solution

The finite-dimensional inner product space has the orthogonal decomposition $U=W\oplus W^{\perp }$. Define $\tau (w+z)=w$ for $w\in W$, $z\in W^{\perp }$. This is a linear map with image $W$, satisfies ${\tau }^2=\tau$, and is self-adjoint by taking inner products between decomposed vectors. Hence it is an orthogonal projection.

Any other orthogonal projection with image $W$ has kernel $W^{\perp }$ by part (a); it is identity on $W$ by idempotence and zero on its kernel. Its values are therefore forced to agree with $\tau$, proving uniqueness.

If ${\tau }_2{\tau }_1=0$, every $w_1\in W_1$ has $w_1={\tau }_1{w}_1$, so ${\tau }_2{w}_1=0$ and $w_1\in W_2^{\perp }$. Conversely, if $W_1\subset W_2^{\perp }$, every image ${\tau }_{1}u$ is killed by ${\tau }_2$. Therefore