Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 19G (Quadratic Mathematics)

For any idempotent linear map $\tau$, each vector decomposes asIf $w\in \ker \tau \cap \mathrm{im}\tau$, write $w=\tau v$; then $0=\tau w={\tau }^{2}v=\tau v=w$. Thus the kernel of a linear map and image of a linear map give a direct sum.

For the self-adjoint map in the question, $k\in \ker \tau$ and $w=\tau v$ satisfy $⟨k,w⟩=⟨k,\tau v⟩=⟨\tau k,v⟩=0$. Hence the direct sum is an orthogonal decomposition:

For a normal operator $\varphi$, the adjoint operator identity and ${\varphi }^{\ast }\varphi =\varphi {\varphi }^{\ast }$ givePositive definiteness implies $\ker \varphi =\ker {\varphi }^{\ast }$. Also $(\mathrm{im}\varphi {)}^{\perp }=\ker {\varphi }^{\ast }$, since orthogonality to every $\varphi v$ means $⟨{\varphi }^{\ast }u,v⟩=0$ for every $v$. In finite dimension this gives $\mathrm{im}\varphi =(\ker \varphi {)}^{\perp }$.

The idempotent linear map decomposes $U=\ker \varphi \oplus \mathrm{im}\varphi$ as in part (a), now orthogonally, and acts as zero on its kernel and identity on its image. Writing $u=k+w$ and $v=k^{\prime }+w^{\prime }$ in these orthogonal summands,Thus $\varphi$ is self-adjoint, and together with idempotence this proves $\varphi$ is an orthogonal projection.

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

Take any positive definite inner product $⟨,{⟩}_0$ on $U$ and define the inner product adapted to an idempotent linear mapIt is symmetric and bilinear. If its quadratic value at $u$ is zero, positive definiteness of the original inner product forces both $\varphi u=0$ and $(I-\varphi )u=0$, hence $u=0$. Thus it is positive definite.

Using ${\varphi }^2=\varphi$ and $(I-\varphi )\varphi =0$ givesThe map is therefore self-adjoint for this new inner product and remains idempotent, so it is an orthogonal projection for the explicitly constructed inner product. Geometrically the construction makes its complementary kernel and image orthogonal.