Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 4

If $Av=\lambda v$ for a nonzero eigenvector, then $A^{2}v={\lambda }^{2}v$. Since $A^2=A$, $\lambda (\lambda -1)=0$: every eigenvalue is zero or one. Let $m$ be the number of ones, counted with algebraic multiplicity. The trace is $m$ and the determinant is the product of these eigenvalues. ThereforeEvery output $y=Ax$ is fixed by a second application:The image of a linear map is thus the eigenspace for the eigenvalue one. To establish its dimension without assuming diagonalizability, decompose every input asThe first term lies in the image of a linear map $A$ and the second in its kernel. Their intersection is zero, because a vector in the image is fixed by $A$ whereas a vector in the kernel is killed by $A$. Thus the space is their direct sum, and an adapted basis puts $A$ into the form $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}A$. Comparing traces gives $r=m$:This is an idempotent linear map, and trace of an idempotent equals its rank. If $m=0$, no nonzero output exists. The dimension in the question refers to the image subspace, including its zero vector.

The first three output components equal $x_1,x_2,x_3$, while the fourth is $x_4-{\sum }_{k=1}^4{x}_k=-x_1-x_2-x_3$. HenceIts outputs have component sum zero, and applying $P$ to any such output leaves it fixed. ThusThe first three columns give a basis of the image of a linear map:They are linearly independent because their first three coordinates are the standard basis. Their span is precisely the three-dimensional hyperplane $y_1+y_2+y_3+y_4=0$.

Apply the Gram-Schmidt process. First $q_1=v_1/\sqrt{2}$. Subtracting the $q_1$ projection from $v_2$ gives $(-1,2,0,-1{)}^T/2$, of length $\sqrt{3/2}$. Subtracting both preceding projections from $v_3$ gives $(-1,-1,3,-1{)}^T/3$, of length $2/\sqrt{3}$. Therefore an orthonormal basis isTheir pairwise dot products vanish and each has length one. Although this image admits an orthonormal basis, $P$ itself is an oblique projection: its kernel is the fourth coordinate axis, which is not perpendicular to the image, and $P\ne P^T$.