Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 1

The identity $T^2=T$ gives the decompositionThe first summand lies in the kernel of a linear map $\ker T$, and the second in the image of a linear map $\mathrm{im}T$. If $w$ belongs to their intersection, write $w=Tu$; then $Tw=T^{2}u=Tu=w$, while $Tw=0$. Thus $w=0$, proving the direct sumOn these two invariant subspaces, $T$ is respectively zero and the identity map. Joining a basis of each produces a basis of eigenvectors, with matrixHence every linear projection is diagonalizable, over the original field, including in characteristic two.

Put $n=\dim V>0$ and $r=\mathrm{rank}T$. The direct sum in part (i) givesThe eigenvalues are the members of $\{0,1\}$ actually present: zero occurs exactly when $r<n$, and one exactly when $r>0$. Since $T$ is diagonalizable, its minimal polynomial contains each present eigenvalue once:In particular, a nonzero vector space does not force both eigenvalues to occur. The zero and identity maps are important special cases.

The same invariant subspaces work for $S=\alpha I+\beta T$: it acts by $\alpha$ on $\ker T$ and by $\alpha +\beta$ on $\mathrm{im}T$. Its minimal polynomial is therefore the product of the distinct linear polynomials belonging to the nonzero summands. Explicitly,When $\beta \ne 0$, the two candidate eigenvalues are distinct in any field. This is the minimal polynomial of an affine function of a projection.

Expand the square using $E^2=E$ and $F^2=F$:ThusThis sum of two linear projections criterion holds in every characteristic; it uses an equality of linear maps, not cancellation of factors in the noncommutative associative algebra of linear endomorphisms.

Suppose first that $E+F$ is a linear projection. Part (i) gives $EF=-FE$. Multiplying on the left by $E$ gives $EF=-EFE$, and multiplying on the right by $E$ gives $EFE=-FE=EF$. Combining them yieldsIn a field of characteristic different from two, the nonzero scalar $2$ is invertible. Hence $EF=0$, and $FE=-EF=0$ too. Conversely these two zero products make the expansion in part (i) equal zero. ThereforeThe argument cancels a nonzero scalar, never a possibly singular linear map.

No. On any nonzero vector space over a field of characteristic two, take $E=F=I$. Both are linear projections, andThe correct condition from part (i) in this characteristic is $EF=FE$, because minus and plus coincide. Vanishing of both products remains sufficient, but is no longer necessary.

The proposed equivalence is false. On a nonzero vector space over any field of characteristic different from two, chooseThese linear projections commute and satisfy $EFG=0$. Nevertheless their sum is $2I$, andThus the proposed conditions are not sufficient. For commuting linear projections the actual obstruction is the sum of the pairwise products:The single condition on the triple product does not remove this obstruction.