Bourbaki

Exercise 21, § 8

Algebra, III, 642

Bourbaki marks this one as harder than the rest.

Let KK be an infinite commutative field, m,n,rm, n, r three integers ≥0\geq 0 such that r≤n≤mr \leq n \leq m, Mm,n(K)\mathbf{M}_{m,n}(K) the vector KK-space (of dimension mnmn) consisting of the matrices of type (m,n)(m, n) over KK and VV a vector subspace of Mm,n(K)\mathbf{M}_{m,n}(K) such that rr is the greatest value of the rank of matrices X∈VX \in V. We propose to prove the inequality

(*) dim⁡(V)≤mr.\dim(V) \leq mr.

(a) Show that we can assume that m=nm = n and that the matrix

X0=(Ir000) X_0 = \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}

belongs to V. Deduce that every matrix X∈VX \in V is then of the form

(X11X12X210) \begin{pmatrix} X_{11} & X_{12} \\ X_{21} & 0 \end{pmatrix}

(where X11X_{11} is of order rr) with the condition X21X12=0X_{21}X_{12} = 0 (use the fact that for all ξ∈K\xi \in K, the matrix ξX0+X\xi X_0 + X is of rank ≤r\leq r).

(b) Deduce from (a) that, for two matrices X,YX, Y in V,

X21Y12+Y21X12=0. X_{21}Y_{12} + Y_{21}X_{12} = 0.

(c) For every matrix X∈VX \in V, we write F(X)=(X11X12)∈Mr,n(K);fF(X) = (X_{11}X_{12}) \in M_{r,n}(K); f is a linear mapping and dim⁡(V)=dim⁡(Im⁡(f))+dim⁡(Ker⁡(f))\dim(V) = \dim(\operatorname{Im}(f)) + \dim(\operatorname{Ker}(f)). On the other hand, let uxu_x denote the linear form (Y11Y12)↦Tr⁡(X21Y12)(Y_{11}Y_{12}) \mapsto \operatorname{Tr}(X_{21}Y_{12}) on Mr,n(K)M_{r,n}(K). Show that the linear mapping X↦uXX \mapsto u_X of Ker⁡(f)\operatorname{Ker}(f) into the dual (Mr,n(K))∗(M_{r,n}(K))^* is injective; complete the proof of (*) by noting that the image under X↦uXX \mapsto u_X of Ker⁡(f)\operatorname{Ker}(f) is contained in the orthogonal of Im⁡(f)\operatorname{Im}(f).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.