442 Algebra
=‖(U−In)(V−In)y−(V−In)(U−In)y‖
≤‖(U−In)(V−In)y‖+‖(V−In)(U−In)y‖.The claim follows if we prove that‖(U−In)(V−In)y‖and‖(V−In)(U−In)y‖are
both less than^14 , and because of symmetry, it suffices to check this for just one of them.
If(V−In)y=0, then‖(U−In)(V−In)y‖= 0 <^14. Otherwise, using the properties
of vector length, we proceed as follows:
‖(U−In)(V−In)y‖=∥
∥∥
∥(U−In)‖(V−In)y‖(V−In)y
‖(V−In)y‖∥
∥∥
∥
=‖(V−In)y‖×‖(U−In)z‖,wherezis the length one vector‖(V−^1 In)y‖(V−In)y. By the hypothesis, each factor in
the product is less than^12. This proves the claim and completes the solution.
257.The equality for generalkfollows from the casek=n, when it is the well-known
det(AB)=det(BA). Apply this to
(
In A
OnIn
)(
λIn−ABOn
B In)
=
(
λInA
B In)
=
(
InOn
B In)(
In A
OnλIn−BA)
to obtain
det(λIn−AB)=det(λIn−BA).The coefficient ofλkin the left-hand side isφk(AB), while the coefficient ofλkin the
right-hand side isφk(BA), and they must be equal.
Remark.From the many applications of the functionsφk(A), we mention the construction
of Chern classes in differential geometry.
258.From
I 2 =(uI 2 +vA)(u′I 2 +v′A)=uu′I 2 +(uv′+vu′)A+vv′A^2 ,using the Cayley–Hamilton Theorem, we obtain
I 2 =(uu′−vv′detA)I 2 +(uv′+vu′+vv′trA)A.Thusu′andv′should satisfy the linear system
uu′−(vdetA)v′= 1 ,
vu′+(u+vtrA)v′= 0.