414 Algebra
204.First solution: Computed by hand, the second, third, and fourth powers ofJ 4 (λ)are
⎛
⎜
⎜
⎝λ^22 λ 10
0 λ^22 λ 1
00 λ^22 λ
000 λ^2⎞
⎟
⎟
⎠,
⎛
⎜
⎜
⎝
λ^33 λ^23 λ 1
0 λ^33 λ^23 λ
00 λ^33 λ^2
00 0λ^3⎞
⎟
⎟
⎠,
⎛
⎜
⎜
⎝
λ^44 λ^36 λ^24 λ
0 λ^44 λ^36 λ^2
00 λ^44 λ^3
00 0λ^4⎞
⎟
⎟
⎠.
This suggest that in general, theijth entry ofJm(λ)nis(Jm(λ)n)ij=
( i
j−i)
λn+i−j, withthe convention
(k
l)
=0ifl<0. The proof by induction is based on the recursive formula
for binomial coefficients. Indeed, fromJm(λ)n+^1 =Jm(λ)nJm(λ), we obtain
(Jm(λ)n+^1 )ij=λ(Jm(λ)n)ij+(Jm(λ)n)i,j− 1=λ(
n
j−i)
λn+i−j+(
n
j− 1 −i)
λn+i−j+^1 =(
n+ 1
j−i)
λn+^1 +i−j,which proves the claim.
Second solution: DefineSto be then×nmatrix with ones just above the diagonal
and zeros elsewhere (usually called a shift matrix), and note thatSkhas ones above the
diagonal at distancekfrom it, and in particularSn=On. Hence
Jm(λ)n=(λIn+S)n=n∑− 1k= 0(
n
k)
λn−kSk.The conclusion follows.
Remark.The matrixJm(λ)is called a Jordan block. It is part of the Jordan canonical
form of a matrix. Specifically, given a square matrixAthere exists an invertible matrix
Ssuch thatS−^1 ASis a block diagonal matrix whose blocks are matrices of the form
Jmi(λi). The numbersλiare the eigenvalues ofA. As a consequence of this problem,
we obtain a standard method for raising a matrix to thenth power. The idea is to write
the matrix in the Jordan canonical form and then raise the blocks to the power.
205.There is one property of the trace that we need. For ann×nmatrixXwith real
entries, tr(XXt)is the sum of the squares of the entries ofX. This number is nonnegative
and is equal to 0 if and only ifXis the zero matrix. It is noteworthy to mention that
‖X‖=
√
tr(CCt)is a norm known as the Hilbert–Schmidt norm.
We would like to apply the above-mentioned property to the matrixA−Btin order
to show that this matrix is zero. Writing
tr[(A−Bt)(A−Bt)t]=tr[(A−Bt)(At−B)]=tr(AAt+BtB−AB−BtAt)
=tr(AAt+BtB)−tr(AB+BtAt),