(a) (6%) Please continue the argument to derive the result rank(A) = rank(R).
(接下來前段是背景知識介紹,之後才是提問)Insthq
given
, instead of using the elementary row operations (i.e. the Gauss eliminations) to
manipulate the equation, we may also apply the QR factorization to the equation to get QRx = b,
which implies further QT QRx = QTb. Since QTQ = In, it gives Rx = QTb. Thus, according to the
result of (a), when all columns of A are linearly independent, the square matrix R is nonsingular and
so the solution x =
b is obtained.
It seems that we may summarize the above argument as the following statement:
Given
, where all columns of A are assumed linearly independent, then
solution to the equation Ax = 6 can always be computed from x =
, where Q and R are
matrices obtained from the QR factorization of A.
However, the simple example
shows that the summary is incorrect because, according
to the summary, the solution is x =
=0 and obviously it does not
satisfy the original equation.
阿摩線上測驗
登入
where the Fourier transform of f1(t) is given in the following figure (a).
where the Fourier transform of f2(t) is given in the following figure (b).
. Suppose
4 0 and u ≡ 0. Find the
solution of x1 and x2 for the initial conditions x1(0) = 
.
Find the
range of k: such that the solution of x1 and x2 converges to zero for any initial condition.
Find the
range of k such that the solution of at and ta exhibits oscillatory behavior for any nonzero
initial condition.
,
calculate the steady-state response of
.
, where C is the straight line going from the points (2,2,1) to the point (1,-1,2).