申論題

3.(19%)下面的問題共有三個子题,(a)子题要清楚地寫出證明,(b)子题只要簡短扼要地回答 提问即可,(c)子題除提出答案外、還需寫出答案背後的推導。 Let A be any matrix in. Then, since rank(A) = dim(R(A)), the dimension of range of A, and R(A) = R(AA^T), we have the result rank(A) = rank(AA^T). Therefore, when replacing A by its QR factorization, we get rank(A) = rank(QRRT^TQ^TT).

接下來前段是背景知識介紹,之後才是提問)In solving the linear equation Ax=b for a igiven , instead of using the elemnentary 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 Q^TQRx = Q^Tb. Since Q^TQ = I_n, it gives Rx = Q^Tb. 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 = o can always be computed from x = , where Q and R are matrices obtained from the @R factorization of A.

 However, the simple example shows that the summary is incorrect because, according to the summary, the solution is x=  and obviously it does not satisfy the original equation. (b) (5%) What is the error (or are the errors) in the argument right before the summary to make it incorrect?