Re: [理工] 線代
※ 引述《lxy351 (kklin)》之銘言：
: Find the determinant of the n*n matrix
: ┌ ┐
: │ a+b ab 0 0 … 0 0 0 │
: │ 1 a+b ab 0 … 0 0 0 │
: │ 0 1 a+b ab … 0 0 0 │
: A = │ … … … … … … … … │
: │ 0 0 0 0 … 1 a+b ab │
: │ 0 0 0 0 … 0 1 a+b │
: └ ┘
: Let T and U be linear transformations with standard matrices A and
: (A^T)A, respectively. Let T be onto.
: (a) Show whether U is onto, on-to-one, or invertible.
: (b) What is the dimension of the null space of U? Explain your answer.
: 這二題不知道怎麼下手~ 麻煩各位高手幫忙 謝謝
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1.
let A_n denote the n by n matrix A
and let D(n) = det(A_n)
then we can use the cofactor expansion of the matrix property
to get the recurrence relation of D(n) :
D(n) = (a+b)*D(n-1) - ab*D(n-2) with D(1) = a+b
D(2) = a^2 + ab + b^2
this can be solved by using generating function :
a^(n+1) - b^(n+1)
det(A_n) = D(n) = ┌ ───────── if a≠b
│ a - b
│
└ (n+1)*a^n if a=b
2.
(a)
assume T: W → V , and W = span{ w1, w2, ..., wn}
V = span{ v1, v2, ..., vm}
since T is onto
for any yεV , there exists xεW , s.t. y = T(x)
or [y]_V = A([x]_W)
it means that the collection { [y]_V | yεV }
can be spanned by the column vectors of A
so A is at least a full-row rank's matrix
→ r(A) = m≦n
→ (A^T)A is ┌ invertible if m=n
└ not invertible if m＜n
hence the mapping U :
┌ does not belong to that three be mentioned if ker{T} ≠ {0}
└ is invertible if ker{T} = {0}
(b)
rank{U} = rank{(A^T)A}
= rank{A}
= rank{T}
hence nullity{U} = dim{Im{T}} - rank{U}
= dim{Im{T}} - rank{T}
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有錯煩請指正︿︿
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04/13 11:51, , 1F
04/13 11:51, 1F
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