推 adu :請問第三題的第一小題，為何rank(q)一定會為1 01/31 05:30
→ adu :而不是任一小於等於n的數呢？ 01/31 05:31
→ armopen :The space spanned by one nonzero elements isdim=1 01/31 12:37
1.A,B是Matrix, A 為 m*n, B為n*k項
證明 Rank(AB) <= min(Rank(A), Rank(B))
proof: By CS(A) and RS(A), we mean the column space and the row space
of a m by n matrix A over a field F.
It is not hard to see that CS(AB), RS(AB) are subsets
of CS(A) and RS(B), respectively. By theorem, we have
rank(M) = dim CS(M) = dim RS(M) for any m by n matrices M.
Hence,
rank(AB) = dim RS(AB) <= dim RS(B) = rank(B)
and
rank(AB) = dim CS(AB) <= dim CS(A) = rank(A).
This gives rank(AB) <= min{rank(A), rank(B)}.
2.計算det, n>=2
x 1 ... 1
1 x 1 ..1
A = . . . . .
. . 1 x 1
1 ... 1 x^2 (畫的不清楚@@)
A的對角線，除了最後一項是x^2 其他都是x
非對角線的都為1
(Hint: consider the case for smaller n, you may find its recursive relation.)
3.q and p are Hom(V,K), q,p/=0, V is n-dim vector space.
證明1. dim(Ker(q))=dim(Ker(p))=n-1
2. q,p非線性獨立<=>Ker(q)=Ker(p)
For 1, since q and p are non-trivial linear functionals on V, there are
non-zero vectors u, v in V such that q(u), p(v) are non-zero elements in K.
Therefore, by rank theorem, dim(Ker(q)) = n - rank(q) = n-1. The other case
is similar, so we omit it.
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