Research Space for Linear Algebra & Discrete Mathematics
5. 設 [I+uu^T]^-1 = I+auu^T=> I + auu^T + uu^T + u(u^T)auu^T = I=> (a + 1 + a(u^T)u)uu^T = 0=> a = -1/(1+(u^T)u)則 (I+2uu^T)[(I+uu^T)^-1]u= (I + 2uu^T)(I + -uu^T/(1+(u^T)u))u= (I + 2uu^T)(u + -(u^T)uu/(1+(u^T)u))= (I + 2uu^T)(1 - (u^T)u/(1+(u^T)u))u= (1 - (u^T)u/(1+(u^T)u))(u + 2(u^T)uu)= (1 - (u^T)u/(1+(u^T)u))(1 + 2(u^T)u)u= (1 - 5/6)(1 + 10)u= (11/6)u6. (設 (v^T)u 不為零)因為 (I+uv^T)u = u+(v^T)u = (1+(v^T)u)u所以 u 為 I+uv^T 相對於 1+(v^T)u 之一 eigenvector令 W = span{v}, per(W) = span{x_1,x_2,...,x_(n-1)}=> (I+uv^T)x_i = x_i+u(v^T)(x_i) = x_i+0 = x_i=> x_i 為 I+uv^T 相對於 1 之 eigenvector, i=1,...,n-1因為 dim({u,x_1,x_2,...,x_(n-1)}) = n所以所有的eigenvalue就是 1+(v^T)u, 還有n-1個1如果 (v^T)u = -1, 則 rank(I+uv^T) = n-1否則 I+uv^T 為可逆, rank(I+uv^T) = n
感謝助教，原來第六題可以這樣求@@
張貼留言
3 則留言:
5.
設 [I+uu^T]^-1 = I+auu^T
=> I + auu^T + uu^T + u(u^T)auu^T = I
=> (a + 1 + a(u^T)u)uu^T = 0
=> a = -1/(1+(u^T)u)
則 (I+2uu^T)[(I+uu^T)^-1]u
= (I + 2uu^T)(I + -uu^T/(1+(u^T)u))u
= (I + 2uu^T)(u + -(u^T)uu/(1+(u^T)u))
= (I + 2uu^T)(1 - (u^T)u/(1+(u^T)u))u
= (1 - (u^T)u/(1+(u^T)u))(u + 2(u^T)uu)
= (1 - (u^T)u/(1+(u^T)u))(1 + 2(u^T)u)u
= (1 - 5/6)(1 + 10)u
= (11/6)u
6. (設 (v^T)u 不為零)
因為 (I+uv^T)u = u+(v^T)u = (1+(v^T)u)u
所以 u 為 I+uv^T 相對於 1+(v^T)u 之一 eigenvector
令 W = span{v}, per(W) = span{x_1,x_2,...,x_(n-1)}
=> (I+uv^T)x_i = x_i+u(v^T)(x_i) = x_i+0 = x_i
=> x_i 為 I+uv^T 相對於 1 之 eigenvector, i=1,...,n-1
因為 dim({u,x_1,x_2,...,x_(n-1)}) = n
所以所有的eigenvalue就是 1+(v^T)u, 還有n-1個1
如果 (v^T)u = -1, 則 rank(I+uv^T) = n-1
否則 I+uv^T 為可逆, rank(I+uv^T) = n
感謝助教，
原來第六題可以這樣求@@
張貼留言