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Math Help - Krylov Spaces

  1. #1
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    Krylov Spaces

    Let A be the nxn matrix whose elements in position (i, i+1) are equal to 1, 1\leq i \leq n-1, while all other elements are zero. Let b = e_{n-1}. Show that the solution to Ax=b does not lie in K_j(A,b) for any j\geq 1.

    We've shown in class that if Ax=b with nonsingular matrix A, the solution x always lies in some Krylov space. The A in this example, however, is being used to show that this is not always true when the matrix A is singular. I am so lost on where to begin. Does anyone have any experience with Krylov Subspaces? Thank you so much.
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  2. #2
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    Quote Originally Posted by azdang View Post

    Let A be the nxn matrix whose elements in position (i, i+1) are equal to 1, 1\leq i \leq n-1, while all other elements are zero. Let b = e_{n-1}. Show that the solution to Ax=b does not lie in K_j(A,b) for any j\geq 1.

    We've shown in class that if Ax=b with nonsingular matrix A, the solution x always lies in some Krylov space. The A in this example, however, is being used to show that this is not always true when the matrix A is singular. I am so lost on where to begin. Does anyone have any experience with Krylov Subspaces? Thank you so much.
    see that Ab=e_{n-2}, \ A^2b=Ae_{n-2}=e_{n-3}, \cdots, A^{n-2}b=e_1, \ A^{n-1}b=A^nb= \cdots = 0. now if x \in \text{span} \{b,Ab,A^2b, \cdots, A^{n-2}b \}, then:

    x=c_1e_1 + c_2e_2 + \cdots + c_{n-1}e_{n-1}. but then Ax=\begin{cases} 0 & \text{if} \ \ n=2 \\ c_2e_1 + c_3e_2 + \cdots + c_{n-1}e_{n-2} & \text{if} \ \ n > 2 \end{cases}, and obviously in either case Ax \neq e_{n-1}=b. Q.E.D.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    see that Ab=e_{n-2}, \ A^2b=Ae_{n-2}=e_{n-3}, \cdots, A^{n-2}b=e_1, \ A^{n-1}b=A^nb= \cdots = 0. now if x \in \text{span} \{b,Ab,A^2b, \cdots, A^{n-2}b \}, then:

    x=c_1e_1 + c_2e_2 + \cdots + c_{n-1}e_{n-1}. but then Ax=\begin{cases} 0 & \text{if} \ \ n=2 \\ c_2e_1 + c_3e_2 + \cdots + c_{n-1}e_{n-2} & \text{if} \ \ n > 2 \end{cases}, and obviously in either case Ax \neq e_{n-1}=b. Q.E.D.

    Thank you very much for your response. I'm a little confused at this part:
    \ A^{n-1}b=A^nb= \cdots = 0. Is this because A is nilpotent?
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  4. #4
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    Alright, I've tried to go over your work carefully, but I'm still confused about this last part.

    x=c_1e_1 + c_2e_2 + \cdots + c_{n-1}e_{n-1}. but then Ax=\begin{cases} 0 & \text{if} \ \ n=2 \\ c_2e_1 + c_3e_2 + \cdots + c_{n-1}e_{n-2} & \text{if} \ \ n > 2 \end{cases}, and obviously in either case Ax \neq e_{n-1}=b. Q.E.D.

    I am confused as to why the subscripts on the c's change in the very last part when you say Ax = {...

    Thank you again!
    Last edited by azdang; April 3rd 2009 at 07:40 AM.
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  5. #5
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    I think I got it. Thanks again!!
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