# Krylov Spaces

• Apr 2nd 2009, 12:34 PM
azdang
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.
• Apr 2nd 2009, 04:07 PM
NonCommAlg
Quote:

Originally Posted by azdang

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.
• Apr 2nd 2009, 06:34 PM
azdang
Quote:

Originally Posted by NonCommAlg
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?
• Apr 2nd 2009, 06:42 PM
azdang
$x=c_1e_1 + c_2e_2 + \cdots + c_{n-1}e_{n-1}.$ 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.