- #1
twoski
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Homework Statement
Given the tridiagonal matrix T[itex]_{n}[/itex]
2+x 1 0 0
1 2+x 1 0
0 1 2+x 1 etc.
Derive an upper bound on the infinity norm ||T[itex]_{n}^{-1}[/itex]|| and also derive an upper bound on the condition number of the matrix
The Attempt at a Solution
This is not actually homework, but rather a question from a study exam. I have the solution but the steps taken to get to the solution are really hard to follow so i was hoping to get some clarity.
First we note T[itex]_{n}[/itex] = (2+x)(I+B) where B contains all off-diagonal entries in [itex]_{n}[/itex]. This much makes sense to me.
Next we show that the inverse of [itex]_{n}[/itex] = 1 / (2+x) (I+B)[itex]^{-1}[/itex]. Generally speaking, do we put the off diagonal value from B in this equation? ie. If A = (9-x)(I+B) then i would say the inverse is 1/(9-x) (I+B)[itex]^{-1}[/itex], right?
The upper bound for the infinity norm of the inverse is ||T[itex]_{n}^{-1}[/itex]|| ≤ 1/(2+x) * 1/(1-(2/(2+x))) = 1/x, this doesn't really help me learn anything since these notes don't show how it's actually calculated. If anyone could show how i might go about doing this it would be appreciated.
The condition number is bounded by the infinity norm of the matrix multiplied by the infinity norm of its inverse which is less than or equal to (4+x)/x.