Find a diagonal matrix D such that the tridiagonal matrix T

[crocomut]

Homework Statement

Homework Equations

For a symmetric matrix B=B' where ' is the transpose.

The Attempt at a Solution

Since we know that for a symmetric matrix,
B = B'

I attempted to substitude that in and tried to solve for D.

DTD^-1 = (D^-1)'T'D
DT = D^-1T'DD
D = D^-1T'DDT^-1

At this point I am stumped since there are D's on both sides, what am I suppose to do?

Thanks a lot for your help.

Croco.

 

[HallsofIvy]

Seems to me you could just "write it out". Let x[subn[/sub] be the nth diagonal entry in D. Then 1/xⁿ is the corresponding entry in D^-1. Now, actually write out the entries for the product DTD^-1. What must be true in order for that to be symmetric? If you are not sure how to do that, try it with 2 by 2 and 3 by 3 matrices first.

Notice, by the way, that saying "b_ic_i+1> 0" is exactly the same as saying "b_i/c_i+1> 0" or even "c_i+1/b_i> 0" because the only thing that is relevant is the sign: all three just say that neither of b_i nor c_i+1 is 0 and they have the same sign.

 

[crocomut]

Hi HallsofIvy,

Thanks so much for your answer, I actually did it last night. Maybe you can have a look at my result and confirm:

In the end I got that the diagonal entries of D are:

d_i+1 = d_i$$\sqrt{b_i/c_{i+1}}$$

The square root is why b_ic_i+1 > 0


Croco