MATLAB: How to create a symmetric Toeplitz matrix with bounds on eigenvalues

eigenvalueshomeworksymmetrictoeplits

Is there a way to creat a symmetic Toeplitz matrix of size 400 X 400 with real entries and its largest eigenvalue is 5 and the smallest eigenvalue is -5?

Best Answer

R=fft(eye(400))/sqrt(400);
e=zeros(1,400);
e(2)=-10; e(end-1)=+10;
e=ifftshift(e);
T=(R'*diag(e)*R);
T=real((T+T.')/2);

>> min(eig(T))
ans =
   -5.0000
>> max(eig(T))
ans =
    5.0000
>> norm(T-T.')
ans =
     0

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MATLAB: Does the function chol correctly indicates that a Matrix is positive definite

For the 10 millionth time, det is a bad way to check is a matrix is singular. (Not your fault for asking though.) In fact, it is NEVER a good way to do so, except in homework, where teachers love to tell you to use det, but not explain why det is bad so often. Sadly, that propagates, because their students will do the same thing. And if nobody ever tells you, then your potential students would learn the same fallacies.

Next, giving us a matrix of elements that are only written in 9 decimal digits of precision for a matrix of doubles tells us nothing.

You are confusing the use of chol to test for a positive definite matrix, with testing for singularity. As well, the matrix you have shown is not even symmetric. That tells me it will usually have complex eigenvalues.

I am a bit surprised that chol does not test to see if the metrix is symmetric. But it looks as if chol only uses the upper triangle of the input array. We can verify that by the following test:

[R,P] = chol(A);
A - R'*R
ans =
 1.1102e-16   5.5511e-17   1.1102e-16   1.1102e-16
   0.063614            0   6.9389e-18   6.9389e-18
    0.14454       0.7565            0            0
   -0.51385      0.92378     0.084864            0

So chol is indeed only looking at the upper triangle. (I think that chol used to test for symmetry too. So that may have changed in some recent release.)

If you want to test to see if A is SINGULAR, NEVER use det. Chol is completely inappropriate too.

rank(A)
ans =
     3

A non-singular 4x4 matrix would have a rank of 4. So A is rank deficient.

Understanding the SVD would be even better for some people, but rank is perfectly adequate in this respect.

MATLAB: Matlab code to transform linear systems to strictly diagonally dominant matrix

I'm sorry, but this is not possible to do. In fact, you can prove the impossibility, by a simple counterexample.

Let the matrix A be ones(3,3). This matrix is singular, worse, it has a rank of 1. No linear transformation that you can apply to A is sufficient to make A STRICTLY diagonally dominant, since a strictly diagonally dominant matrix would be NON-SINGULAR.

Sorry, but mathematics (in this case, linear algebra) is simple, and unrelenting. It does not allow you to do just anything.

In your case, we have a non-singular matrix A, so there are many (all trivial) solutions.

A=[6 5 7; 4 3 5; 2 3 4];
rank(A)
ans =
     3

As I said, there are trivial solutions, and many of them, IF A is non-singular. For example, all you need do is multiply both sides by the inverse of A. Thus, if we start with the linear system:

A*x = b

then as long as A is not singular (and square) then we can write it as

B*x = eye(size(A))*x = inv(A)*b

So, your new coefficient matrix is

B = eye(size(A))

and the transformed right hand side (b) is now

inv(A)*b

That satisfies the requirements of this question, as long as A is non-singular. Since you cannot get any more diagonally dominant than an identity matrix, this is the answer, and no answer can be better.

Another simple answer is to use pinv, which for your purposes is again only valid if A is non-singular, if the created matrix B is to be strictly diagonally dominant. As I said, no linear transformation of A will suffice for the bad case. So, again we have

B = eye(size(A));
bhat = pinv(A)*b;

so the new linear system is essentially

eye(size(A))*x = pinv(A)*b

As it turns out, for a non-singular matrix A, pinv(A) is mathematically equivalent to inv(A). pinv is arguably a little better behaved for some nearly singular matrices, but if the matrix is nearly singular, you are in deep trouble anyway with any approach.

Alternatively, one can use a QR factorization of A to do the transformation. It will take slightly more effort to do (but really only a few extra characters.) If you want the system to be stable, then a pivoted QR would be a better choice than a simple QR, but a pivoted QR will implicitly re-sequence the unknown vector x, in order to get the form you seem to desire.

If you do not re-sequence the vector x, then the solution may be less numerically stable for SOME problems. You can resolve the re-sequencing simply enough. For example...

[Q,R,E] = qr(A);

Here, E is a permutation matrix. The necessary linear transformation is now

B = eye(size(A));
bhat = E'*(R\Q')*b;

Or you could use an LU decomposition. Again, a pivoted LU is a better choice for stability.

You could also use an eigenvalue decomposition in a similar way, as long as A has a complete set of eigenvalues and eigenvectors.

Do I need to go on writing what could be a basic linear algebra text?

Best Answer

Related Solutions

MATLAB: Does the function chol correctly indicates that a Matrix is positive definite

MATLAB: Matlab code to transform linear systems to strictly diagonally dominant matrix

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