# Find s, such that det(A+s*D) = d.

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Feng Cheng Chang 2014년 10월 28일
댓글: Feng Cheng Chang 2014년 10월 31일
Given a non-sigular matrix A, an arbitrary matrix D, and a real or complex number d, I want to see any short-cut to find the numerical number s, such that M = A+s*D, and det(M) = d. Specially, the matrix M is singular if d = 0.
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Feng Cheng Chang 2014년 10월 31일
I would like to know what your simple example is. Any way I still believe the following:
Given a nxn non-singular matrix A, and D = ones(n) in
det(A+D*s) = 0
s is uniquely found to be
s = -1/r(1),
r = eig(D*inv(A))
where r(1) is the only non-zero element in a vector r.
For example,
r = eig(ones(7)*inv(magic(7)))'
=> 0.02439 0 0 0 0 0 0
By the way I found that
eig(ones(n)) => [ n, 0, 0, ......., 0 ].

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### 채택된 답변

SK 2014년 10월 29일
편집: SK 2014년 10월 30일
Since A is non-singular, we can write (warning: below is math, not code)
M = A + s*D
=> M/A = I + sD/A (where D/A stands for D*inv(A))
=> lambda*M/A = lambda*I - D/A (where lambda = -1/s)
Therefore you need to solve the eigenvalue problem:
(lambda*I - D/A)*v = 0
for lambda. You can find lambda using:
lambda = eig(D/A)
setting s = -1./lambda, you have n solutions (with multiplicity) for d = 0, where n is the size of A.
NOTE: Don't try to find eigen vectors (you dont need them here anyway) using Matlab's eig(), since it wont work if eigenvalues are repeated.
Now for non-zero d, we want:
d/det(A) = det(I + s*D/A)
since det(D/A) = product(lambda), we have to choose s so that
product_over_i(1 + s*lambda(i)) = d/det(A)
1. Find vector of eigenvalues, lambda, of D/A
2. Solve prod(1 + s*lambda) = d/det(A)
(1) should not be a problem even for very large matrices.
I'm not sure about (2) unless n is not too large. You could try the roots() or fzero() functions in Matlab. Note that if you use roots(), you will have to compute the coefficients of the polynomial first from the product in (2) and you may lose a lot of precision in computing the coefficients themselves. I think it would be better to use fzero().
Also note that there are n solutions in general, so you'll have to find a way to choose the solution based on your needs and also make an appropriate initial guess.
[edited to correct sign error].
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Feng Cheng Chang 2014년 10월 31일
Thank you, SK, for correcting a very critical typing error.
Also I would like to thank RS for his nice suggestion.

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### 추가 답변(1개)

Roger Stafford 2014년 10월 30일
If d is real and only real roots are being sought for s, you can simply write a function that accepts a scalar value s and computes det(A+s*D)-d, and then call on matlab's 'fzero' to adjust s to seek a zero value.
If d is complex or you are also seeking complex roots, you can instead call on 'fsolve', calling on your function with the separate real and imaginary parts of s and giving the separate real and imaginary parts of det(A+s*D)-d as a result.
Note that in general there will be n solutions to your equation where your matrices are n-by-n in size, since your equation is actually an nth-degree polynomial equation in s. However, it does not look easy to compute the coefficients of such a polynomial, so it would be difficult to use the 'roots' function.

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