Determinant of a symmetric matrix

[krindik]

Hi,

Is there a simplification for the determinant of a symmetric matrix? For example, I need to find the roots of $$\det [A(x)]$$
where
$$A(x) = \[ \left( \begin{array}{ccc} f(x) & a_{12}(x) & a_{13}(x) \\ a_{12}(x) & f(x) & a_{23}(x) \\ a_{13}(x) & a_{23}(x) & f(x) \end{array} \right)\]$$

Really appreciate if you could point me in the correct directions. Thanks in advance,

Krindik

 

[tiny-tim]

Hi Krindik!

If we define a vector B = (B₁, B₂, B₃) = (a₂₃, a₃₁, a₁₂),

then the determinant is f(x)³ - B²f(x)

 

[krindik]

Thanks :)

 

[yiorgos]

Hi Krindik!

If we define a vector B = (B₁, B₂, B₃) = (a₂₃, a₃₁, a₁₂),

then the determinant is f(x)³ - B²f(x)

How is this generalized to nxn matrices?

 

[blue_raver22]

.

Hello Krindik,

Thank you for your question. The determinant of a symmetric matrix can indeed be simplified in certain cases. In general, the determinant of a symmetric matrix can be found using the same methods as for any other square matrix. This includes using cofactor expansion, row reduction, or using the properties of determinants (such as the fact that the determinant of a diagonal matrix is equal to the product of its diagonal elements).

In your specific example, the matrix A(x) is symmetric, meaning that a_{ij}(x) = a_{ji}(x) for all i and j. This allows us to simplify the determinant using the property that the determinant of a symmetric matrix is equal to the product of its eigenvalues. In other words, if \lambda_1, \lambda_2, and \lambda_3 are the eigenvalues of A(x), then

\det [A(x)] = \lambda_1 \lambda_2 \lambda_3.

To find the eigenvalues of A(x), you can use techniques such as diagonalization or the characteristic polynomial. Once you have the eigenvalues, you can then find the roots of the determinant by setting \det [A(x)] = 0 and solving for x.

I hope this helps guide you in the right direction. Best of luck with your research!

Sincerely,