1
$\begingroup$

while reading this article I came across following:

$[C]$ is a symmetric and not positive definite matrix. However a positive definite matrix $[C']$ can be obtained by equation:

$[C'] = [\phi]*[\lambda '] * [\phi]^{T}$

$[\phi]$ is the square matrix consisting of the eigenvectors of $[C]$. $[\lambda ']$ is the diagonal matrix containing the eigenvalues of $[C]$, however all negative eigenvalues are set to zero.

The result is the matrix $[C']$ which is positive definite.

Why does this work?

$\endgroup$
  • 2
    $\begingroup$ Is $C'$ positive definite or positive semidefinite? Setting eigenvalues to $0$ should only give you semidefinite. $\endgroup$ – Michael Burr Feb 26 '17 at 17:58
  • $\begingroup$ the article says $C'$ is a positive definite matrix $\endgroup$ – noobprogrammer Feb 26 '17 at 18:06
  • $\begingroup$ I agree with @Michael Burr. $\endgroup$ – Jean Marie Feb 26 '17 at 18:22
0
$\begingroup$

In the following I will use the standard notations.

Let's say our matrix in question is $A$ and it has the eigenvectors $v_1, v_2,...,v_n$ with the corresponding eigenvalues $\lambda_1,\lambda_2,...,\lambda_n$. Then we can write (as this is the definition of eigenvalues and eigenvectors) for each $i\in[1,n]$:

$$ A\cdot v_i = \lambda_i\cdot v_i $$

(see https://en.wikipedia.org/wiki/Eigenvalue#Overview).

If we take the matrix $V$ where each column is a $v_i$, we can write

$$ A\cdot V = V\cdot diag(\lambda) $$

where $diag(\lambda)$ is the diagonal matrix, where all the $\lambda_i$ are on the diagonal.

$V$ is an orthogonal matrix, meaning that its columns and rows are orthogonal to each other, so (with $I$ being the identiy matrix)

\begin{align} V^{T}\cdot V &= I\\ V^{T} &= V^{-1} \end{align}

So we can write

\begin{align} A\cdot V &= V\cdot diag(\lambda)\\ A &= V\cdot diag(\lambda)\cdot V^{-1} = V\cdot diag(\lambda)\cdot V^{T} \end{align}

See also at https://en.wikipedia.org/wiki/Symmetric_matrix#Decomposition the point that a complex symmetric matrix can be diagonalized by unitary congruence.

Here the authors modified $diag(\lambda)$ and rebuild the whole thing to $A'$.

An alternative is given by http://animalbiosciences.uoguelph.ca/~lrs/ELARES/PDforce.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.