# Transforming a matrix to a positive-definite matrix?

$[C]$ 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?

• Is $C'$ positive definite or positive semidefinite? Setting eigenvalues to $0$ should only give you semidefinite. Feb 26, 2017 at 17:58
• the article says $C'$ is a positive definite matrix Feb 26, 2017 at 18:06
• I agree with @Michael Burr. Feb 26, 2017 at 18:22

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 $$i\in[1,n]$$:

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

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