Jtdylktuy

Given a matrix, is there a way, by changing its singular values, to transform it into a positive definite matrix?

I do a Singular Value Decomposition. I look at the singular values. If the matrix is not positive definite, then I change them to create a positive definite matrix.

How would I change them, and is there any theory saying how close the resulting matrix would be from the original matrix? No preference for norm.

A comment suggested that to the diagonal matrix with the singular values, I add $mu I$, with $mu>0$. This was my original proceadure, however for the following matrix this doesn't seem to work.

$$A=left(
begin{array}{cccc}
1.24215 & 1.6337 & -0.650968 & 0.635641 \
0.183443 & 0.0432606 & 0.130042 & -0.318818 \
-0.204386 & -0.435348 & -0.966572 & 0.506948 \
-0.321633 & 0.286963 & 0.587576 & 1.37028 \
end{array}
right)$$

I add the identity matrix to the Diagonal matrix with singular values, i.e. I do $U(Diag+I)V^intercal$ and I get:

$$left(
begin{array}{cccc}
1.79872 & 2.36397 & -0.976203 & 0.861808 \
1.00304 & -0.442107 & 0.434236 & -0.331093 \
-0.197866 & -0.914375 & -1.73106 & 0.938273 \
-0.457474 & 0.327607 & 1.05367 & 2.24358 \
end{array}
right)$$

Which is not Pos.Def.

2nd Addition:

I take the original matrix $A$, and compute the diagonal decomposition of $(A+A^T)/2 =P text{Diag}(lambda_i) P^{-1}$. I transform all the non-positive eigenvalues into positive eigenvalues, by $max(lambda_i,mu)$ where $mu$ is a positive number, and $lambda_i$ is the i-th eigenvalue. Then I give as output the new matrix $P text{Diag}(max(lambda_i,mu)) P^{-1}$.

This seems to give a PD matrix.

Among other things, it depends what definition of positive definite you use. One is that all the eigenvalues are positive, another is that $langle A v, v rangle geq 0$ for vectors $v$ over the complex field.

In either case, you need positive eigenvalues. Singular values are nonnegative numbers, the $i$th largest of which is greater than or equal to the absolute value of the $i$th largest eigenvalue. Your singular values can't possibly detect if your matrix has positive eigenvalues, and you can't modify them to get positive eigenvalues.

The second definition requires that your matrix is self-adjiont. In this case, the eigenvalues are very useful. Every self-adjoint matrix is unitarily diagonalizable as $A=UDU^*$, where $D$ is a diagonal matrix with the eigenvalues of $A$ as entries. Simply replace the negative entries of $D$ with zero and you get the best positive (semi)-definite approximation to $A$ (in the two norm).

In the non-self-adjoint case you can also modify your eigenvalues but, its not as clear to me how you would get a best approximation. One great thing about unitaries is that the two norm is invariant under conjugation by them.

Note that no negative definite matrix has a best strictly positive definite approximation since the set of strictly positive definite matrices is not closed.