# Thread: "Maximally diagonal" representation for a symmetric matrix

1. ## "Maximally diagonal" representation for a symmetric matrix

In a given representation for a real symmetric positive-definite NxN matrix C, express it in the form

C = S + K,

where S is a diagonal positive-definite matrix of rank N, and K is a semipositive-definite matrix of the _smallest_ possible rank M < N. Do the solutions for S and K exist? If yes, are they unique? If yes, outline an algorithm to find S and K.

2. Originally Posted by pmn1
In a given representation for a real symmetric positive-definite NxN matrix C, express it in the form

C = S + K,

where S is a diagonal positive-definite matrix of rank N, and K is a semipositive-definite matrix of the _smallest_ possible rank M < N. Do the solutions for S and K exist? If yes, are they unique? If yes, outline an algorithm to find S and K.
The answer certainly won't be unique. Take a very simple numerical example. If $C = \begin{bmatrix}5&2\\2&3\end{bmatrix}$ then $\begin{bmatrix}5&2\\2&3\end{bmatrix} = \begin{bmatrix}1&0\\0&2\end{bmatrix} + \begin{bmatrix}4&2\\2&1\end{bmatrix} = \begin{bmatrix}3&0\\0&1\end{bmatrix} + \begin{bmatrix}2&2\\2&2\end{bmatrix}$.

In fact, for any x in the interval 4/3 < x < 5 we can write $\begin{bmatrix}5&2\\2&3\end{bmatrix} = \begin{bmatrix}5-x&0\\0&3-\tfrac4x\end{bmatrix} + \begin{bmatrix}x&2\\2&\tfrac4x\end{bmatrix}$, with the last matrix having rank 1.

In fact, I suspect that the only case where the answer is unique will be when C is a multiple of the identity. In that case, the unique solution is C = C + 0. For any other positive definite matrix C, we can write $C = \lambda I + (C-\lambda I)$, where $\lambda$ is the smallest eigenvalue of C. Then $\lambda I$ is positive definite, and $C-\lambda I$ is positive semi-definite. Of course, it may not have minimal rank. But even if it does, then (as in the numerical example above) I wouldn't expect that to be the unique solution.