I have the following equation:

$$ \left( \begin{array}{c} \mathbf{I}\\ \mathbf{K} \end{array} \right)\mathbf{x} = \mathbf{y} $$

where $\mathbf{K}$ is a symmetric $2\times 2$ matrix, $\mathbf{I}$ is the identity matrix, $\mathbf{x}$ is a $2\times 1$ vector and $\mathbf{y}$ is a $4\times 1$ vector.

$\mathbf{y}=[y_1,\ldots,y_4]^T$ is a known quantity. Can I use the fact that $\mathbf{x} = [y_1, y_2]^T$ and that $\mathbf{K}$ is symmetric to obtain $\mathbf{K}$ using just matrix operations (i.e. without manually setting up a set of equations)?

  • $\begingroup$ If I understand correctly, you want to find a $2$-by-$2$ symmetric matrix $K$ given $Kx = y$ when $x$ and $y$ are given $2$-by-$1$ vectors. That means you have $2$ equations, but $3$ unknowns (because $K$ is determined by $3$ numbers). Your solution cannot be unique. $\endgroup$ – Tunococ Aug 2 '13 at 21:24

Notice that $K$ is not uniquely defined. For example,

$$K = \mathop{\rm diag}(y_3/y_1, y_4/y_2)$$

is one viable solution whenever $y_1,y_2 \ne 0$.

If $y_1 = y_2 = 0$, then any $K$ is a solution for $y = 0$ and no $K$ is a solution when $y \ne 0$.

It remains to see that happens if $y_1 = 0$ and $y_2 \ne 0$ (or vice versa). In this case, any $K$ of the form

$$K = \frac{1}{y_2} \begin{bmatrix} x & y_3 \\ y_3 & y_4 \end{bmatrix}$$

is a solution ($x \in \mathbb{R}$ is arbitrary). It's similar when $y_2 = 0$ and $y_1 \ne 0$.

  • $\begingroup$ Thanks. I suppose then, that you could solve explicitly if $\mathbf{K}=[K_1,K_2;K_2,K_1]$ or $\mathrm{Det}(\mathbf{K})=0$? $\endgroup$ – user88764 Aug 2 '13 at 21:30
  • $\begingroup$ Yes. The only case worth observing here is $y_1,y_2 \ne 0$ (as others are already like that, given an appropriate choice of $x$). Then you get a system $$K_1y_1 + K_2y_2 = y_3, \quad K_2y_1 + K_1y_2 = y_4,$$which you can easily solve on paper to obtain formulas. Notice that in case if $y_3 = y_4$ and $y_1 = y_2$, you still don't get a unique solution. $\endgroup$ – Vedran Šego Aug 2 '13 at 21:34
  • $\begingroup$ Ok, thanks a lot! $\endgroup$ – user88764 Aug 2 '13 at 21:43

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