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Let $\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n$ column vectors, each with the same $n$ components. So: \begin{equation} \mathbf{v}_i = \left[\begin{array}{c}v_i\\ v_i \\\vdots \\ v_i\end{array}\right] \end{equation} Also let \begin{equation} \mathbf{b} = \left[\begin{array}{c}b\\ b \\\vdots \\ b\end{array}\right] \end{equation} and \begin{equation} \mathbf{C} = D(c_1,c_2,\ldots,c_n) \end{equation} where $D(c_1,c_2,\ldots,c_n)$ denotes the diagonal square matrix with $c_1,c_2,\ldots,c_n$ on the diagonal.

All the $v$, $b$ and $c$ are known to be positive.

How would you solve the equation \begin{equation} (\,(\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n)+\mathbf{C}\,)\,\mathbf{x} = \mathbf{b} \end{equation} Any pointers in the right direction would be appreciated. I'm essentially looking for a formula for each of the $x_i$ (each component of the solution $\mathbf{x}$).

Is the determinant of $(\,(\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n)+\mathbf{C}\,)$ at least positive? That would at least guarantee that a solution always exists.

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If you define $e=[1,1,1,\ldots,1]$ and $u=[v_1,v_2,\ldots,v_n]$, then the coefficient matrix can be written as $$ C + eu^T.$$

Since $C$ is diagonal, it is invertible if all $c_i\neq0$. With the assumption of invertibility, you can apply Sherman-Morrison formula to obtain the inverse of the coefficient matrix, also assuming that $1+uC^{-1}e\neq 0$. The solution of a linear system is fairly straightforward once you know the inverse of a matrix.

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  • $\begingroup$ Ah! I had seen this formula a while back, but had forgotten its name. Thank you! $\endgroup$ – cont-math Sep 18 '12 at 1:24

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