I need to solve the following problem:

Suppose a non-sparse, non-singular complex matrix $\mathbf{P}$. If I want to force all rows in $\mathbf{P}$ to present unitary Euclidean norms by multiplying $\mathbf{P}$ with a diagonal matrix ($\mathbf{A}$) from the right side (it has to be on the right), such that:

\begin{align} \mathbf{P} = \left[ \begin{array}{cc} a & b\\ c & d \end{array} \right] \end{align}

\begin{align} \mathbf{A} = \left[ \begin{array}{cc} x_1 & 0\\ 0 & x_2 \end{array} \right] \end{align}

\begin{align} \mathbf{P A} = \left[ \begin{array}{cc} a & b\\ c & d \end{array} \right] \left[ \begin{array}{cc} x_1 & 0\\ 0 & x_2 \end{array} \right] = \left[ \begin{array}{cc} ax_1 & bx_2\\ cx_1 & dx_2 \end{array} \right] \end{align}

then, the following system must be solved:

\begin{cases} \left| a \right|^2 x_1^2 + \left| b \right|^2 x_2^2 = 1\\ \left| c \right|^2 x_1^2 + \left| d \right|^2 x_2^2 = 1\\ \end{cases}

In matrix notation

\begin{align} \left[ \begin{array}{cc} \left| a \right|^2 & \left| b \right|^2\\ \left| c \right|^2 & \left| d \right|^2 \end{array} \right] \left[ \begin{array}{c} x_1^2\\ x_2^2 \end{array} \right] &= \left[ \begin{array}{c} 1\\ 1 \end{array} \right]\\ \mathbf{T} \left[ \begin{array}{c} x_1^2\\ x_2^2 \end{array} \right] &= \left[ \begin{array}{c} 1\\ 1 \end{array} \right] \end{align} where $\mathbf{T} = \left| \mathbf{P} \right|^2$, considering $\left| \cdot \right|$ to be element-wise.

The solution for $x_1^2$ and $x_2^2$ is easy to state:

\begin{align} \left[ \begin{array}{c} x_1^2\\ x_2^2 \end{array} \right] &= \mathbf{T}^{-1} \left[ \begin{array}{c} 1\\ 1 \end{array} \right]\\ &= \left[ \begin{array}{c} \sum \limits_{j=0}^1 t_{0j}\\ \sum \limits_{j=0}^1 t_{1j} \end{array} \right] \end{align} where $t_{ij}$ is the element from $\mathbf{T}^{-1}$ in the $i$-th line and $j$-th column.

However, if any of $x_1^2$ or $x_2^2$ happen to be negative, $x_1$ or $x_2$ will be complex and the unitary Euclidean norm fails.

Am I making any mistake?


One way to look at this would be: $$ |a|^2 x_1^2 + |b|^2 x_2^2 = 1 $$ is an ellipse. So is the other equation. Now, you have a valid solution when these ellipses intersect in a matching way. Since I am lazy, I used Wolfram Alpha to do this. In essence, the ellipses must intersect, and the resulting points' coordinates must provide a valid solution for the original system. That is, you may get a solution for an instance where $(x_1, x_2)$ work for $a = 2$, but not for $a = -2$, for example.

The cases where it would not work for any signs of $a, b, c$ and $d$ would be those where the ellipses don't intersect at all.

If someone can come up with a better intuition, I'd be happy to learn.

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  • $\begingroup$ Thank your for the answer. I would appreciate also if someone can come up with an optimization strategy to find solutions (in case the ellipses don't intersect). $\endgroup$ – igorauad Aug 2 '13 at 12:13
  • $\begingroup$ I don't believe there would exist real solutions if the ellipses don't intersect. $\endgroup$ – kumanna Aug 3 '13 at 0:42

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