MHB Integrating a diagonal 2x2 matrix

[Dustinsfl]

\(r = r(x, t)\), \(q = q(x, t)\), \(\sigma_3 =
\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}
\)
I have the equation
\begin{align}
\frac{\partial\mathbf{V}_{-1}^{(D)}}{\partial x} &= \frac{i}{2}\begin{bmatrix}
-(qr)_t & 0\\
0 & (qr)_t
\end{bmatrix}\\
\mathbf{V}_{-1}^{(D)} &= \alpha\sigma_3 + c\mathbb{I}\qquad (*)
\end{align}
where \(\alpha\) is a function of \(x\) and \(t\) related to \(q\) and \(r\) and
\[
\alpha_x + \frac{1}{2}i(qr)_t = 0
\]

How is \((*)\) obtained? I don't see it. I know that \(c\mathbb{I}\) is the matrix constant of integration so I am only focused on how \(\alpha\sigma_3\) comes from integrating the diagonal matrix.

I am trying to figure out the last page of
http://math.arizona.edu/~mcl/Miller/MillerLecture06.pdf

 

[Ackbach]

\(r = r(x, t)\), \(q = q(x, t)\), \(\sigma_3 =
\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}
\)
I have the equation
\begin{align}
\frac{\partial\mathbf{V}_{-1}^{(D)}}{\partial x} &= \frac{i}{2}\begin{bmatrix}
-(qr)_t & 0\\
0 & (qr)_t
\end{bmatrix}\\
\mathbf{V}_{-1}^{(D)} &= \alpha\sigma_3 + c\mathbb{I}\qquad (*)
\end{align}
where \(\alpha\) is a function of \(x\) and \(t\) related to \(q\) and \(r\) and
\[
\alpha_x + \frac{1}{2}i(qr)_t = 0
\]

How is \((*)\) obtained? I don't see it. I know that \(c\mathbb{I}\) is the matrix constant of integration so I am only focused on how \(\alpha\sigma_3\) comes from integrating the diagonal matrix.

I am trying to figure out the last page of
http://math.arizona.edu/~mcl/Miller/MillerLecture06.pdf

Hmm. Interesting. Well, we can write
$$\frac{\partial\mathbf{V}_{-1}^{(D)}}{\partial x}=\frac{i}{2}\begin{bmatrix}
-(qr)_t & 0 \\ 0 & (qr)_t \end{bmatrix}=- \frac{i(qr)_{t}}{2} \begin{bmatrix} 1 &0 \\ 0 &-1 \end{bmatrix}= - \frac{i(qr)_{t}}{2} \sigma_{3}.$$
At the very least, if we take
$$\mathbf{V}_{-1}^{(D)} = \alpha \sigma_3 + c\mathbb{I} \qquad (*),$$
where
$$\alpha_x + \frac{1}{2}i(qr)_t = 0,$$
then if we differentiate $(*)$ w.r.t. $x$, we get that
$$\frac{\partial\mathbf{V}_{-1}^{(D)}}{\partial x}= \alpha_{x} \sigma_{3}=
- \frac{i(qr)_{t}}{2} \sigma_{3},$$
which is what we had before. Given that the author has not specified $\alpha$ explicitly, but only given a DE that it satisfies, it looks to me as though he's merely substituted one DE for another.

 

[Dustinsfl]

Hmm. Interesting. Well, we can write
$$\frac{\partial\mathbf{V}_{-1}^{(D)}}{\partial x}=\frac{i}{2}\begin{bmatrix}
-(qr)_t & 0 \\ 0 & (qr)_t \end{bmatrix}=- \frac{i(qr)_{t}}{2} \begin{bmatrix} 1 &0 \\ 0 &-1 \end{bmatrix}= - \frac{i(qr)_{t}}{2} \sigma_{3}.$$
At the very least, if we take
$$\mathbf{V}_{-1}^{(D)} = \alpha \sigma_3 + c\mathbb{I} \qquad (*),$$
where
$$\alpha_x + \frac{1}{2}i(qr)_t = 0,$$
then if we differentiate $(*)$ w.r.t. $x$, we get that
$$\frac{\partial\mathbf{V}_{-1}^{(D)}}{\partial x}= \alpha_{x} \sigma_{3}=
- \frac{i(qr)_{t}}{2} \sigma_{3},$$
which is what we had before. Given that the author has not specified $\alpha$ explicitly, but only given a DE that it satisfies, it looks to me as though he's merely substituted one DE for another.

Thanks, I actually figured everything out but forgot to mark the thread as solved.