# [SOLVED]integrating a diagonal 2x2 matrix

#### dwsmith

##### Well-known member
$$r = r(x, t)$$, $$q = q(x, t)$$, $$\sigma_3 = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$$
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

Last edited:

#### Ackbach

##### Indicium Physicus
Staff member
$$r = r(x, t)$$, $$q = q(x, t)$$, $$\sigma_3 = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$$
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.

#### dwsmith

##### Well-known member
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.