3

I have a large matrix

\begin{align}
    \begin{bmatrix} 
    \bigg(m\Gamma_1(\kappa_1^1)\omega_1\mathbf{h}_0\mathbf{P}_1^\varepsilon + m\Gamma_2(\kappa_1^1)\sin\bigg(\frac{2\omega_1\bar\zeta x_3}{\varepsilon}\bigg)\omega_1\mathbf{h}_1\mathbf{P}^\varepsilon_1 \\
    + m\Gamma_2(\kappa_1^1)\cos\bigg(\frac{2\omega_1\bar\zeta x_3}{\varepsilon}\bigg)\omega_1\mathbf{h}_2\mathbf{P}_1^\varepsilon\bigg) & \mathbf{0} & \dots & \mathbf{0} \\
    \mathbf{0} & \ddots & \ddots & \vdots \\
    \vdots & \ddots & \ddots & \mathbf{0} \\
    \mathbf{0} & \dots & \mathbf{0} & X
    \end{bmatrix}
\end{align}

where I want to insert

\begin{align}
    &\bigg(m\Gamma_1(\kappa_1^M)\omega_M\mathbf{h}_0\mathbf{P}_1^\varepsilon + m\Gamma_2(\kappa_1^M)\sin\bigg(\frac{2\omega_M\bar\zeta x_3}{\varepsilon}\bigg)\omega_M\mathbf{h}_1\mathbf{P}^\varepsilon_M \\
    &+ m\Gamma_2(\kappa_1^1)\cos\bigg(\frac{2\omega_M\bar\zeta x_3}{\varepsilon}\bigg)\omega_M\mathbf{h}_2\mathbf{P}_M^\varepsilon\bigg)
\end{align}

where the "X" is located. What is the best procedure on how to achieve this>?>?

enter image description here

3
  • align is a top level display use aligned (or perhaps more naturally, matrix) also note it should be \biggl( \biggr) not \bigg Oct 26 '21 at 9:24
  • Is this by chance a block-diagonal matrix (with a total of M blocks)? Please advise
    – Mico
    Oct 26 '21 at 10:07
  • @Mico yes it is Oct 26 '21 at 10:08
7

I would place both the top-left and bottom-right elements of the bmatrix environment on separate lines.

I would also use inline-fraction notation for the trigonometric terms, saving a lot of vertical space.

enter image description here

\documentclass{article}
\usepackage{mathtools}

\begin{document}
\begin{align*}
\mathbf{A} &=
    \begin{bmatrix} \,
      \mathbf{U} & \mathbf{0} & \dots      & \mathbf{0} \\
      \mathbf{0} & \ddots     & \ddots     & \vdots     \\
      \vdots     & \ddots     & \ddots     & \mathbf{0} \\
      \mathbf{0} & \dots      & \mathbf{0} & \mathbf{V}
    \end{bmatrix} \\
\shortintertext{where}
\mathbf{U}
  &= m\Gamma_1(\kappa_1^1)\omega_1\mathbf{h}_0\mathbf{P}_1^\varepsilon 
   + m\Gamma_2(\kappa_1^1)\sin(2\omega_1\bar\zeta x_3/\varepsilon)\omega_1\mathbf{h}_1\mathbf{P}^\varepsilon_1 \\
  &\quad + m\Gamma_2(\kappa_1^1)\cos(2\omega_1\bar\zeta x_3/\varepsilon) \omega_1\mathbf{h}_2\mathbf{P}_1^\varepsilon \\
\shortintertext{and}
\mathbf{V}
  &= m\Gamma_1(\kappa_1^M)\omega_M\mathbf{h}_0\mathbf{P}_1^\varepsilon + m\Gamma_2(\kappa_1^M)\sin(2\omega_M\bar\zeta x_3/\varepsilon)\omega_M\mathbf{h}_1\mathbf{P}^\varepsilon_M \\
  &\quad+ m\Gamma_2(\kappa_1^1)\cos(2\omega_M\bar\zeta x_3/\varepsilon) \omega_M\mathbf{h}_2\mathbf{P}_M^\varepsilon
\end{align*}
\end{document}

Addendum: The OP has indicated in a comment that the big matrix is block-diagonal. Incorporating this information, one may re-state the material much more succinctly as

enter image description here

\documentclass{article}
\usepackage{mathtools} % for '\shortintertext' macro
\DeclareMathOperator{\diag}{diag}
\begin{document}
\begin{align*}
\mathbf{A} &= \diag(\mathbf{U}_1,\dots,\mathbf{U}_M)\,,
\shortintertext{where}
\mathbf{U}_j
  &= \,\phantom{+} m\Gamma_1(\kappa_1^j)\omega_j\mathbf{h}_0\mathbf{P}_j^\varepsilon \\
  &\phantom{{}=} + m\Gamma_2(\kappa_1^j)\sin(2\omega_j\bar\zeta x_3/\varepsilon)\omega_j\mathbf{h}_1\mathbf{P}_j^\varepsilon \\
  &\phantom{{}=} + m\Gamma_2(\kappa_1^j)\cos(2\omega_j\bar\zeta x_3/\varepsilon) \omega_j\mathbf{h}_2\mathbf{P}_j^\varepsilon 
\end{align*}
for $j=1,\dots,M$.
\end{document}
2
  • 1
    Thanks for your very detailled answer. Oct 26 '21 at 10:23
  • @rami_salazar - You're most welcome. Please also see the addendum I just posted, in which I show how one may state things more succinctly by incorporating the information that the big matrix is block-diagonal.
    – Mico
    Oct 26 '21 at 10:25

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