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Using phantom in align environment

dwsmith

Well-known member
Feb 1, 2012
1,673
The extra align row at the top contributes a lot of white space but I need phantom equal at spacing of a previous align above it. How can I keep the spacing but remove the white space gap?
Code:
\begin{alignat*}{3}
\phantom{\sigma_1:} & \phantom{\begin{bmatrix}
3 - \sigma_1 & -10 & 0\\
-10 & 0 - \sigma_1 & 30\\
0 & 30 & -27 - \sigma_1
\end{bmatrix}} & \phantom{=} &\phantom{\begin{bmatrix}
3 + 47 & -10 & 0\\
-10 & 47 & 30\\
0 & 30 & -27 + 47
\end{bmatrix}}\\
\sigma_3: & \begin{bmatrix}
3 - \sigma_3 & -10 & 0\\
-10 & 0 - \sigma_3 & 30\\
0 & 30 & -27 - \sigma_3
\end{bmatrix} & = & \begin{bmatrix}
3 - 23 & -10 & 0\\
-10 & -23 & 30\\
0 & 30 & -27 - 23
\end{bmatrix}\\
 & & = & \begin{bmatrix}
-20 & -10 & 0\\
-10 & -23 & 30\\
0 & 30 & -50
\end{bmatrix}\\
 & & = & \begin{bmatrix}
-2 & -1 & 0\\
-10 & -23 & 30\\
0 & 3 & -5
\end{bmatrix}\\
 & & = & \begin{bmatrix}
-2 & -1 & 0\\
0 & -18 & 30\\
0 & 3 & -5
\end{bmatrix}\\
 & & = & \begin{bmatrix}
-2 & -1 & 0\\
0 & -3 & 5\\
0 & 3 & -5
\end{bmatrix}\\
 & & = & \begin{bmatrix}
-2 & -1 & 0\\
0 & -3 & 5\\
0 & 0 & 0
\end{bmatrix}\\
\end{alignat*}
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
I would use the aligned environment rather than alignat*.
Code:
\begin{aligned}
\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3 \end{bmatrix} 
& = \begin{bmatrix}3 - 23 & -10 & 0 \\-10 & -23 & 30 \\0 & 30 & -27 - 23\end{bmatrix} \\ 
& = \begin{bmatrix}-20 & -10 & 0 \\-10 & -23 & 30 \\0 & 30 & -50\end{bmatrix} \\ 
& = \begin{bmatrix}-2 & -1 & 0 \\-10 & -23 & 30 \\0 & 3 & -5\end{bmatrix} \\
& = \begin{bmatrix}-2 & -1 & 0 \\0 & -18 & 30 \\0 & 3 & -5\end{bmatrix} \\
& = \begin{bmatrix}-2 & -1 & 0 \\0 & -3 & 5 \\0 & 3 & -5\end{bmatrix} \\
& = \begin{bmatrix}-2 & -1 & 0 \\0 & -3 & 5 \\0 & 0 & 0\end{bmatrix} \\ 
\end{aligned}
That produces $$\begin{aligned}\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3
\end{bmatrix} & = \begin{bmatrix}3 - 23 & -10 & 0 \\-10 & -23 & 30 \\0 & 30 & -27 - 23\end{bmatrix} \\ & = \begin{bmatrix}-20 & -10 & 0 \\-10 & -23 & 30 \\0 & 30 & -50\end{bmatrix} \\ & = \begin{bmatrix}-2 & -1 & 0 \\-10 & -23 & 30 \\0 & 3 & -5\end{bmatrix} \\& = \begin{bmatrix}-2 & -1 & 0 \\0 & -18 & 30 \\0 & 3 & -5\end{bmatrix} \\& = \begin{bmatrix}-2 & -1 & 0 \\0 & -3 & 5 \\0 & 3 & -5\end{bmatrix} \\& = \begin{bmatrix}-2 & -1 & 0 \\0 & -3 & 5 \\0 & 0 & 0\end{bmatrix} \\ \end{aligned}$$

Edit. If you want the left side of the top row to occupy the same space as $\sigma_1: \begin{bmatrix}3 - \sigma_1 & -10 & 0 \\ -10 & 0 - \sigma_1 & 30\\0 & 30 & -27 - \sigma_1 \end{bmatrix}$ then (in the above code) you can replace
Code:
\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3 \end{bmatrix}
by
Code:
\phantom{\sigma_1: \begin{bmatrix}3 - \sigma_1 & -10 & 0 \\-10 & 0 - \sigma_1 & 30 \\0 & 30 & -27 - \sigma_1 \end{bmatrix}}
 \llap{$\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3 \end{bmatrix}$}
That material will be aligned close to the $=$ sign, without unwanted white space.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
I would use the aligned environment rather than alignat*.
Code:
\begin{aligned}
\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3 \end{bmatrix} 
& = \begin{bmatrix}3 - 23 & -10 & 0 \\-10 & -23 & 30 \\0 & 30 & -27 - 23\end{bmatrix} \\ 
& = \begin{bmatrix}-20 & -10 & 0 \\-10 & -23 & 30 \\0 & 30 & -50\end{bmatrix} \\ 
& = \begin{bmatrix}-2 & -1 & 0 \\-10 & -23 & 30 \\0 & 3 & -5\end{bmatrix} \\
& = \begin{bmatrix}-2 & -1 & 0 \\0 & -18 & 30 \\0 & 3 & -5\end{bmatrix} \\
& = \begin{bmatrix}-2 & -1 & 0 \\0 & -3 & 5 \\0 & 3 & -5\end{bmatrix} \\
& = \begin{bmatrix}-2 & -1 & 0 \\0 & -3 & 5 \\0 & 0 & 0\end{bmatrix} \\ 
\end{aligned}
That produces $$\begin{aligned}\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3
\end{bmatrix} & = \begin{bmatrix}3 - 23 & -10 & 0 \\-10 & -23 & 30 \\0 & 30 & -27 - 23\end{bmatrix} \\ & = \begin{bmatrix}-20 & -10 & 0 \\-10 & -23 & 30 \\0 & 30 & -50\end{bmatrix} \\ & = \begin{bmatrix}-2 & -1 & 0 \\-10 & -23 & 30 \\0 & 3 & -5\end{bmatrix} \\& = \begin{bmatrix}-2 & -1 & 0 \\0 & -18 & 30 \\0 & 3 & -5\end{bmatrix} \\& = \begin{bmatrix}-2 & -1 & 0 \\0 & -3 & 5 \\0 & 3 & -5\end{bmatrix} \\& = \begin{bmatrix}-2 & -1 & 0 \\0 & -3 & 5 \\0 & 0 & 0\end{bmatrix} \\ \end{aligned}$$

Edit. If you want the left side of the top row to occupy the same space as $\sigma_1: \begin{bmatrix}3 - \sigma_1 & -10 & 0 \\ -10 & 0 - \sigma_1 & 30\\0 & 30 & -27 - \sigma_1 \end{bmatrix}$ then (in the above code) you can replace
Code:
\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3 \end{bmatrix}
by
Code:
\phantom{\sigma_1: \begin{bmatrix}3 - \sigma_1 & -10 & 0 \\-10 & 0 - \sigma_1 & 30 \\0 & 30 & -27 - \sigma_1 \end{bmatrix}}
 \llap{$\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3 \end{bmatrix}$}
That material will be aligned close to the $=$ sign, without unwanted white space.
Thanks it did remove the white space, but it doesn't align correct then with the align environment above it.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
The tabbing environment provides more direct control over where things align, and you can make them consistent even when interrupted by text in-between environments. See this thread. Also see this thread.