Using phantom in align environment

[Dustinsfl]

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?

\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]

I would use the aligned environment rather than alignat*.

\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

\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3 \end{bmatrix}

by

\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.

 

[Dustinsfl]

I would use the aligned environment rather than alignat*.

\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

\sigma_3: \begin{bmatrix}3 - \sigma_3 & -10 & 0 \\-10 & 0 - \sigma_3 & 30 \\ 0 & 30 & -27 - \sigma_3 \end{bmatrix}

by

\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]

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 http://www.mathhelpboards.com/f26/possibly-tricky-alignment-problem-1493/. Also see http://www.mathhelpboards.com/f26/align*-environment-inside-tabbing-environment-2150/.