Spatially invariant systems I was watching this video on spatially invariant systems:

And I found the definition a bit confusing, or seems vacuous, because isn't $T[x(n_1 - k_1, n_2 - k_2)] = y(n_1 - k_1, n_2 - k_2)$ always true by definition?
 A: Spatial invariance simply means that shifting the input signal results in an equally shifted output signal. So if the response to an arbitrary signal $x(n_1,n_2)$ is $y(n_1,n_2)$, and the response to a shifted version of the same input signal $x(n_1-k_1,n_2-k_2)$ is $\tilde{y}(n_1,n_2)$, then the system is shift invariant if and only if
$$\tilde{y}(n_1,n_2)=y(n_1-k_1,n_2-k_2)\tag{1}$$
Any system that can be described by the following convolution sum is shift-invariant:
$$y(n_1,n_2)=\sum_{l_1}\sum_{l_2}h(l_1,l_2)x(n_1-l_1,n_2-l_2)\tag{2}$$
where $h(n_1,n_2)$ is the system's impulse response.
EDIT: As requested in the comments, here are some simple examples of shift-variant systems:


*

*$y(n_1,n_2)=x(2n_1,n_2)$

*$y(n_1,n_2)=n_1\cdot x(n_1,n_2)$

*$y(n_1,n_2)=x(-n_1,-n_2)$


Let's look at the examples above in more detail:


*

*\begin{align*}
y(n_1 - d_1, n_2 - d_2) &= x(2(n_1 - d_1), n_2 - d_2) \\
& \ne x(2n_1 - d_1, n_2 - d_2)
\end{align*}

*\begin{align*}
y(n_1 - d_1, n_2 - d_2) &= (n_1 - d_1) & \cdot x(n_1 - d_1, n_2 - d_2) \\
&\ne n_1 & \cdot x(n_1 - d_1, n_2 - d_2)
\end{align*}

*\begin{align*}
y(n_1 - d_1, n_2 - d_2) &= x(-(n_1 - d_1), -(n_2 - d_2)) \\
& \ne x(-n_1 - d_1, -n_2 - d_2)
\end{align*}
A: My answer is too late but anyway ...
I was watching this exact slide and got as confused as you! Then I came across your question. I don't fully get Matt L's answer. After some thinking I realized that the main cause of the confusion is how we view the system $T[]$.
I came up with the following explanatory example:
If $T[x(n)]=nx(n)$, what does it mean? It means that the system will multiply the input by $n$. Thus, if we give the system another input function $z(n)$, it will output $nz(n)$.
Thus:
For $x(n)=n$, $T[x(n)]=nx(n)=n\cdot n$.
For $z(n)=n-\tau$, $T[z(n)]=nz(n)=n\cdot(n-\tau)$.
In addition, if $T[]$ is spatially invariant, then $T[x(n-\tau)]=(n-\tau)(n-\tau)$.
However, note that $z(n)$ is indeed $x(n-\tau)$, and the output is $n\cdot(n-\tau)$ rather than $(n-\tau)(n-\tau)$. Thus, $T[]$ is spatial variant.
In general, if the modification applied by the system to the input involves the time $n$, then the system is spatially varying.
