A simple example might help..
Let us define the set of points to be clustered as A = {1,2,3,4}
.
Say you're trying to find 2 appropriate clusters for A (2-means). There are (at least) two different settings which satisfy the stationary condition of k-means.
Setting 1:
Center1 = 1, Cluster1 = {1}
Center2 = 3, Cluster1 = {2,3,4}
Here the objective is 2.
As a matter of fact this is a saddle point (try center1 = 1 + epsilon
and center1 = 1 - epsilon
)
Setting 1:
Center1 = 1.5, Cluster1 = {1,2}
Center2 = 3.5, Cluster1 = {3,4}
here the objective is 1/4.
If k-means would be initialized as the first setting then it would be stuck.. and that's by no means a global minimum.
You can use a variant of previous example to create two different local minima.
For A = {1,2,3,4,5}
, setting cluster1={1,2}
and cluster2={3,4,5}
would results in the same objective value as cluster1={1,2,3}
and cluster2={4,5}
Finally, what would happen if you choose
A = {1,2,3,4,6}
center1={2.5} cluster1={1,2,3,4} and
center1={6} cluster1={6}
vs
center1={2} cluster1={1,2,3} and
center1={5} cluster1={4,6}
?