Why is weight vector orthogonal to decision plane in neural networks

The weights are simply the coefficients that define a separating plane. For the moment, forget about neurons and just consider the geometric definition of a plane in N dimensions:

w1*x1 + w2*x2 + ... + wN*xN - w0 = 0

You can also think of this as being a dot product:

w*x - w0 = 0

where w and x are both length-N vectors. This equation holds for all points on the plane. Recall that we can multiply the above equation by a constant and it still holds so we can define the constants such that the vector w has unit length. Now, take out a piece of paper and draw your x-y axes (x1 and x2 in the above equations). Next, draw a line (a plane in 2D) somewhere near the origin. w0 is simply the perpendicular distance from the origin to the plane and w is the unit vector that points from the origin along that perpendicular. If you now draw a vector from the origin to any point on the plane, the dot product of that vector with the unit vector w will always be equal to w0 so the equation above holds, right? This is simply the geometric definition of a plane: a unit vector defining the perpendicular to the plane (w) and the distance (w0) from the origin to the plane.

Now our neuron is simply representing the same plane as described above but we just describe the variables a little differently. We'll call the components of x our "inputs", the components of w our "weights", and we'll call the distance w0 a bias. That's all there is to it.

Getting a little beyond your actual question, we don't really care about points on the plane. We really want to know which side of the plane a point falls on. While w*x - w0 is exactly zero on the plane, it will have positive values for points on one side of the plane and negative values for points on the other side. That's where the neuron's activation function comes in but that's beyond your actual question.

Intuitively, in a binary problem the weight vector points in the direction of the '1'-class, while the '0'-class is found when pointing away from the weight vector. The decision boundary should thus be drawn perpendicular to the weight vector.

See the image for a simplified example: You have a neural network with only 1 input which thus has 1 weight. If the weight is -1 (the blue vector), then all negative inputs will become positive, so the whole negative spectrum will be assigned to the '1'-class, while the positive spectrum will be the '0'-class. The decision boundary in a 2-axis plane is thus a vertical line through the origin (the red line). Simply said it is the line perpendicular to the weight vector.

Lets go through this example with a few values. The output of the perceptron is class 1 if the sum of all inputs * weights is larger than 0 (the default threshold), otherwise if the output is smaller than the threshold of 0 then the class is 0. Your input has value 1. The weight applied to this single input is -1, so 1 * -1 = -1 which is less than 0. The input is thus assigned class 0 (NOTE: class 0 and class 1 could have just been called class A or class B, don't confuse them with the input and weight values). Conversely, if the input is -1, then input * weight is -1 * -1 = 1, which is larger than 0, so the input is assigned to class 1. If you try every input value then you will see that all the negative values in this example have an output larger than 0, so all of them belong to class 1. All positive values will have an output of smaller than 0 and therefore will be classified as class 0. Draw the line which separates all positive and negative input values (the red line) and you will see that this line is perpendicular to the weight vector.

Also note that the weight vector is only used to modify the inputs to fit the wanted output. What would happen without a weight vector? An input of 1, would result in an output of 1, which is larger than the threshold of 0, so the class is '1'.

The second image on this page shows a perceptron with 2 inputs and a bias. The first input has the same weight as my example, while the second input has a weight of 1. The corresponding weight vector together with the decision boundary are thus changed as seen in the image. Also the decision boundary has been translated to the right due to an added bias of 1.