Question

A linear transformation of the form z = Γx was applied to the data, where Γ is a 2 × 2 matrix. The decision boundary associated with the BDR is now the hyperplane of normal w = (1/ √ 2, −1/ √ 2)T which passes through the origin.

Answer 1

Part a: The transformation matrix is the clockwise rotation matrix of π/4.

Part b: The hyperplane would move towards the mean of class 1.

Part c: The distance will remain in the Euclidean Space due to the rotation transformation only.

Explanation:

As the complete question is not available, the question is searched online and a reference question is obtained which has 3 parts as follows:

Part a:

The decision boundary after transformation coincides with the line x1 = x2, the two class means must lie on a line that is normal to the decision boundary, i.e. on x1 = −x2. This implies that the transformation matrix Γ is a clockwise rotation transformation of π/4, given as

`\Gamma=\left[\begin{array}{cc}(√(2))/(2)&(√(2))/(2)\\(-√(2))/(2)&(√(2))/(2)\end{array}\right]`

Part b:

If the prior probability of class 0 was increased after transformation, then the decision boundary of BDR would still have the same normal as before, i.e.,
`w=(1/√(2),-/√(2))^T`, but move toward the mean of class 1.

Part c:

Noting that

`||\bold{T}_x-\bold{T}_y||^2=(x-y)^T \bold{T}^T\bold{T}(x-y)\\||\bold{T}_x-\bold{T}_y||^2=(x-y)^T(x-y)\\||\bold{T}_x-\bold{T}_y||^2=||x-y||^2`

This indicates that the distance is still the same and is in Euclidean space. This is due to the fact that rotation transformations does not affect the distances between the points.