Friday, November 30, 2012

Projection in Multiple Dimensions

In this section, we extend from the one-dimensional subspace to the more general two-dimensional subspace. This means that there are two vectors, $\mathbf{v}_1$ and $\mathbf{v}_2$ that are not colinear and that span the subspace. In the previous case, we had only one vector ( $\mathbf{v}$), so we had a one-dimensional subspace, but now that we have two vectors, we have a two-dimensional subspace (i.e. a plane). The extension from the two-dimensional subspace to the n-dimensional subspace follows the same argument but introduces more notation than we need so we'll stick with the two-dimensional case for awhile. For the two-dimensional case, the optimal MMSE solution has the form

$$ \hat{\mathbf{y}} = \alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 \in \mathbb{R}^m $$

where $\mathbf{y}$ exists in the m-dimensional space of real numbers. We want to project this vector onto the two m-dimensional $\mathbf{v}_i$ vectors. Here, the orthogonality requirement extends as

$$ \langle \mathbf{y} - \alpha_1 \mathbf{v}_1 -\alpha_2 \mathbf{v}_2 , \mathbf{v}_1\rangle= 0 $$

and

$$ \langle \mathbf{y} - \alpha_1 \mathbf{v}_1 -\alpha_2 \mathbf{v}_2 , \mathbf{v}_2\rangle= 0 $$

Recall that for vectors, we have

$$ \langle \mathbf{x} , \mathbf{y}\rangle = \mathbf{x}^T \mathbf{y} \in \mathbb{R}$$

This leads to the linear system of equations:

$$ \begin{eqnarray} \langle \mathbf{y}, \mathbf{v}_1\rangle = & \alpha_1 \langle \mathbf{v}_1, \mathbf{v}_1\rangle & +\alpha_2 \langle \mathbf{v}_1, \mathbf{v}_2\rangle \\ \langle \mathbf{y}, \mathbf{v}_2\rangle = & \alpha_1 \langle \mathbf{v}_1, \mathbf{v}_2\rangle & +\alpha_2 \langle \mathbf{v}_2, \mathbf{v}_2\rangle \end{eqnarray} $$

which can be written in matrix form as

$$ \left[ \begin{array}{c} \langle \mathbf{y}, \mathbf{v}_1\rangle \\ \langle \mathbf{y}, \mathbf{v}_2\rangle \\ \end{array} \right] = \left[ \begin{array}{cc} \langle \mathbf{v}_1, \mathbf{v}_1\rangle & \langle \mathbf{v}_1, \mathbf{v}_2\rangle \\ \langle \mathbf{v}_1, \mathbf{v}_2\rangle & \langle \mathbf{v}_2, \mathbf{v}_2\rangle \\ \end{array} \right] \left[ \begin{array}{c} \alpha_1 \\ \alpha_2 \\ \end{array} \right]$$

which can be further reduced by stacking the columns into

$$ \mathbf{V} = \left[ \mathbf{v}_1, \mathbf{v}_2 \right] \in \mathbb{R}^{m \times 2} $$

and

$$ \boldsymbol{\alpha}= \left[ \alpha_1, \alpha_2\right]^T \in \mathbb{R}^{2}$$

which gives

$$ \mathbf{V}^T \mathbf{y} = (\mathbf{V}^T \mathbf{V}) \boldsymbol{\alpha} $$

Note that by writing this using vector notation, we have implicitly generalized beyond two dimensions since there is nothing to stop from stacking $\mathbf{V}$ with more column vectors to create a larger subspace. By solving we obtain,

$$ \boldsymbol{\alpha} = (\mathbf{V}^T \mathbf{V})^{-1} \mathbf{V}^T \mathbf{y} $$

and so the optimal solution is then,

$$ \hat{\mathbf{y}} = \mathbf{V} \boldsymbol{\alpha} \in \mathbb{R}^m $$

Note that the existence of the inverse is guaranteed by the non-co-linearity of the $\mathbf{v}_i$ vectors. Whether or not that inverse is numerically stable is another issue.

Then, we can combine these to obtain

$$ \hat{\mathbf{y}} = \mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1} \mathbf{V}^T \mathbf{y} $$

when then makes the projection operator for this case:

$$ \mathbf{P}_{V}= \mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1} \mathbf{V}^T \in \mathbb{R}^{m \times m} $$

As a quick check, we can see this reduce to the 1-dimensional case by setting

$$ \mathbf{V}= \mathbf{v} \in \mathbb{R}^m$$

so then,

$$ \mathbf{P}_{v}= \mathbf{v} \frac{1}{\mathbf{v}^T \mathbf{v}} \mathbf{v}^T $$

which matches our previous result. The point of all these manipulations is that we can construct another projection operator with all the MMSE properties we had before, but now in a bigger subspace. We can further verify the idempotent property of projection matrices by checking that

$$ \mathbf{P}_V \mathbf{P}_V = \mathbf{P}_V$$

The following graphic shows that when we project the three dimensional $\mathbf{y}$ vector onto the plane, which is spanned by the two $\mathbf{v}_i$ vectors, we obtain the MMSE solution where the sphere is tangent to the plane. The point of tangency is the point $\hat{\mathbf{y}} $ which is the MMSE solution.

In [2]:
#http://stackoverflow.com/questions/10374930/matplotlib-annotating-a-3d-scatter-plot

from mpl_toolkits.mplot3d import proj3d
import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure()
fig.set_size_inches([8,8])

ax = fig.add_subplot(111, projection='3d')

ax.set_aspect(1)
ax.set_xlim([0,2])
ax.set_ylim([0,2])
ax.set_zlim([0,2])
ax.set_aspect(1)
ax.set_xlabel('x-axis',fontsize=16)
ax.set_ylabel('y-axis',fontsize=16)
ax.set_zlabel('z-axis',fontsize=16)

y = matrix([1,1,1]).T 
V = matrix([[1,0.25], # columns are v_1, v_2
            [0,0.50],
            [0,0.00]])

alpha=inv(V.T*V)*V.T*y # optimal coefficients
P = V*inv(V.T*V)*V.T
yhat = P*y         # approximant


u = np.linspace(0, 2*np.pi, 100)
v = np.linspace(0, np.pi, 100)

xx = np.outer(np.cos(u), np.sin(v))
yy = np.outer(np.sin(u), np.sin(v))
zz = np.outer(np.ones(np.size(u)), np.cos(v))

sphere=ax.plot_surface(xx+y[0,0], yy+y[1,0], zz+y[2,0],  
                       rstride=4, cstride=4, color='gray',alpha=0.3,lw=0.25)

ax.plot3D([y[0,0],0],[y[1,0],0],[y[2,0],0],'r-',lw=3)
ax.plot3D([y[0,0]],[y[1,0]],[y[2,0]],'ro')

ax.plot3D([V[0,0],0],[V[1,0],0],[V[2,0],0],'b-',lw=3)
ax.plot3D([V[0,0]],[V[1,0]],[V[2,0]],'bo')
ax.plot3D([V[0,1],0],[V[1,1],0],[V[2,1],0],'b-',lw=3)
ax.plot3D([V[0,1]],[V[1,1]],[V[2,1]],'bo')

ax.plot3D([yhat[0,0],0],[yhat[1,0],0],[yhat[2,0],0],'g--',lw=3)
ax.plot3D([yhat[0,0]],[yhat[1,0]],[yhat[2,0]],'go')


x2, y2, _ = proj3d.proj_transform(y[0,0],y[1,0],y[2,0], ax.get_proj())
ax.annotate(
    "$\mathbf{y}$", 
    xy = (x2, y2), xytext = (-20, 20), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

x2, y2, _ = proj3d.proj_transform(yhat[0,0],yhat[1,0],yhat[2,0], ax.get_proj())
ax.annotate(
    "$\hat{\mathbf{y}}$", 
    xy = (x2, y2), xytext = (-40, 10), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

x2, y2, _ = proj3d.proj_transform(V[0,0],V[1,0],V[2,0], ax.get_proj())
ax.annotate(
    "$\mathbf{v}_1$", 
    xy = (x2, y2), xytext = (120, 10), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

x2, y2, _ = proj3d.proj_transform(V[0,1],V[1,1],V[2,1], ax.get_proj())
ax.annotate(
    "$\mathbf{v}_2$", 
    xy = (x2, y2), xytext = (-30, 30), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

plt.show()

Weighted Distances

As before, we can easily extend this projection operator to cases where the measure of distance between $\mathbf{y}$ and the subspace $\mathbf{v}$ is weighted (i.e. non-uniform). We can accomodate these weighted distances by re-writing the projection operator as

$$ \mathbf{P}_{V} = \mathbf{V} ( \mathbf{V}^T \mathbf{Q V})^{-1} \mathbf{V}^T \mathbf{Q} $$

where $\mathbf{Q}$ is positive definite matrix. Earlier, we started with a point $\mathbf{y}$ and inflated a sphere centered at $\mathbf{y}$ until it just touched the plane defined by $\mathbf{v}_i$ and this point was closest point on the subspace to $\mathbf{y}$. In the general case with a weighted distance except now we inflate an ellipsoid, not a sphere, until the ellipsoid touches the line.

The code and figure below illustrate what happens using the weighted $ \mathbf{P}_v $. It is basically the same code we used above. You can download the IPython notebook corresponding to this post and try different values on the diagonal of $\mathbf{Q}$.

In [9]:
from mpl_toolkits.mplot3d import proj3d
import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure()
fig.set_size_inches([8,8])

ax = fig.add_subplot(111, projection='3d')
ax.set_xlim([0,2])
ax.set_ylim([0,2])
ax.set_zlim([0,2])
ax.set_aspect(1)
ax.set_xlabel('x-axis',fontsize=16)
ax.set_ylabel('y-axis',fontsize=16)
ax.set_zlabel('z-axis',fontsize=16)

y = matrix([1,1,1]).T 
V = matrix([[1,0.25], # columns are v_1, v_2
            [0,0.50],
            [0,0.00]])

Q = matrix([[1,0,0],
            [0,2,0],
            [0,0,3]])

P = V*inv(V.T*Q*V)*V.T*Q
yhat = P*y         # approximant


u = np.linspace(0, 2*np.pi, 100)
v = np.linspace(0, np.pi, 100)

xx = np.outer(np.cos(u), np.sin(v))
yy = np.outer(np.sin(u), np.sin(v))
zz = np.outer(np.ones(np.size(u)), np.cos(v))

xx,yy,yz=map(squeeze,split(tensordot(dstack([xx,yy,zz]),Q,axes=1),3,axis=2))

ellipsoid=ax.plot_surface(xx+y[0,0], yy+y[1,0], zz+y[2,0],  
                           rstride=4, cstride=4, color='gray',alpha=0.3,lw=0.25)

ax.plot3D([y[0,0],0],[y[1,0],0],[y[2,0],0],'r-',lw=3)
ax.plot3D([y[0,0]],[y[1,0]],[y[2,0]],'ro')

ax.plot3D([V[0,0],0],[V[1,0],0],[V[2,0],0],'b-',lw=3)
ax.plot3D([V[0,0]],[V[1,0]],[V[2,0]],'bo')
ax.plot3D([V[0,1],0],[V[1,1],0],[V[2,1],0],'b-',lw=3)
ax.plot3D([V[0,1]],[V[1,1]],[V[2,1]],'bo')

ax.plot3D([yhat[0,0],0],[yhat[1,0],0],[yhat[2,0],0],'g--',lw=3)
ax.plot3D([yhat[0,0]],[yhat[1,0]],[yhat[2,0]],'go')

x2, y2, _ = proj3d.proj_transform(y[0,0],y[1,0],y[2,0], ax.get_proj())
ax.annotate(
    "$\mathbf{y}$", 
    xy = (x2, y2), xytext = (-20, 20), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

x2, y2, _ = proj3d.proj_transform(yhat[0,0],yhat[1,0],yhat[2,0], ax.get_proj())
ax.annotate(
    "$\hat{\mathbf{y}}$", 
    xy = (x2, y2), xytext = (40, 30), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

x2, y2, _ = proj3d.proj_transform(V[0,0],V[1,0],V[2,0], ax.get_proj())
ax.annotate(
    "$\mathbf{v}_1$", 
    xy = (x2, y2), xytext = (120, 10), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

x2, y2, _ = proj3d.proj_transform(V[0,1],V[1,1],V[2,1], ax.get_proj())
ax.annotate(
    "$\mathbf{v}_2$", 
    xy = (x2, y2), xytext = (-30, 30), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

plt.show()

Summary

In this section, we extended the concept of a projection operator beyond one dimension and showed the corresponding geometric concepts that tie the projection operator to MMSE problems in more than one dimension.

References

This post was created using the nbconvert utility from the source IPython Notebook which is available for download from the main github site for this blog. The projection concept is masterfully discussed in the classic Strang, G. (2003). Introduction to linear algebra. Wellesley Cambridge Pr. Also, some of Dr. Strang's excellent lectures are available on MIT Courseware. I highly recommend these as well as the book.

Appendix

Below is some extra code to handle the more general case where there is a rotation as well as a weighting of the axes. Note the projection operator is constructed exactly the same way.

In [15]:
from mpl_toolkits.mplot3d import proj3d
import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure()
fig.set_size_inches([8,8])

ax = fig.add_subplot(111, projection='3d')
ax.set_xlim([0,2])
ax.set_ylim([0,2])
ax.set_zlim([0,2])
ax.set_aspect(1)
ax.set_xlabel('x-axis',fontsize=16)
ax.set_ylabel('y-axis',fontsize=16)
ax.set_zlabel('z-axis',fontsize=16)

y = matrix([1,1,1]).T 
V = matrix([[1,0.25], # columns are v_1, v_2
            [0,0.50],
            [0,0.00]])

def rotation_matrix(angle,axis='z'):                  
    angle = angle/180.*pi
    if axis=='z':
        return matrix([[cos(angle),sin(angle),0],
                  [sin(-angle),cos(angle),0],
                  [0,0,1]])
    elif axis=='y':
        return matrix([[cos(angle),sin(angle),0],
                       [0,1,0],
                       [sin(-angle),cos(angle),0]])
    elif axis=='x':
        return matrix([[1,0,0],
                       [cos(angle),sin(angle),0],
                       [sin(-angle),cos(angle),0]])
          
S = matrix([[3,0,0],
            [0,2,0],
            [0,0,1]])

R = rotation_matrix(30)*rotation_matrix(30,'x')*rotation_matrix(40,'y')

Q = R.T*S*R # apply 3-D rotations

P = V*inv(V.T*Q*V)*V.T*Q # build projection matrix
yhat = P*y         # approximant


u = np.linspace(0, 2*np.pi, 100)
v = np.linspace(0, np.pi, 100)

xx = np.outer(np.cos(u), np.sin(v))
yy = np.outer(np.sin(u), np.sin(v))
zz = np.outer(np.ones(np.size(u)), np.cos(v))

xx,yy,yz=map(squeeze,split(tensordot(dstack([xx,yy,zz]),Q,axes=1),3,axis=2))

ellipsoid=ax.plot_surface(xx+y[0,0], yy+y[1,0], zz+y[2,0],  
                           rstride=4, cstride=4, color='gray',alpha=0.3,lw=0.25)

ax.plot3D([y[0,0],0],[y[1,0],0],[y[2,0],0],'r-',lw=3)
ax.plot3D([y[0,0]],[y[1,0]],[y[2,0]],'ro')

ax.plot3D([V[0,0],0],[V[1,0],0],[V[2,0],0],'b-',lw=3)
ax.plot3D([V[0,0]],[V[1,0]],[V[2,0]],'bo')
ax.plot3D([V[0,1],0],[V[1,1],0],[V[2,1],0],'b-',lw=3)
ax.plot3D([V[0,1]],[V[1,1]],[V[2,1]],'bo')

ax.plot3D([yhat[0,0],0],[yhat[1,0],0],[yhat[2,0],0],'g--',lw=3)
ax.plot3D([yhat[0,0]],[yhat[1,0]],[yhat[2,0]],'go')

x2, y2, _ = proj3d.proj_transform(y[0,0],y[1,0],y[2,0], ax.get_proj())
ax.annotate(
    "$\mathbf{y}$", 
    xy = (x2, y2), xytext = (-20, 20), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

x2, y2, _ = proj3d.proj_transform(yhat[0,0],yhat[1,0],yhat[2,0], ax.get_proj())
ax.annotate(
    "$\hat{\mathbf{y}}$", 
    xy = (x2, y2), xytext = (40, 30), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

x2, y2, _ = proj3d.proj_transform(V[0,0],V[1,0],V[2,0], ax.get_proj())
ax.annotate(
    "$\mathbf{v}_1$", 
    xy = (x2, y2), xytext = (120, 10), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

x2, y2, _ = proj3d.proj_transform(V[0,1],V[1,1],V[2,1], ax.get_proj())
ax.annotate(
    "$\mathbf{v}_2$", 
    xy = (x2, y2), xytext = (-30, 30), fontsize=24,
    textcoords = 'offset points', ha = 'right', va = 'bottom',
    bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
    arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

plt.show()

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