Introduction Matrix Action Perpframes, Aligners and Hangers Stretchers Coordinates Projections SVD Matrix Subspaces Linear Systems, PseudoInverse Condition Number Matrix Norm, Rank One Data Compression Noise Filtering
Todd Will

Multiple
Hits
Projecting to a line 
Projections 
Projecting
to a plane
Simplified formula Exercise 
Hitting with one matrix and then anotherHere's a perpframe and specified by angle .Plot an ellipse with a minor axis of 0.5 in the direction of
and major axis of 1.2 in the direction of . Answer: You know how to plot the unit circle: You know how to use an xystretcher to stretch the circle into an ellipse with the desired axes lengths:
All you need to do is hang the ellipse on the perpendicular frame. To do that you hit the curve
with the
matrix.
An ellipse with a minor axis of 0.5 in the direction of and major axis of 1.2 in the direction of . Just what the doctor ordered.
In the problem above you used two matrix hits to plot an ellipse. Plot the same ellipse again, but this time use only one matrix hit. Answer: Above, you defined a stretcher matrix and a hanger matrix and then plotted the results of a hit with the stretcher followed by a hit with the hanger: If you multiply the out (hanger) (stretcher)
you get:
Now hit the unit circle with the product: Apparently, the action of the matrix product: product = (hanger) (stretcher) is the same as the action of the stretcher followed
by the action of the hanger.
Key point:
The action of a matrix product: A=B C is the same as the action
of the C followed by the action of B.
The point on a line through {0,0} closest to P.Here's a plot of the point .What point on the xaxis is closest to the point ? Answer: It's gotta be the point . Click here only if you're in doubt about this. What point on the yaxis is closest to the point ? Answer: It's gotta be the point . Click here only if you're in doubt about this.
How can you get these closest points with matrix hits? Answer: To get the closest point on the xaxis, hit with the xystretcher that squashes the ycoordinate but leaves the xcoordinate unchanged. The xystretcher you want is . Check: =
To get the closest point on the yaxis hit with the xystretcher that squashes the xcoordinate but leaves the ycoordinate unchanged. The xystretcher you want is . Check: =
Here's a plot of a perpframe and the point . The plot also shows lines through parallel to and What point on the green line is closest to the point ?
Answer: This is a lot harder than asking what point on the xaxis is closest to . Look at the following plots. S is the point on the green line closest to . Explain how the S got where it did. Answer: Look again at the first two plots. You get from the first plot to the second plot by rotating the perpframe vectors and to the x and y axes. This rotates the green line to the xaxis and the point to the new position . You get this rotation with the .
Look at the second and third plots: R is the point on the xaxis closest to Q. You can get R by hitting Q with the . This squashes Q's ycoordinate, but leaves the xcoordinate unchanged.
Look at the third and fourth plots: Here you rotate the xyaxis back to the perpframe vectors. The point R gets rotated to S. This is a job for the .
Summary: You arrived at S by hitting
P,
This says you can get the coordinates of the point on the green line closest to by computing Compute this point and add it to the plot
Looks good.
Take a look at the original setup again. What point on the blue line is closest to the point ?
Answer: This time you should
try to imagine the picture for each step:
This rotates P to a new position Q. You use for the stretcher since you want to squash the x but leave the y unchanged. Compute and add it to the plot. Looks good.
It's good to be able to imagine all of the matrix action that leads from to the blue dot, but you deserve to see it too. Enjoy!The point on a plane through {0,0,0} closest to P.Here's a plot of a point and a perpframe .The plot also shows the plane through spanned by and What point on the plane is closest to the point ?
Answer this question in stages. First align the plane spanned by and with the xyplane. Do this by hitting with the aligner=. The aligner has rotated the perpframe to xyzaxis. The original plane has been rotated to the xyplane. The point P has been rotated to the point Q. The distances from P to the original plane are the same as distances from Q to the xyplane. But finding the point on the xyplane closest to Q is a snap. Just hit Q with a stretcher matrix that leaves the x and y coordinates
unchanged, but squashes the z coordinate.
Now hang the plane back in its original position to find the point closest to P. Do this with the hanger matrix.
Simplified formulas.You'll see later that the product simplifies to just .More generally, (but without the pictures), if
Exercise1. Here's a plot of a perpframe and the curve{x[t],y[t]}={1+Cos[t] (0.2 Sin[t]) Sin[t], 1.2 +Cos[t] (0.6 Sin[t])} Hitting this curve with the stretcher matrix
reflects the curve across the yaxis. Your job is to define matrices A, B, C, (in terms of
and ),
so that the matrix product ABC reflects the curve across the green
line as shown below: 