Page 305 - 35Linear Algebra

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However, the converse is often false. In fact, the equation MX = V may have
T
T
no solutions at all, but still have least squares solutions to M MX = M V .
T
Observe that since M is an m × n matrix, then M is an n × m matrix.
T
T
T
T
Then M M is an n × n matrix, and is symmetric, since (M M) = M M.
T
T
Then, for any vector X, we can evaluate X M MX to obtain a num-
2
ber. This is a very nice number, though! It is just the length |MX| =
T
T
T
(MX) (MX) = X M MX.
Reading homework: problem 1
Now suppose that ker L = {0}, so that the only solution to MX = 0 is
X = 0. (This need not mean that M is invertible because M is an n × m
T
matrix, so not necessarily square.) However the square matrix M M is
T
invertible. To see this, suppose there was a vector X such that M MX = 0.
2
T
T
Then it would follow that X M MX = |MX| = 0. In other words the
vector MX would have zero length, so could only be the zero vector. But we
are assuming that ker L = {0} so MX = 0 implies X = 0. Thus the kernel
T
of M M is {0} so this matrix is invertible. So, in this case, the least squares
T
solution (the X that solves M MX = MV ) is unique, and is equal to
T
−1
T
X = (M M) M V.
In a nutshell, this is the least squares method:
T
T
• Compute M M and M V .
T
T
• Solve (M M)X = M V by Gaussian elimination.
Example 153 Captain Conundrum falls off of the leaning tower of Pisa and makes
three (rather shaky) measurements of his velocity at three different times.

t s v m/s
1 11
2 19
3 31

1
Having taken some calculus , he believes that his data are best approximated by
a straight line
v = at + b.
1
In fact, he is a Calculus Superhero.

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