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Unformatted text preview: Solutions to mock midterm test . Subtracting the rst equation from the second leads to − x = , while subtracting
the rst equation from the third leaves x = , so the system is obviously inconsistent.
e leastsquares approximation x ∗ to Ax = b satis es AT Ax ∗ = AT b ; in this
case rank AT A = rank A = , so the solution is unique:
x∗
=
x∗
AT ( AT A . ⇒ x∗ = ⁄
⁄ )T e determinant of any × matrix
ab
de
gh c
f
i is aei + b f g + cdh − ceg − bdi − a f h. We compute det A = ⋅ ⋅ + ⋅ (− ) ⋅
− ⋅ (− ) ⋅ − ⋅ (− ) ⋅ = . Since det A is nonzero, A in invertible. Since
det(AT A) = det(A) = ≠ , AT A is also invertible.
. Each of the matrices A and B has determinant , so, in particular, they are both
invertible, hence rowequivalent to the identity matrix, and to each other. However,
since trace(A) = ≠ = trace(B), they are not similar.
. In order to check that we have a real inner product space, we need to verify that the
bracket is linear (say in the rst variable), symmetric, and positive. Linearity follows
from basic properties of the de nite integral, symmetry is obvious, and positive
de niteness follows from the fact that if f is a nonzero continuous function, then it
will be bounded away from zero on some interval I ⊆ [ , π ], and the same will be
true for the nonnegative function f (t ) , so
f, f = π∫ π f (t ) d t ≥ π ∫ f (t ) d t >
I . If W = span{ , sin t , cos t }, one can verify that the given functions , sin t and
cos t already form an orthogonal basis of W , but to get an orthonormal basis they
need to be normalized. e following table of inde nite integrals will be useful: ∫ sin t dt = − cos t + C ∫ sin t dt = ∫ cos t dt = sin t + C
∫ cos t dt = t + sin t + C
∫ t cos t dt = t sin t + cos t + C . t − sin t + C ∫ t sin t dt = −t cos t + sin t + C
We have ∫ π∫ π d t = π, π (sin t ) d t = π∫ √
so an orthonormal basis of W is given by { ,
orthogonal projection of t onto W equals
π ∫ π t dt + π ∫ π √
t(
+ π (cos t ) d t = ,
√ sin t , cos t }. erefore the √
sin t ) dt (
π∫ π sin t )+ √
t( √
cos t ) dt ( cos t ) = π − sin t . . We calculate
T ( ) = ( t )′ = t T ( t ) = ( t )′ = t T ( t ) = ( t )′ = t . e matrix representation of T is therefore just [T ] = (into each column we put the coordinates of the image of the corresponding basis
vector). Since Ker T = , the empty set forms a basis of Ker T . Next, we observe that
up to scaling the images of the standard basis vectors for P are the standard basis
vectors t , t , and t . erefore, a basis for the range of T is {t , t , t }.
. e characteristic polynomial of A is λ − λ − = ( λ − )( λ + ). An eigenvector for λ = − is − , and an eigenvector for λ = is of a matrix
P=
− , . Putting these into the columns we have P− = and P − AP = − − So, if D denotes this diagonal matrix, then
A = PD P− = (− ) −
= − +
− −
+ . ...
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 Winter '08
 R.Schmidt

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