Math 203 Practice
Please work out each
of the given problems. Credit will
be based on the steps towards the final answer.
Show your work. Do your work
on your own paper. Let S
= {t^{2}, t^{2} + 2t, t^{2} + 3}
and T
= {2t  1, 5t  3, t^{2}}
be subsets of P_{2}
A. Prove that S is a basis for P_{2}. Solution Since the dimension of P_{2} is 3 and S has 3 vectors, we need only prove that the vectors in S are linearly independent. If we let c_{1}t^{2} + c_{2}(t^{2} + 2t) + c_{3}(t^{2} + 3) = 0 then we can combine like terms to obtain (c_{1} + c_{2} + c_{3})t^{2} + 2c_{2}t + 3c_{3} = 0 If a polynomial equals zero then each of the coefficients equal zero. We get c_{1}
+ c_{2} + c_{3} = 0 The matrix for this equation is
has determinant equal to 6. Since the determinant is nonzero, the system has only the trivial solution, hence the three vectors are linearly independent. We can conclude that S is a basis for P_{2}.
B. Find the
change of basis
matrix P_{S<T}
. Solution Let E be a standard basis E = {t^{2}, t, 1} and look at the following diagram
Of the following two
subsets of the vector space of differentiable functions, determine which is a
subspace. For the one that is not a
subspace, demonstrate why it is not. For
the one that is a subspace, prove that it is a subspace. A.
S = {f  f(3) = f '(3)}
B.
T = {f  f(0)f '(0) = 0}
Solution A. This is a subspace. We need to show that the two closure properties hold. f(3) = f '(3) and g(3) = g'(3) then (f + g)(3) = f(3) + g(3) = f '(3) + g'(3) = (f + g)'(3) hence f + g is in the subset. If c is a constant, then (cf)(3) = cf(3) = cf '(3) = (cf)'(3) hence cf is in the subset. Since both closure properties are true, we can conclude that S is a subspace. B. This set is not a subspace. Let f(t) = 1 and g(t) = t. Then f(0)f '(0) = (1)(0) = 0 and g(0)g'(0) = (0)(1) = 0 so both f and g are in T. However (f + g)(0)(f + g)'(0) = [f(0) + g(0)] [f '(0) + g'(0)] = (1 + 0)(0 + 1) = 1 Hence T is not closed under addition.
Let
V be the subspace of differentiable functions
spanned by {e^{x}, e^{2x}, e^{3x}}
and let
L: V
> V
be the linear transformation with
L(f(x)) = f ''(x)  3f '(x) + 2f(x) A.
Write down the matrix A_{L} with
respect to the given basis. We have L((1,0,0) = L(e^{x}) = e^{x}  3e^{x} + 2e^{x} = 0 = (0,0,0) L((0,1,0) = L(e^{2x}) = 4e^{2x}  6e^{2x} + 2e^{2x} = 0 = (0,0,0) L((0,0,1) = L(e^{3x}) = 9e^{3x}  9e^{3x} + 2e^{3x} = 2e^{3x} = (0,0,2) The matrix of L is the matrix whose columns are the vectors above.
B.
Find the a basis for the kernel and range of L. A basis for the kernel corresponds to the null space of A_{L} which has basis {(1,0,0), (0,1,0)} in other words the basis is given by {e^{x}, e^{2x}}
Problem 4 Let W
= Span{(1,1,0,1), (0,1,2,3)}. Find
a basis for the orthogonal complement of W. The orthogonal complement is equal to the null space of the matrix whose rows are the given vectors. If
then the rref(A) is given by
so that a basis for the null space is given by {(2,2,1,0), (2,3,0,1)}
Problem 5 LetFind the following: Solution
The rrefA is shown below
Problem 6 Let L: V > V be a linear transformation. Use the fact that
dim(Ker L) + dim(Range L) = dim(V)
to show that if L
is one to one then L is onto.
Solution dim(Ker L) = 0 The equation tells us that dim(Range L) = dim(V) Since the range is a subspace of V and has the same dimension as V, the range of L equal V. We can conclude that L is onto.
Problem 7
Let A
and B be matrices and let v
be an eigenvector of both A and B.
Prove that v
is an eigenvector of the product AB. Proof If v is an eigenvector of A and B, then there are numbers a and b with Av = av and Bv = bv We then have (AB)v = A(Bv) = A(bv) = b(Av) = bav = abv Hence v is an eigenvector of AB with eigenvalue ab.
Without the use of a calculator, diagonalize the matrix
Solution First find the eigenvalues. Thus l = 11 or l = 2
Plugging in 11 gives
Thus the equations are: x = 2/3
y and y = y.
Setting y = 3 gives the vector
(2,3).
Now plug in 2 to get
Thus the equations are: x = 3/2 y and y = y. Setting y = 2 gives the vector (3,2). The diagonal matrix has the eigenvalues on the diagonal:
and the P matrix is the matrix with its columns as the eigenvectors.
Finally, we just need the inverse of P. We use the adjoint formula.
Problem 9 Answer True of False and
explain your reasoning. A. Let A be a 3x3 matrix such that the columns of A form an orthonormal set of vectors. Then
Solution True, since A is an orthogonal matrix, we have A^{T} = A^{1} so that A^{T}A = A^{1}A = I B. B. If A is an mxn matrix such that the rows of A and the columns of A are both linear independent sets, then n = m.
Solution
True. Since the rows are linear dependent the dimension of the row space is m. Since the columns of A are linear independent, the dimension of the column space of A is n. But the row space and column space always have the same dimension. Thus m = n. C. If u and v are eigenvectors for a matrix A, then u + v is also an eigenvector for A. Solution False. For example if Au = u and Av = 2v, then A(u+v) = Au + Av = u + 2v which is not a multiple of u+v. D. If A is similar to B and B is similar to C, then A is similar to C. Solution True. Since A is similar to B, there is an invertible matrix P with A = PBP^{1}. Since B is similar to C, there is an invertible matrix R with B = RCR^{1}. Substituting gives A = P(RCR^{1})P^{1} = (PR)C(R^{1}P^{1}) = (PR)C(PR)^{1} which is in the correct form. E. If l is an eigenvalue for A, then l^{2} is an eigenvalue for A^{2}. True. If v is the eigenvector corresponding to l, then Av = lvso that
A^{2}v = AAv = A(lv)
= lAv = l*lv
= l^{2}v
F. The vectors (1,1,0), (1,0,1), and (0,0,1) are a basis for P^{2}. False, for vectors to be a basis for a vector space, they must be in the vector space. (1,1,0) is not a polynomial.
