Math 308F Answers to Major Quiz 2

Problem 1

Note: For formatting convenience, a column vector will be written as the transpose of a row vector. for example

[-1 1 0 0 0]^T

Problem 1(a).

Equations from reduced matrix are:

x_5 = 0
x_4 = 0
x_3 = x_3 (free)
x_2 = x_2 (free)
x_1 = -x_2

so solutions are 

x = x_2[-1 1 0 0 0]^T + x_3[0 0 1 0 0]^T

A basis is set of two vectors {[-1 1 0 0 0]^T, [0 0 1 0 0]^T}

Note: there should be no free variables in the basis. It is a set of 2 vectors. The answer is also not a matrix.

Problem 1(b).

Dimension = 2

Problem 1(c).

Choose the pivot columns of the original matrix:

{[1 2 1 2]^T, [0 1 2 2]^T, [2 -1 2 3]^T}

Problem 1(d).

Dimension = 3

Problem 1(e) and 1(f).

dim col + dim nul = n
In this case, 2 + 3 = 5.

Problem 2.

Note: This was taken from a homework problem in 1.9.

Problem 2(a).

The problem is modeled by

X_1 = A X_0,
X_2 = A X_1,
...
X_(n+1) = A X_n,

where, if c_n = city pop at time n. and s_n = suburban pop at time n, then

X_n = [c_n s_n]^T

The matrix A is
| .96 .03 |
| .04 .97 |

Problem 2(b).

To check whether k = 1 is an eigenvalue, check whether A - 1I is invertible using row reduction.

A - I = |-.04  .03 |  ->  | -4 3 | -> | 1 -3/4 |
        | .04 -.03 |      |  0 0 |    | 0    0 |
        

so 1 is an eigenvalue and any non-zero multiple of X = [3 4]^T is a non-zero eigenvector.

Problem 3.

The matrix B is
| .8 .3 |
| .2 .7 |

Problem 3(a).

B - I = |-.2  .3 |  ->  | -2 3 | -> | 1 -3/2 |
        | .2 -.3 |      |  0 0 |    | 0    0 |

B - .5I = | .3  .3 |  ->  | 1 1 | 
          | .2  .1 |      | 0 0 | 

Problem 3(b).

To find eigenvectors, find null space vectors for matrices in (a).

Eigenvector for eigenvalue 1:  [3/2 1]^T  or [3 2]^T.

Eigenvector for eigenvalue 0.5:  [-1 1]^T.

Problem 3(c).

P = | 3  -1 |  , D = |  1  0 | 
    | 2   1 |        |  0 .5 |

Problem 3(d).

B^50 = P D^50 P^(-1) = 
| 3  -1 | | 1     0 | |  .2  .2 |                   
| 2   1 | | 0 .5^50 | | -.4  .6 |

Problem 4.

Problem 4(a).

Eigenvalues are 2, 0, -8 (2 has alg. multiplicity 2). These can be read off without calculation, since the matrix is a triangular matrix.

Problem 4(b).

Compute det(A - kI) = k^2 - 4k + 13. Roots are 2+3i, 2-3i.

Eigenvector for 2+31 is [i 1]^T or [1 -i]^T.
Eigenvector for 2-31 is [-i 1]^T or [1 i]^T.

Problem 5.

Problem 5(a).

Answer is 0 (zero) for any m or n if columns are independent. Compare with 1(e).

Problem 5(b).

Yes this is a subspace. The set of vectors is the set of all linear combinations of [1 0 1]^T and [2 1 -1], so the set is a span, and spans are always subspaces, so V is a subspace.

Since the two vectors are independent, they form a basis: {[1 0 1]^T, [2 1 -1]^T}.

Since the number of vectors in the basis is two, the dimension of V is two.

Note: Some people wrote R^2 as the dimension. The dimension is a number, a non-negative integer like, 0, 1, 2, 3, 4, 5, .... The symbol R^2 stands for the real two-space, the set of all vectors [x y]^T. This is a plane. It is not an integer. However, the dimension of R^2 is 2.

Problem 5(c).

No, W is not a subspace.

Several possible reasons: (1) the zero vector is not in W (2) if x is in the set, then y = 2x is not in W, because when you stick y into the equation for W, this gives a right side of 2 (3) if x is in W and z is in W, then x+z is not in W for the same reason.

(Any one of these is a correct answer. You do not need more than one.)

Problem 5(d).