Math 310: Hour Exam 2 (Solutions)

Prof. S. Smith: Fri 3 Nov 1995

You must SHOW WORK to receive credit. Problem 1:

(a) Are the rows of the matrix ( \begin{matrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 2 & 2 & 4 \end{matrix} ) a spanning set for R^{3} ?

NO: Maybe easiest is det(A) = 0. Or check rref(A) ≠ identity .

(b) In polynomial space, are the “vectors” 1 + x, 1 − x, and 2 + x^{2} linearly independent? YES: The polynomial equation a(1 + x) + b(1 − x) + c(2 + x^{2}) = 0 leads to Ax = 0

where A is ( \begin{matrix}1 & 1 & 2 \\ 1 & −1 & 0 \\ 0 & 0 & 1 \end{matrix} ) and we see det(A) = −2 ≠ 0.

Problem 2:

(a) Find a basis for the nullspace of ( \begin{matrix}1 & 3 & 3 & 2 \\ 2 & 6 & 9 & 5 \\ −1 & −3 & 3 & 0 \end{matrix} ) .

Compute rref(A) as ( \begin{matrix}1 & 3 & 0 & 1 \\ 0 & 0 & 1 & \frac{1}{3} \\ 0 & 0 & 0 & 0 \end{matrix} ) ; notice 2nd and 4th variables free.

So basis (−3, 1, 0, 0)^{T} and (−1, 0, −\frac{1}{3}, 1)^{T}.

(b) Using the basis 1, 1 + x, 1 + x^{2} for the space of polynomials of degree at most 2, give the coordinates of the “vector” 1 + x + x^{2}.

Quickly: (−1, 1, 1) since we see (−1)(1) + (1)(1 + x) + (1)(1 + x^{2}) = 1 + x + x^{2}.

Or in detail: the polynomial equation r(1) + s(1 + x) + t(1 + x^{2}) = 1 + x + x^{2}

leads to the “Ax = b” problem ( \begin{matrix}1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{matrix} ) which has unique solution (−1, 1, 1)^{T}.

Problem 3: Find the matrix, in the standard bases, for the linear transformation

L(( \begin{matrix}x \\ y \\ z \end{matrix} )) = ( \begin{matrix}2x − y − z \\ x − 2y + z \\ −x + 3y + 2z \end{matrix} ) .

Apply L to standard basis, put into columns to get ( \begin{matrix}2 & −1 & −1 \\ 1 & −2 & 1 \\ −1 & 3 & 2 \end{matrix} ) .

Problem 4:

(a) Find an orthonormal basis for the column space of A = ( \begin{matrix}2 & 1 \\ 1 & 1 \\ 2 & 1 \end{matrix} ) .

Let’s use Gram-Schmidt with normalization afterwards. Call initial vectors x_{1},x_{2}. Form y_{2} = x_{2} − \frac{x_{2}·x_{1}}{x_{1}·x_{1}} x_{1} = (1, 1, 1) − \frac{5}{9}(2, 1, 2) = \frac{1}{9}(−1, 4, −1). Now normalize to get u_{1} = \frac{1}{3}(2, 1, 2) and u_{2} = \frac{1}{\sqrt{18}}(−1, 4, −1).

(b) Now use part (a) to give the QR factorization of A. (That is, A = QR where Q has orthonormal columns and R is upper triangular).

We find r_{11} =∥ x_{1} ∥= 3, r_{12} = u_{1}^{T} x_{2} = \frac{5}{3}, r_{22} =∥ y_{2} ∥= \sqrt{\frac{18}{81}} = \frac{\sqrt{2}}{3}.

So Q = ( \begin{matrix}\frac{2}{3} & −\frac{1}{\sqrt{18}} \\ \frac{1}{3} & \frac{4}{\sqrt{18}} \\ \frac{2}{3} & −\frac{1}{\sqrt{18}} \end{matrix} ) and R = ( \begin{matrix}3 & \frac{5}{3} \\ 0 & \frac{\sqrt{2}}{3} \end{matrix} ).

Problem 5: Find the least-squares approximate-solution of the inconsistent system

( \begin{matrix}1 & 1 \\ 2 & −3 \\ 0 & 0 \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix} ) = ( \begin{matrix}3 \\ 1 \\ 2 \end{matrix} ) .

What is the error with this approximation?

Multiply A^{T} by the augmented matrix [A|b] to get

( \begin{matrix}1 & 2 & 0 \\ 1 & −3 & 0 \end{matrix} ) ( \begin{matrix}1 & 1 & 3 \\ 2 & −3 & 1 \\ 0 & 0 & 2 \end{matrix} ) = ( \begin{matrix}5 & −5 & 5 \\ −5 & 10 & 0 \end{matrix} ) .

Row operation A_{2}^{1×1} leads to ( \begin{matrix}5 & −5 & 5 \\ 0 & 5 & 5 \end{matrix} ). This gives y = 1 and then x = 2.

Multiply by A to get approximation ( \begin{matrix}3 \\ 1 \\ 0 \end{matrix} ), so error is ( \begin{matrix}0 \\ 0 \\ 2 \end{matrix} ), of length 2.