Math 310: Hour Exam 2 (Solutions) Prof. S. Smith: Fri 7 Apr 2000

You must SHOW WORK to receive credit.

Wherever you use a calculator, write “used calculator”.

Problem 1:

(a) Find a basis for the kernel of the linear transformation L : R^{3} → R^{2} defined by L((x_{1}, x_{2}, x_{3})^{T}) = (x_{1} − 2x_{2}, x_{2} − x_{3})^{T}.

Equations: x_{1}−2x_{2} = 0 = x_{2}−x_{3}; free variable: x_{3}; solutions (2x_{3}, x_{3}, x_{3})^{T}; so basis (2, 1, 1)^{T}.

(b) Show that the mapping L(A) = A^{T}, where A^{T} is the transpose of a 2 × 2 matrix A, is a linear transformation on the space R^{2×2}.

(addition) L(A + B) = (A + B)^{T} = A^{T} + B^{T} = L(A) + L(B)

(scalar mult.) L(cA) = (cA)^{T} = c(A^{T}) = cL(A).

Problem 2:

(a) Give the matrix A representing (in the standard basis) the linear transformation L : R^{2} → R^{2} defined by L((x_{1}, x_{2})^{T}) = (2x_{1} + x_{2}, 4x_{1} − 5x_{2})^{T}.

Apply L to standard basis, put into columns to get A = ( \begin{matrix}2 & 1 \\ 4 & −5 \end{matrix} )

(b) Now give the matrix B for the same L as in part (a), but using the basis (1, 1)^{T}, (1, −1)^{T}. Either compute directly with respect to this “new” basis; or use change-of-basis matrix

from “new” to “old” basis given by S = ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ), and mutliply out B = S^{−1}AS:

( \begin{matrix}1 & 5 \\ 2 & −4 \end{matrix} ) = \frac{1}{2} ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ) ( \begin{matrix}2 & 1 \\ 4 & −5 \end{matrix} ) ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} )

Problem 3:

(a) In R^{3} with the usual dot product, determine the vector projection of (1, 2, 1) on (2, 1, 2).

This is \frac{(1, 2, 1) · (2, 1, 2)}{(2, 1, 2) · (2, 1, 2)}(2, 1, 2) = \frac{6}{9}(2, 1, 2) = (\frac{4}{3}, \frac{2}{3}, \frac{4}{3}).

(b) Let S be the span of the vectors (1, 2, 1) and (2, 1, 2). Find the orthogonal complement S^{⊥}. Put vectors in as rows of matrix A, solve Ax = 0:

( \begin{matrix}1 & 2 & 1 & 0 \\ 2 & 1 & 2 & 0 \end{matrix} ) →^{A_{2}^{−2×1}} ( \begin{matrix}1 & 2 & 1 & 0 \\ 0 & −3 & 0 & 0 \end{matrix} ) →^{M_{− \frac{1}{3} ×1}} ( \begin{matrix}1 & 2 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{matrix} ) →^{A_{1}^{−2×2}} ( \begin{matrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{matrix} )

to conclude S^{⊥} consists of the vectors (a, 0, −a) = a(1, 0, −1).

Problem 4: The system Ax = b given by:

( \begin{matrix}1 & 2 \\ 2 & 1 \\ 1 & 2 \end{matrix} ) ( \begin{matrix}x_{1} \\ x_{2} \end{matrix} ) = ( \begin{matrix}1 \\ 1 \\ 3 \end{matrix} )

is inconsistent. Find the least-squares solution x̂, and the projection p of b into the column space of A. Also find the error-vector for this approximation, and the size of that error.

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

( \begin{matrix}1 & 2 & 1 \\ 2 & 1 & 2 \end{matrix} ) ( \begin{matrix}1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 2 & 3 \end{matrix} ) = ( \begin{matrix}6 & 6 & 6 \\ 6 & 9 & 9 \end{matrix} ), and solve to get x̂ = ( \begin{matrix}0 \\ 1 \end{matrix} ).

Then p = Ax̂ = ( \begin{matrix}2 \\ 1 \\ 2 \end{matrix} ), so the error is b − p = ( \begin{matrix}−1 \\ 0 \\ 1 \end{matrix} ), of length \sqrt{2}.

Problem 5:

(a) Let S be the subspace of R^{3} spanned by v_{1} = (0, 1, −1) and v_{2} = (1, −1, 1). Use the Gram-Schmidt process to find an orthonormal basis for S.

First get orthogonal: use q_{1} = v_{1} = (0, 1, −1) and then

q_{2} = v_{2} − [(v_{2} · q_{1})/(q_{1} · q_{1})]q_{1} = (1, −1, 1) − [−2/2](0, 1, −1) = (1, 0, 0).

To make orthoNORMAL, divide by lengths to get u_{1} = \frac{1}{\sqrt{2}}(0, 1, −1) and u_{2} = (1, 0, 0).

(b) Now find an orthonormal basis for the orthogonal complement S^{⊥}, for S in (a).

Find S^{⊥} as in problem (3b): ( \begin{matrix}0 & 1 & −1 & 0 \\ 1 & −1 & 1 & 0 \end{matrix} ) →^{E_{1,2}} ( \begin{matrix}1 & −1 & 1 & 0 \\ 0 & 1 & −1 & 0 \end{matrix} ) →^{A_{1}^{1×2}} ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & −1 & 0 \end{matrix} )

Thus solutions have form (0, a, a), with basis (0, 1, 1). Divide by length to get \frac{1}{\sqrt{2}}(0, 1, 1).