Math 310: Hour Exam 2 (Solutions) Prof. S. Smith: Mon 8 Nov 1999

You must SHOW WORK to receive credit.

Problem 1:

(a) Are the columns of the matrix ( \begin{matrix}1 & 1 & 0 \\ 1 & 2 & 1 \\ 1 & 3 & 2 \end{matrix} ) linearly independent? (Why/why not?)

No. One way: det(A) = 0.

(b) Give a basis for the column space of the matrix A from part (a). What is the rank of A ? Notice by part (a) that the rank is not 3. But any pair of columns is LI, and so would give a

basis; or you can find other possible bases. In particular, the rank of A is 2.

Problem 2:

(a) For a 2 × 2 matrix A, define the symmetrization Sym(A) to be \frac{1}{2}(A + A^{T}), where T denotes transpose. (As the name suggests, the symmetrization is in fact a symmetric matrix). Show that Sym is a linear transformation (from the matrix space R^{2×2} into that same space).

(for +:) Sym(A+B) = \frac{1}{2}((A+B)+(A+B)^{T}) = \frac{1}{2}(A+A^{T})+ \frac{1}{2}(B +B^{T}) = Sym(A)+Sym(B). (for sc.mult.:) Sym(cA) = \frac{1}{2}(cA + (cA)^{T}) = c · \frac{1}{2}(A + A^{T}) = c · Sym(A).

(b) Give the matrix representing (in the standard basis) the linear transformation L : R^{3} → R^{3} defined by L(x_{1}, x_{2}, x_{3}) = (2x_{1} + 3x_{2} − x_{3}, −2x_{2} − x_{3}, 4x_{1} + x_{2} − 3x_{3}).

Apply L to standard basis, put into columns to get ( \begin{matrix}2 & 3 & −1 \\ 0 & −2 & −1 \\ 4 & 1 & −3 \end{matrix} )

Problem 3:

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

This is \frac{(1, 1, 1) · (1, 2, 3)}{(1, 2, 3) · (1, 2, 3)}(1, 2, 3) = \frac{6}{14}(1, 2, 3).

(b) Let P_{1} denote the space of linear polynomials (ax + b for a, b ∈ R); with inner product given by ⟨f(x), g(x)⟩ = ∫_{0}^{1} f(x)g(x) dx. Let S be the subspace spanned by the constant function 1; determine the orthogonal complement S^{⊥} to S in P_{∞}.

For general ax + b, compute 0 = ⟨1, ax + b⟩ = ∫_{0}^{1} (1)(ax + b) dx = [a\frac{x^{2}}{2} + bx]_{0}^{1} = \frac{a}{2} + b.

Then a = −2b, so S^{⊥} contains polynomials −2bx + b = b(−2x + 1) for all b.

Problem 4: Find the least-squares solution x̂, and the projection p into the column space of A, and the error b − p; in the inconsistent system Ax = b given by:

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

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

( \begin{matrix}−1 & 0 & 1 \\ 1 & 1 & 1 \end{matrix} ) ( \begin{matrix}−1 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & 1 & 5 \end{matrix} ) = ( \begin{matrix}2 & 0 & 4 \\ 0 & 3 & 8 \end{matrix} ) .

Thus gives x_{1} = 2 and then x_{2} = \frac{8}{3}.

Multiply this vector by A to get projection p = \frac{1}{3} ( \begin{matrix}2 \\ 8 \\ 14 \end{matrix} ), so the error is b − p = \frac{1}{3} ( \begin{matrix}1 \\ −2 \\ 1 \end{matrix} ),

of length \sqrt{\frac{2}{3}}.

Problem 5:

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

First get orthogonal: use x_{1} = (0, 1, 1) and then x_{2} = v_{2} − [(v_{2} · x_{1})/(x_{1} · x_{1})]x_{1} = (1, 0, 2) − [2/2](0, 1, 1) = (1, −1, 1). To make orthoNORMAL, divide by lengths to get u_{1} = \frac{1}{\sqrt{2}}(0, 1, 1) and u_{2} = \frac{1}{\sqrt{3}}(1, −1, 1).

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

To find S^{⊥}, use matrix from original rows ( \begin{matrix}0 & 1 & 1 \\ 1 & 0 & 2 \end{matrix} )

and apply E_{1,2} to get row-reduced form ( \begin{matrix}1 & 0 & 2 \\ 0 & 1 & 1 \end{matrix} )

So x_{3} is free and solutions have form (−2x_{3}, −x_{3}, x_{3}); a basis is given by setting x_{3} = 1 to get (−2, −1, 1). To make orthonormal, divide by length to get u_{3} = \frac{1}{\sqrt{6}}(−2, −1, 1).