Math 310: Hour Exam 2 (Solutions) Prof. S. Smith: Mon 9 Nov 1998

You must SHOW WORK to receive credit.

Problem 1:

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

Yes. One way: det(A) = 10 ≠ 0.

(b) Give a basis for the nullspace of the 1 × 3 matrix A = (1, 1, 1).

Already row-reduced, so x_{2} and x_{3} are free: solutions of form (−x_{2} − x_{3}, x_{2}, x_{3}).

Choosing x_{2} = 1, x_{3} = 0 gives (−1, 1, 0); and x_{2} = 0, x_{3} = 1 gives (−1, 0, 1).

Problem 2:

(a) The trace of a 2 × 2 matrix A is the sum of its diagonal entries; that is, trace(A) = A_{1,1} + A_{2,2}. Show that trace is a linear transformation (from 2x2-matrix space to numbers).

(for +:) trace(A + B)=(A_{1,1} + B_{1,1}) + (A_{2,2} + B_{2,2})

trace(A)+trace(B)=(A_{1,1} + A_{2,2}) + (B_{1,1} + B_{2,2}) ...equal.

(for sc.mult.:) trace(cA)=cA_{1,1} + cA_{2,2}; while c.trace(A)=c(A_{1,1} + A_{2,2}) ...equal.

(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}) = (x_{1} − x_{2} + 2x_{3}, 2x_{1} + x_{2}, −x_{1} − 2x_{2} + 2x_{3}).

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

Problem 3:

(a) In R^{3} with the usual dot product, determine the orthogonal complement S^{⊥} to the subspace S spanned by (1, 1, 1) and (1, 2, 3).

Put in as rows of A and row reduce with A_{2}^{−1×1} to ( \begin{matrix}1 & 1 & 1 \\ 0 & 1 & 2 \end{matrix} ) .

Then x_{3} is free, and solutions have form (x_{3}, −2x_{3}, x_{3})

(b) Determine the orthogonal (vector) projection of the vector b = (1, 2, 3) in the subspace spanned by the two vectors v_{1} = \frac{1}{\sqrt{2}}(1, −1, 0) and v_{2} = \frac{1}{\sqrt{6}}(1, 1, −2).

These vectors are already an orthonormal set,

so the scalar projections are just given by the dot products b · v_{1} and b · v_{2}; then vector p = \frac{−1}{\sqrt{2}}v_{1} + \frac{−3}{\sqrt{6}}v_{2} = −\frac{1}{2}(1, −1, 0) − \frac{1}{2}(1, 1, −2) = (−1, 0, 1).

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

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

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

( \begin{matrix}1 & 0 & 2 \\ 2 & 1 & 0 \end{matrix} ) ( \begin{matrix}1 & 2 & 5 \\ 0 & 1 & −3 \\ 2 & 0 & 1 \end{matrix} ) = ( \begin{matrix}5 & 2 & 7 \\ 2 & 5 & 7 \end{matrix} ) .

Row operation A_{2}^{− \frac{2}{5} ×1} leads to ( \begin{matrix}5 & 2 & 7 \\ 0 & \frac{21}{5} & \frac{21}{5} \end{matrix} ). This gives x_{2} = 1 and then x_{1} = 1.

Multiply by A to get projection p = ( \begin{matrix}3 \\ 1 \\ 2 \end{matrix} ), so error is b − p = ( \begin{matrix}2 \\ −4 \\ −1 \end{matrix} ), of length \sqrt{21}.

Problem 5:

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

First get orthogonal: use x_{1} = (1, 0, 1) and then x_{2} = v_{2} − [(v_{2} · x_{1})/(x_{1} · x_{1})]x_{1} = (1, 1, 0) − [1/2](1, 0, 1) = \frac{1}{2}(1, 2, −1). To make orthoNORMAL, divide by lengths to get u_{1} = \frac{1}{\sqrt{2}}(1, 0, 1) and u_{2} = \frac{1}{\sqrt{6}}(1, 2, −1).

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

To ( \begin{matrix}1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} ) apply A_{2}^{−1×1} to get ( \begin{matrix}1 & 0 & 1 \\ 0 & 1 & −1 \end{matrix} )

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