FREQUENTLY ASKED QUESTIONS, MATH 340

JIM MORROW

(1) My office is Padelford, C-439 and my office hours are MW 8:30-9:30 or by appointment.

(2) My email address is morrow@math.washington.edu
(3) My homepage is http://www.math.washington.edu/~morrow/personal.html

(4) Assignment IV is due on April 17

(5) Assignment V is due on April 19

(6) Assignment VI is due on April 21

(7) The first midterm is scheduled for April 24

(8) The test will cover through §5.4

(9) More details on §4.3, # 1a. Look at Example 3, page 90 to see how to work this problem. Here is a slightly different explanation. We want to compute M_{B′}^{B}(id). Let B = {⌊_{∞}, ⋯, ⌊_{\}}, B′ = {⌊′_{∞}, ⋯, ⌊′_{\}}. Then

b_{j} = a_{1j}b′_{1} + ⋯ a_{nj}b′_{n}

Notice the difference between this notation and that of Example

3. Consider the vectors b_{j}, b′_{j} to be column vectors in R^{n}. Let B = [b_{1}, ⋯, b_{n}], B′ = [b′_{1}, ⋯, b′_{n}]. Also let A = (a_{ij}). Then the relationship between B and B′ is B = B′A. Hence

M_{B′}^{B}(id) = A = (B′)^{−1}B

Let me say this in words. Find id(b_{1}) = b_{1} as a linear combination of the b′_{j}’s. Put the coefficients of this combination in the first column of A. Find id(b_{2}) = b_{2} as a linear combination of the b′_{j}’s. Put the coefficients of this combination in the second column of A. Continue doing this to find A. It is easier to remember what to do than to remember the formula.

(10) Here is the argument that {(u_{1}, 0), ⋯, (u_{n}, 0), (0, v_{1}), ⋯, (0, v_{m})} is a a basis for U × V if {u_{1}, ⋯, u_{n}} is a basis for U and {v_{1}, ⋯, v_{m}} is a basis for V. Let (u, v) be a vector in U × V. Then u = ∑ a_{i}u_{i}, v = ∑ b_{j}v_{j}, so (u, v) = ∑ a_{i}(u_{i}, 0) + ∑ b_{j}(0, v_{j}). This proves that the vectors span. If ∑ a_{i}(u_{i}, 0) + ∑ b_{j}(0, v_{j}) = 0, then ∑ a_{i}u_{i} = 0, ∑ b_{j}v_{j} = 0, hence a_{i} = 0, i = 1, ⋯, n, b_{j} = 0, j = 1, ⋯, m, since the u_{i} and v_{j} are linearly independent.

(11) Some of you are confused (not without reason) by the book’s notation X_{B}(v). This notation means the column vector whose components are the coordinates β_{1}, ⋯, β_{n} of the representation of v as a sum, ∑ β_{j}b_{j}, of the elements of B. Suppose v ∈ R^{n} and b_{j} ∈ R^{n}, j = 1, ⋯, n. Let B = [b_{1}, ⋯, b_{n}] be the n × n matrix whose columns are the b_{j}’s. Then it is easy to check that

X_{B}(v) = B^{−1}v

Thus corollary 3.2 on page 88 states that

(B′)^{−1}v = M_{B′}^{B}(id)B^{−1}v

Hence we get another derivation of the relation

M_{B′}^{B}(id) = (B′)^{−1}B

(12) There are many easy examples of scalar products that are \underline{not} positive definite. For example use the matrix

A = [\begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix}]

This corresponds to the scalar product −x_{1}y_{1} − x_{2}y_{2}.

(13) More on projections. Let w be a vector in a vector space V. Let W be a subspace with basis {v_{1}, ⋯, v_{k}}. Let <, > be a positive definite scalar product on V. Then we can define the orthogonal projection, P, onto W as follows. P will be of the form P (v) = ∑_{1}^{k} a_{k}(v)v_{k} where v − P (v) is orthogonal to all vectors in W. This is equivalent to < v − P (v), w >= 0 for all w ∈ W. This in turn is equivalent to

0 =< v − P (v), v_{j} >=< v, v_{j} > − ∑_{i=1}^{k} a_{i} < v_{i}, v_{j} >, j = 1, ⋯, k

This gives a system of equations for the a_{i}’s,

∑_{i=1}^{k} a_{i} < v_{i}, v_{j} >=< v, v_{j} >, j = 1, ⋯, k

This is a k × k system which is always invertible if {v_{1}, ⋯, v_{k}} is a linearly independent set (exercise, left to you to verify). Thus the a_{j}’s are uniquely determined and in turn, determined the projection P. If {v_{1}, ⋯, v_{k}} is an orthonormal set then the system of equations is very simple. It is just a_{j}(v) < v_{j}, v_{j} >=< v, v_{j} >, j = 1, ⋯, k. In other words when {v_{1}, ⋯, v_{k}} is an orthonormal set,

P (v) = ∑ a_{i}v_{i}, a_{i} = \frac{< v, v_{i} >}{< v_{i}, v_{i} >}

The moral is that orthogonal projection is easy to describe in an orthonormal basis, but complicated in a general basis.

(14) The words scalar product and inner product mean the same thing (at least in this course).

(15) The midterm is closed book. No notes are allowed.