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Winter 2012 Final CC-BY-NC

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Winter 2012 disastrously hard closed book MATH 223 final, by Wilbur Jonsson. No solutions yet, feel free to fill them in.

The original PDF is available through Docuum.

1Question 1

1.1Question

For each of the following subsets, decide whether it is or is not a subspace of the given vector space (justify your answers using the three part subspace criterion):

  • (a) The subset $\left \{z \in \mathit{C} \mid \left \vert z \right \vert \leq 1 \right \}$ of the complex vector space $\mathit{C}$.
  • (b) The subset of the real vector space of real valued functions of one variable consisting of differentiable functions.
  • (c) The subset of the complex vector space of polynomials with complex coefficients consisting of those polynomials all of whose roots in $\mathit{C}$ are distinct.
  • (d) The subset of the real vector space of polynomials with complex coefficients consisting of those polynomials of degree at most 10.
  • (e) The intersections of any collection of subspaces of a vector space V.

1.2Solution

Recall that the three part subspace criterion for $W \leq V$ with $V, W $ over $ \mathbb{F}$ is the following:

  1. $0 \in W$
  2. $u,v \in W \to u+v \in W$
  3. $u \in W, c \in \mathbb{F} \to cu \in W$

(a)

Let $z_1, z_2 \in \mathbb{C}$ such that $z_1, z_2 \in W$. Then,

$$\begin{cases} |z_1| \leq 1\\ |z_2| \leq 1 \end{cases} \qquad \to \qquad |z_1|+|z_2| \leq 2 \quad \text{ but }\quad |z_1|+|z_2| \leq 1 \quad \text{not guaranteed}$$

For example, take a counterexample $z_1 = z_2 = 0.7$. Then, $z_1, z_2 \in W$ since $|z_1| \leq 1$ and $|z_2| \leq 1$, but $|z_1| + |z_2| = 1.4 > 1$ meaning $|z_1| + |z_2| \not \in W$. This violates criterion 2.

$W$ is not a subspace

(b)

$W = \{f(x) \in \mathbb{F} \ | \ f'(x) \in \mathbb{F}\}$ where $\mathbb{F} =$ real valued functions of one variable

  1. $f(x) = 0 \in W$
  2. Let $f(x), g(x) \in W$, then $(f(x) + g(x))' = f'(x) + g'(x)$. Both $f'(x)$ and $g'(x)$ are defined given $f, g$ are differentiable, so their sum is as well. Therefore, $f(x) + g(x) \in W$.
  3. Let $f(x) \in W, c \in \mathbb{R}$, then $(cf(x))' = cf'(x)$. $f'(x)$ is defined and so is any of its multiples. Therefore, $cf(x) \in W$.

$W$ is a subspace

(c)

$W = \{f(x) = k(x-z_1)(x-z_2)\ldots(x - z_k) \in \mathbb{F} \ | \ z_1 \neq z_2 \neq \ldots \neq z_k, \quad k, z_1, z_2, \ldots z_k \in \mathbb{C}\}$ where $\mathbb{F} =$ complex vector space of polynomials with complex coefficients

Let $z_1 \neq z_2 \neq z_3$ be complex numbers and $f(x) = (x-z_1)(x-z_2), g(x) = (x-z_1)(x-z_3)$ so that $f(x), g(x) \in W$ since their roots are distinct.

Then,
$$ f(x) + g(x) = (x-z_1)(x-z_2)+(x-z_1)(x-z_3) = (x-z_1)(x-z_2 +x -z_3) = 2(x-z_1)(x - \frac{z_2 + z_3}{2}) $$

So, it is sufficient to choose $z_2 \neq z_3$ such that $z_1 = \frac{z_2 + z_3}{2}$ to show $f(x) + g(x) \not \in W$ and violate criterion 2.

$W$ is not a subspace

(d)

$W = \{p(x) = a_0 + a_1 x + \ldots + a_{10}x^{10} \ | a_i \in \mathbb{R}\}$

  1. $p(x) = 0 \in W$ (zero polynomial)
  2. (I'll combine it with criterion 3)

Let $f(x), g(x) \in W$, then $cf(x) + g(x) = c(a_0 + a_1 x + \ldots + a_{10}x^{10}) + (b_0 + b_1 x + \ldots + b_{10}x^{10}) = (ca_0 + b_0) + (ca_1 + b_1)x + \ldots + (ca_{10} + b_{10})x^{10} \in W$ since the coefficients are real.

$W$ is a subspace

(e)

$W = \{\cap_{i \in S} U_i \ | \ U_i \leq V \}$ with $S = \{1,2, \ldots \}$

Note: $\leq$ is Pr. Jonsson's notation for subspace.

  1. $\ 0 \in U_i \quad \forall i \in S$ since this is part of a subspace's definition, therefore $0$ is also in their intersection, so $0 \in W$
  2. Let $u, v \in W$, therefore $u$ and $v$ are in every subspace $U_i$ of V. Since subspaces are closed under addition (from criterion 2 precisely), then $u+v$ is also in every subspace $U_i$ of V hence, $u+v \in W$
  3. A similar argument can be made here.

$W$ is a subspace

1.3Accuracy and discussion

Solution written by @eleyine. I still need someone to verify it.

Looks right - @dellsystem

2Question 2

2.1Question

Consider the polynomial $f(x)=x^3-x^2-5x-3=(x-3)(x+1)^2$ and define $U_f = \left \{ f(x)g(x) \mid g(x) \in \mathit{F} \left [ x \right ] \right \}$ whereby $\mathit{F} \left [ x \right ]$ consists of all polynomials in the indeterminate x with coefficients from the field $\mathit{F}$.

  • (a) Find polynomials $a(x)$, $b(x)$ such that $a(x)(x-3) + b(x)(x+1)^2 = 1$
  • (b) Show that the quotient space $V/U_f$ is the internal direct sum of $\text{Image} \hspace{2mm} T_{x-3}$ and $\text{Image} \hspace{2mm} T_{(x+1)^2}$ with the notation used in class. Why is $a(T)$ invertible on $\text{Image} \hspace{2mm} T_{(x-3)}$ ?
  • (c) Find a basis for the factor space such that the matrix of the induced linear transformation has the form

$$\begin{pmatrix} 3 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{pmatrix}$$

2.2Solution

None

2.3Accuracy and discussion

N/A

3Question 3

3.1Question

  • (a) Consider the set $\mathit{V}$ of all sequences of elements of the field $\mathit{F}$ as vectors with countably many components, e.g. $x= \left ( x_1, x_2, ..., x_k, ... \right )$ with all $x_j \in \mathit{F}$ is a typical such vector.
    Show that the sequences satisfying the recurrence relation $x_{n+2} = ax_{n+1} + bx_n$ for a two dimensional subspace of $\mathit{V}$.
  • (b) With the two dimensional subspace of the previous part of this question in mind, solve the following difference equation explicitly by reducing it to a problem about the eigenvalues of a two by two matrix. (That is, find an explicit, nonrecursive formula for $x_n$.)

$$x_1 = 5, \hspace{5mm} x_2 = 3, \hspace{5mm} x_{n+2} = 3x_{n+1} + 4x_n \hspace{2mm} \text{for} \hspace{2mm} n\geq1$$

3.2Solution

(a)

We first note that $\mathbf{Z} = (1,0,x_3,...), \mathbf{Y} = (0,1,x_3,...)$ are a basis for sequences satisfying $x_{n+2} = \alpha x_{n+1}+ \beta x_n$ since $x_1 Z+x_2Y = (x_1, x_2, x_3, \ldots) = \mathbf{x}$. Now, to prove such sequences (that we call $W$) form a subspace of $V$, we use the three part criterion.

  1. $\mathbf{0} = (0,0,\ldots) \in W$ obviously because it satisfies the recurrence relation $x_{n+2} = \alpha x_{n+1}+ \beta x_n \leftrightarrow 0 = \alpha \cdot 0+ \beta \cdot 0$.
  2. Let $\mathbf{l} = (l_1, l_2, \ldots , l_n, \ldots) \in W$ and $ \mathbf{m} = (m_1, m_2, \ldots , m_n, \ldots) \in W$. Then, $l_{n+2} = \alpha l_{n+1} + \beta l_{n}, \quad m_{n+2} = \alpha m_{n+1} + \beta m_{n}$. What of $(l_{n+2} + m_{n+2})?$
    $$l_{n+2} + m_{n+2} = (\alpha l_{n+1} + \beta l_{n}) + (\alpha m_{n+1} + \beta m_{n}) = \alpha(l_{n+1}+ m_{n+1}) + \beta (l_n + m_{n}) \qquad \to \mathbf{l + m} = (l_1+m_1, l_2+m_2, \ldots , l_n+m_n, \ldots) \in W$$
  3. Similarly,
    $$cl_{n+2} = c(\alpha l_{n+1} + bl_{n})= \alpha (cl_{n+1})+ \beta (cl_n) \qquad \to \mathbf{cl} = (cl_1, cl_2, \ldots , cl_n, \ldots) \in W \qquad \blacksquare$$

(b)

We first express the recurrence relation as a matrix transformation.

$$ \begin{bmatrix}x_{n+2} \\ x_{n+1} \end{bmatrix} = \begin{bmatrix}3 & 4 \\ 1 & 0 \end{bmatrix} \begin{bmatrix}x_{n+1} \\ x_{n} \end{bmatrix} $$

We then find the eigenvalues of the corresponding matrix.

$$ det(A - \lambda I) = \begin{vmatrix}3-\lambda & 4 \\ 1 & -\lambda \end{vmatrix} = (3-\lambda)(-\lambda) - 4 = \lambda^2 - 3\lambda - 4 = (\lambda + 1)(\lambda -4) $$

Then, the explicit formula is given by $x_n = \alpha \lambda_1^n + \beta \lambda_2^n = \alpha (-1)^n + \beta (4)^n$ (magic, I guess?).

We find the values $\alpha, \beta$ using $x_1, x_2$ by solving the system:

$$ \begin{cases} x_1 = \alpha \lambda_1^1 + \beta \lambda_2^1 = \alpha (-1) + \beta (4) \\ x_2 = \alpha \lambda_1^2 + \beta \lambda_2^2 = \alpha (-1)^2 + \beta (4)^2 \end{cases} \quad \leftrightarrow \quad \begin{cases} 5 = -\alpha + 4 \beta \\ 3 = \alpha + 16 \beta \end{cases} $$

which has solutions $\alpha = 5, \beta = \frac{5}{2}$.

So, $x_n = 5 \lambda_1^n + \frac{5}{2} \lambda_2^n$

3.3Accuracy and discussion

Solution by @eleyine. Someone should check accuracy.

4Question 4

4.1Question

  • (a) State and prove Cramer's rule for solving a system on $n$ equations in $n$ unknowns based on the three axioms given in class for the definition of the determinant function.
  • (b) Given two three by three matrices $\mathit{A}$ and $\mathit{B}$ with entries from a field $\mathit{F}$, define
    $$\mathit{C} = \begin{pmatrix} \mathit{A} & \mathbf{0} \\ \mathit{-I} & \mathit{B} \end{pmatrix}$$
    whereby $\mathbf{0}$ is a three by three block of zeros and $\mathit{I}$ is the tree by three identity matrix.
    • i. Using only column and row operations, show that $\det \mathit{C} = -\det \begin{pmatrix} \mathit{-I} & \mathit{B} \\ \mathbf{0} & \mathit{AB} \end{pmatrix}$.
    • ii. Using column operations and/or the row cofactor expansion, conclude $\det \mathit{A} \hspace{1mm} \det \mathit{B} = \det \mathit{C} = \det \left (\mathit{AB} \right )$

4.2Solution

(a) To prove Cramer's rule, use two properties:

  1. adding (or subtracting) a multiple of a column to another does not change the determinant
  2. $\det\begin{bmatrix} a_1, & \ldots, & b a_j, & \ldots, a_n \end{bmatrix} = b \det(A)$ where $A = \begin{bmatrix} a_1, & \ldots, & a_j, & \ldots, a_n \end{bmatrix}$

$$\begin{aligned} \begin{pmatrix} a_{1,1} & a_{1,2} & \dots & a_{1,j} & \dots & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & a_{2,j} & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n,j} & \dots & a_{n,n} \end{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ \vdots \\ x_n \end{pmatrix} &= \begin{pmatrix} b_1 \\ \vdots \\ \vdots \\ b_n \end{pmatrix} \\ \begin{vmatrix} a_{1,1} & a_{1,2} & \dots & b_{1} & \dots & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & b_{2} & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & b_{n} & \dots & a_{n,n} \end{vmatrix} &= \begin{vmatrix} a_{1,1} & a_{1,2} & \dots & a_{1,1}x_1 + a_{1,2}x_2 + \dots + a_{1,j}x_j+ \dots + a_{1,n}x_n & \dots & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & a_{2,1}x_1 + a_{2,2}x_2 + \dots + a_{2,j}x_j+ \dots + a_{2,n}x_n & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n,1}x_1 + a_{n,2}x_2 + \dots + a_{n,j}x_j+ \dots + a_{n,n}x_n & \dots & a_{n,n} \end{vmatrix} \\ &= \begin{vmatrix} a_{1,1} & a_{1,2} & \dots & a_{1,j}x_j & \dots & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & a_{2,j}x_j & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n,j}x_j & \dots & a_{n,n} \end{vmatrix} \qquad \text{using property 1} \\ &= x_j \begin{vmatrix} a_{1,1} & a_{1,2} & \dots & a_{1,j} & \dots & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & a_{2,j} & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n,j} & \dots & a_{n,n} \end{vmatrix} \qquad \text{using property 2} \end{aligned} $$

We obtain Cramer's rule by dividing each side by $\det \ A$.

4.3Accuracy and discussion

Cramer's rule proof by @eleyine.

5Question 5

5.1Question

  • (a) Let $\mathit{V} = \mathit{C}^4$ as an inner product over the complex numbers $\mathit{C}$ with the usual inner product and define $\mathit{W} = \text{Span} \left \{ \left ( 1, i, -1, i \right )^t, \left ( -i, 1, i, -1 \right )^t \right \}$
    • i. Find an orthonormal basis for $\mathit{W}^\perp$, the orthogonal complement of W.
    • ii. Extend this basis to an othonormal basis of $\mathit{V}$.
  • (b) Let $\mathit{U}$ be an inner product space over the complex numbers $\mathit{C}$ and $\mathit{H}$ a Hermitian opertor $\mathit{H} : \mathit{U} \to \mathit{U}$. Prove that eigenvectors of $\mathit{H}$ for distince eigenvalues are orthogonal.

5.2Solution

None

5.3Accuracy and discussion

N/A