## Tag Archives: real numbers

### Minkowski’s Criterion

Suppose $A$ is an $n \times n$ matrix with real entries such that the diagonal entries are all positive, off diagonal entries are all negative, and the row sums are all positive. Prove that $\mathsf{det}(A) \neq 0$.

Suppose to the contrary that $\mathsf{det}(A) = 0$. Then there must exist a nonzero solution $X$ to the matrix equation $AX = 0$. Let $X = [x_1\ \cdots\ x_n]^\mathsf{T}$ be such a solution, and choose $k$ such that $|x_k|$ is maximized. Using the triangle inequality, we have $|\sum_j |a_{i,j}|x_j| \leq \sum_j |a_{i,j}||x_j| \leq \sum_j|a_{i,j}||x_k|$. Recall that a consequence of the triangle inequality is that $|a|-|b| \leq |a-b|$ for all $a$ and $b$. Here, we have $|a_{k,k}x_k| - \sum_{j \neq k} |a_{k,j}||x_k| \leq |a_{k,k}x_k| - |\sum_{j \neq k} |a_{k,j}|x_j| \leq |a_{k,k}x_k - \sum_{j \neq k} |a_{k,j}|x_j| = \sum_j a_{k,j}x_j$. On the other hand, $|a_{k,k}x_k| - \sum_{j \neq k} |a_{k,j}||x_k| = |x_k|(\sum_j a_{k,j}) > 0$. Thus $\sum_j a_{k,j}x_j > 0$, a contradiction since $AX = 0$.

Thus $\mathsf{det}(A) \neq 0$.

### Compute the matrix of a linear transformation

Consider $V = \mathbb{R}^2$ as an $\mathbb{R}$-vector space in the usual way. Let $\varphi$ be the linear transformation $V \rightarrow V$ which rotates the plane counterclockwise about the origin by an angle $\theta$. Compute the matrix of $\varphi$ with respect to the standard basis on $V$.

Note that we can write $\theta$ as $\theta^\prime + k\frac{\pi}{2}$, where $k$ is in $\{0,1,2,3\}$ and $0 \leq \theta < \frac{\pi}{2}$. Moreover, if we let $\theta_\alpha$ be the transformation which rotates by $\alpha$, then $\varphi_{\alpha+\beta} = \varphi_\alpha + \varphi_\beta$. Thus we will consider separately the cases $0 \leq \theta < \frac{\pi}{2}$ and $\theta = \frac{\pi}{2}$. First suppose $0 \leq \theta < \frac{\pi}{2}$.

To compute $M(\varphi_\theta)$, we express $\varphi_\theta(1,0)$ and $\varphi_\theta(0,1)$ in terms of the standard basis. Using some basic trig, we see that $\varphi_\theta(1,0) = (\cos \theta, \sin \theta)$ and $\varphi_\theta(0,1) = (\sin \theta, \cos \theta)$. (See the diagram below. Note that the angles are not to scale.)

a cruddy diagram (click for full size)

Thus $M(\varphi_\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.

It is easy to see that $\varphi_{\pi/2}(1,0) = (0,1)$ and $\varphi_{\pi/2}(0,1) = (-1,0)$. Thus $M(\varphi_{\pi/2}) = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.

Suppose $\zeta = \theta + \frac{\pi}{2}$, where $0 \leq \theta < \frac{\pi}{2}$. Now $M(\varphi_\zeta) = M(\varphi_\theta) \cdot M(\varphi_{\pi/2})$ $= \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \cdot \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ $= \begin{bmatrix} -\sin \theta & -\cos \theta \\ \cos \theta & -\sin \theta \end{bmatrix}$ $= \begin{bmatrix} \cos \zeta & -\sin \zeta \\ \sin \zeta & \cos \zeta \end{bmatrix}$, where we made use of the trig identities $\sin(\theta + \frac{\pi}{2}) = \cos \theta$ and $\cos(\theta + \frac{\pi}{2}) = -\sin \theta$.

Similarly, $M(\varphi_\zeta) = \begin{bmatrix} \cos \zeta & -\sin \zeta \\ \sin \zeta & \cos \zeta \end{bmatrix}$ for all $0 \leq \zeta < 2\pi$.

### As vector spaces over QQ, RR is isomorphic to any finite direct power of itself

Prove that as vector spaces over $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{R}^n$ are isomorphic for any positive natural number $n$.

Note that $\mathbb{Q}$ is countable. By this previous exercise, any basis of $\mathbb{R}$ over $\mathbb{Q}$ must have cardinality $\mathsf{card}\ \mathbb{R}$. Likewise, any basis of $\mathbb{R}^n$ over $\mathbb{Q}$ must have cardinality $\mathsf{card}\ \mathbb{R}^n = \mathsf{card}\ \mathbb{R}$.

Thus, as vector spaces over the rationals, $\mathbb{R} \cong_\mathbb{Q} \bigoplus_\mathbb{R} \mathbb{Q} \cong_\mathbb{Q} \mathbb{R}^n$. In particular, $\mathbb{R}$ and $\mathbb{R}^n$ are isomorphic as abelian groups.

### Prove that a subset of a vector space is a subspace and find a basis

Let $V = \mathbb{R}^n$ and let $(a_i) \in V$ be a fixed vector. Prove that the set $W = \{ (x_i) \in V \ |\ \sum a_ix_i = 0 \}$ is a subspace of $V$. Determine the dimension of $W$ and find a basis.

First we show that $W$ is a subspace of $V$. Note that $0 \in W$, since $\sum a_i0 = 0$. Now let $(x_i), (y_i) \in W$ and let $r \in \mathbb{R}$. Now $\sum a_ix_i = 0$ and $\sum y_ia_i = 0$. Consider $(x_i) + r(y_i) = (x_i + ry_i)$; we have $\sum (x_i+ry_i)a_i = (\sum x_ia_i) + r(\sum y_ia_i) = 0+r0 = 0$, so that $(x_i) + r(y_i) \in W$. By the submodule criterion, $W \subseteq V$ is an $\mathbb{R}$-subspace.

If $(a_i) = 0$, then clearly $W = V$, so that $\mathsf{dim}\ W = n$ and we may take the standard basis for $W$.

Suppose now that $(a_i) \neq 0$, with $a_k \neq 0$. Now define vectors $w_i \in V$, where $i \neq k$, as follows: $w_i = (w_{i,j})$, where $w_{i,j} = 1$ if $j = i$, $\frac{-a_j}{a_k}$ if $j = k$, and 0 otherwise. We claim that the set $E = \{e_i \ |\ 1 \leq i \leq n, i \neq k \}$ is a basis for $W$.

First, note that for each $i$, $\sum a_je_{i,j} = a_i - a_k\frac{a_i}{a_k} = 0$, so that $E \subseteq W$. Now suppose $\sum r_ie_i = 0$. Now $0 = \sum_i r_ie_i$ $= \sum_i r_i(e_{i,j})$ $= (\sum_i r_ie_{i,j})$. Consider the $i$th component of this vector, where $i \neq k$; $r_ie_{i,j}$ is 0 if $j \neq i$, so that $r_i = 0$. Thus the $e_i$ are linearly independent.

Finally, suppose $(x_i) \in W$. Then $\sum a_ix_i = 0$. Rearranging, we see that $x_k = \frac{-1}{a_k} \sum_{i \neq k} a_ix_i$. Evidently, then, $(x_i) = \sum_{i \neq k} x_ie_i$.

$E$ is a linearly independent generating set for $W$, hence a basis. In particular, $W$ has dimension $n-1$.

### An inequality involving sqrt(2)

Prove that for all $a,b \in \mathbb{Z}^+$, $| \sqrt{2} - \frac{a}{b} | \geq \frac{1}{3b^2}$.

Let $p(x) = x^2 - 2$.

First we will show that the inequality holds for $b = 1$. Note that $|\sqrt{2} - \frac{1}{1}| = 0.414 + \epsilon > 1/3$ and $|\sqrt{2} - \frac{2}{1}| = 0.585 + \epsilon > 1/3$. For $a > 2$, we have $|\sqrt{2} - \frac{a}{1}| > 1 > 1/3$. So the inequality holds for $b = 1$. Henceforth, we will assume that $b \geq 2$.

Suppose $|\sqrt{2} - \frac{a}{b}| \geq \frac{3 - 2\sqrt{2}}{2}$. Note that since $b \geq 2$, $b^2 \geq 4 > 3.88 + \epsilon = \frac{2}{9 - 6\sqrt{2}}$; so $\frac{3 - 2\sqrt{2}}{2} > \frac{1}{3b^2}$. Hence $|\sqrt{2} - \frac{a}{b}| \geq \frac{1}{3b^2}$.

Now suppose $|\sqrt{2} - \frac{a}{b}| \leq \frac{3 - 2\sqrt{2}}{2}$. By the Mean Value Theorem from calculus (which we will assume to be valid), there exists an element $\xi$ between $\sqrt{2}$ and $a/b$ such that $p^\prime(\xi) = \frac{p(\sqrt{2}) - p(a/b)}{\sqrt{2} - a/b}$, and hence $|p^\prime(\xi)| = \frac{|p(\sqrt{2}) - p(a/b)|}{|\sqrt{2} - a/b|}$. Now $\xi \in [\sqrt{2} - \frac{3-2\sqrt{2}}{2}, \sqrt{2} + \frac{3-2\sqrt{2}}{2}] = [1.328+\epsilon, 1.5]$. Since $p^\prime(x) = 2x$ is strictly increasing, we have $|p^\prime(\xi)| \leq 3$. Since $p(\sqrt{2}) = 0$, we have $|p(a/b)| \leq 3|\sqrt{2} - \frac{a}{b}|$.

Note that $p(a/b) \neq 0$, since (for example) $p(x)$ is irreducible over $\mathbb{Q}$. Now $|p(a/b)| = |\frac{a^2}{b^2} - 2|$ $= \frac{|a^2 - 2b^2|}{b^2}$. Since $p(a/b) \neq 0$, $a^2 - 2b^2$ is a nonzero integer. In particular, we have $|a^2 - 2b^2| \geq 1$, so that $|p(a/b)| \geq \frac{1}{b^2}$.

Hence $|\sqrt{2} - \frac{a}{b}| \geq \frac{1}{3b^2}$.

### As a ring, the ZZ-tensor product of ZZ[i] and RR is isomorphic to CC

Let $\mathbb{Z}[i]$ and $\mathbb{R}$ be $\mathbb{Z}$-algebras via the inclusion map. Prove that as rings, $\mathbb{Z}[i] \otimes_\mathbb{Z} \mathbb{R}$ and $\mathbb{C}$ are isomorphic.

Define $\varphi : \mathbb{Z}[i] \times \mathbb{R} \rightarrow \mathbb{C}$ by $\varphi(z,x) = zx$. Certainly this mapping is $\mathbb{Z}$-bilinear, and so induces a $\mathbb{Z}$-algebra homomorphism $\Phi : \mathbb{Z}[i] \otimes_\mathbb{Z} \mathbb{R} \rightarrow \mathbb{C}$ such that $\Phi(z \otimes x) = zx$.

Note that every simple tensor (hence every element) of $\mathbb{Z}[i] \otimes_\mathbb{Z} \mathbb{R}$ can be written in the form $1 \otimes x + i \otimes y$, where $x,y \in \mathbb{R}$. Now $\Phi(1 \otimes x + i \otimes y) = x+iy$, so that $\Phi$ is surjective. Suppose now that $0 = \Phi(1 \otimes x + i \otimes y) = x+iy$; then $x = y = 0$, so that $1 \otimes x + i \otimes y = 0$. Thus $\mathsf{ker}\ \Phi = 0$, and so $\Phi$ is injective. Thus $\Phi$ is a ring isomorphism.

### Every polynomial in RR[x] which takes only nonnegative values is a sum of two squares

Let $p(x) \in \mathbb{R}[x]$ be a polynomial such that $p(c) \geq 0$ for all $c$. Prove that $p(x) = a(x)^2 + b(x)^2$ for some polynomials $a,b \in \mathbb{R}[x]$.

Note that $p(x)$ must have even degree. We proceed by induction on the degree of $p(x)$. Note as a lemma that if $h = a^2+b^2$ and $k = c^2 + d^2$ are sums of squares, then so is $hk$ since (evidently) $hk = (ac-bd)^2 + (ad+bc)^2$. We find this identity by rearranging and partially simplifying the factorization $hk = (a+bi)(a-bi)(c+di)(c-di)$.

The base case $p(x) = c$ is trivial; $p(x) = 0^2 + 0^2$ if $c = 0$ and $p(x) = (\sqrt{c}/2)^2 + (\sqrt{c}/2)^2$ if $c \neq 0$.

For the inductive step, suppose the result holds for all polynomials of degree $n$ and let $p(x)$ have degree $n+2$. Suppose $p(x)$ has a real root $c$. Now $p(x)$ is concave up on a sufficiently small neighborhood about $c$, so that $p^\prime(c) = 0$. In particular, $c$ is a root of $p$ of multiplicity at least 2. Say $p(x) = q(x)(x-c)^2$. Now $q(x)$ is a sum of two squares, and $(x-c)^2 = (x-c)^2 + 0^2$ is as well, so that by the lemma $p(x)$ is a sum of two squares. Suppose instead that $p(x)$ has no real roots. Instead, we have a complex root $z$. Since conjugation is a ring homomorphism, $\overline{z}$ is also a root of $p$. Letting $z = a+bi$, we see that $x^2+2ax + a^2+b^2 = (x+a)^2 + b^2$ is a factor of $p(x)$; say $p(x) = q(x)((x+a)^2 + b^2)$. Again, $p(x)$ is a sum of two squares.

### The real numbers contain a maximal subring not containing 1/2

Prove that the real numbers $\mathbb{R}$ contain a subring $A$ with $1 \in A$ and such that $A$ is inclusion-maximal with respect to the property that $1/2 \notin A$.

Let $\mathcal{C}$ be the set of all subrings of $\mathbb{R}$ containing 1 and not containing $1/2$; since $\mathbb{Z}$ contains 1 and not 1/2, $\mathcal{C}$ is not empty. Now let $\{C_i\}_\mathbb{N}$ be a chain in $\mathcal{C}$, and let $C = \bigcup C_i$. Recall that $C$ is a subring of $\mathbb{R}$. Note that $1 \in C$ since $1 \in C_0$. Suppose $1/2 \in C$; then we have $1/2 \in C_i$ for some $i$, a contradiction. Thus $C \in \mathcal{C}$, and thus $C$ is an upper bound for the chain $\{C_i\}_\mathbb{N}$. By Zorn’s Lemma, $\mathcal{C}$ contains a maximal element $A$ as desired.

### Every subfield of the real numbers must contain the rational numbers

Prove that any subfield of $\mathbb{R}$ must contain $\mathbb{Q}$.

By the previous exercise, $\mathbb{R}$ contains a unique inclusion-smallest subfield which is isomorphic either to $\mathbb{Z}/(p)$ for a prime $p$ or to $\mathbb{Q}$.

Suppose the unique smallest subfield of $\mathbb{R}$ is isomorphic to $\mathbb{Z}/(p)$, and let $a$ in this subfield be nonzero. Then $pa = 0$ in $\mathbb{R}$, and since $p \in \mathbb{R}$ is a unit, $a = 0$, a contradiction.

Thus the unique smallest subfield of $\mathbb{R}$ is isomorphic to $\mathbb{Q}$. In particular, any subfield of $\mathbb{R}$ contains a subfield isomorphic to $\mathbb{Q}$.

### Embed the Hamiltonian quaternions in a ring of real matrices

Prove that the ring $M_4(\mathbb{R})$ contains a subring isomorphic to the real Hamiltonian quaternions $\mathbb{H}$.

Define $\varphi : \mathbb{H} \rightarrow M_4(\mathbb{R})$ as follows.

 $a+bi+cj+dk \mapsto \begin{bmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{bmatrix}$

We will show that this mapping is an injective ring homomorphism. To that end, let $\alpha = a_1 + b_1i + c_1j + d_1k$ and $\beta = a_2 + b_2i + c_2j + d_2k$. Then we have the following.

 $\varphi(\alpha + \beta)$ = $\varphi((a_1 + b_1i + c_1j + d_1k)+(a_2 + b_2i + c_2j + d_2k))$ = $\varphi((a_1 + a_2) + (b_1+b_2)i + (c_1+c_2)j + (d_1+d_2)k)$ = $\begin{bmatrix} a_1+a_2 & b_1+b_2 & c_1+c_2 & d_1+d_2 \\ -b_1-b_2 & a_1+a_2 & -d_1-d_2 & c_1+c_2 \\ -c_1-c_2 & d_1+d_2 & a_1+a_2 & -b_1-b_2 \\ -d_1-d_2 & -c_1-c_2 & b_1+b_2 & a_1+a_2 \end{bmatrix}$ = $\begin{bmatrix} a_1 & b_1 & c_1 & d_1 \\ -b_1 & a_1 & -d_1 & c_1 \\ -c_1 & d_1 & a_1 & -b_1 \\ -d_1 & -c_1 & b_1 & a_1 \end{bmatrix} + \begin{bmatrix} a_2 & b_2 & c_2 & d_2 \\ -b_2 & a_2 & -d_2 & c_2 \\ -c_2 & d_2 & a_2 & -b_2 \\ -d_2 & -c_2 & b_2 & a_2 \end{bmatrix}$ = $\varphi(a_1 + b_1i + c_1j + d_1k) + \varphi(a_2 + b_2i + c_2j + d_2k)$ = $\varphi(\alpha) + \varphi(\beta)$
 $\varphi(\alpha\beta)$ = $\varphi((a_1 + b_1i + c_1j + d_1k)(a_2 + b_2i + c_2j + d_2k))$ = $\varphi((a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2)$ $+ (a_1b_2 + b_1a_2 + c_1d_2 - d_1c_2)i$ $+ (a_1c_2 - b_1d_2 + c_1a_2 + d_1b_2)j$ $+ (a_1d_2 + b_1c_2 - c_1b_2 + d_1a_2)k)$ = $\begin{bmatrix} a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2 & a_1b_2 + b_1a_2 + c_1d_2 - d_1c_2 & a_1c_2 - b_1d_2 + c_1a_2 + d_1b_2 & a_1d_2 + b_1c_2 - c_1b_2 + d_1a_2 \\ -a_1b_2 - b_1a_2 - c_1d_2 + d_1c_2 & a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2 & -a_1d_2 - b_1c_2 + c_1b_2 - d_1a_2 & a_1c_2 - b_1d_2 + c_1a_2 + d_1b_2 \\ -a_1c_2 + b_1d_2 - c_1a_2 - d_1b_2 & a_1d_2 + b_1c_2 - c_1b_2 + d_1a_2 & a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2 & -a_1b_2 - b_1a_2 - c_1d_2 + d_1c_2 \\ -a_1d_2 - b_1c_2 + c_1b_2 - d_1a_2 & -a_1c_2 + b_1d_2 - c_1a_2 - d_1b_2 & a_1b_2 + b_1a_2 + c_1d_2 - d_1c_2 & a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2 \end{bmatrix}$ = $\begin{bmatrix} a_1 & b_1 & c_1 & d_1 \\ -b_1 & a_1 & -d_1 & c_1 \\ -c_1 & d_1 & a_1 & -b_1 \\ -d_1 & -c_1 & b_1 & a_1 \end{bmatrix} \cdot \begin{bmatrix} a_2 & b_2 & c_2 & d_2 \\ -b_2 & a_2 & -d_2 & c_2 \\ -c_2 & d_2 & a_2 & -b_2 \\ -d_2 & -c_2 & b_2 & a_2 \end{bmatrix}$ = $\varphi(a_1 + b_1i + c_1j + d_1k) \cdot \varphi(a_2 + b_2i + c_2j + d_2k)$ = $\varphi(\alpha) \cdot \varphi(\beta)$

Thus $\varphi$ is a ring homomorphism.

Suppose now that $\alpha = a+bi+cj+dk \in \mathsf{ker}\ \varphi$; clearly, then, we have $a = b = c= d = 0$, so that $\alpha = 0$. Thus $\varphi$ is injective.

By Proposition 5 in the text, $\mathsf{im}\ \varphi$ is a subring of $M_4(\mathbb{R})$ which is isomorphic to $\mathbb{H}$.