Recently, I have been thinking and learning a lot about algebraic geometry. I decided to write about the Zariski topology in my final project for my introductory algebraic geometry class. You can find the paper here.
I mainly discuss some of the basic topological properties of the Zariski topology on affine space. These are all elementary properties of the Zariski topology, but it is difficult to find a source anywhere actually listing these properties with proof, so I found it to still be instructive to think about the proofs. One remarkable thing is that one can use the Zariski topology to prove the Cayley-Hamilton theorem, which is something I wrote about. I essentially filled in the details from here. The proof essentially shows that the collection of operators that have distinct eigenvalues is dense with respect to the Zariski topology. Since affine space is compact with that topology, the result really tells us that the collection of operators with distinct eigenvalues is precompact. This is a result reminiscent of the Arzelà-Ascoli theorem, which led me to wonder if affine space is sequentially compact under the Zariski topology. My hunch is that this is false. I also briefly allude to the Zariski topology on the spectrum of a ring, but I don't really say much of substance about this. Over the summer, I plan on fleshing out this section and making precise the relationship between the topology on affine space and the topology on the spectrum of a ring. Based on what I've worked through in Atiyah-MacDonald, this relationship is not very easy to state.
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Here is a problem from Chapter 1 of B. C. Hall. Lie groups, Lie algebras and Representations that I am working on for my REU reading.
Problem: A subset \(E\) of a matrix Lie group \(G\) is called discrete if for each \(A\) in \(E\) there is a neighborhood \(U\) of \(A\) in \(G\) such that \(U\) contains no point in \(E\) except for \(A\). Suppose that \(G\) is a path-connected matrix Lie group and \(N\) is a discrete normal subgroup of \(G\). Show that \(N\) is contained in the center of \(G\). Solution: To solve this problem, I was first immediately reminded of the map \(\Phi_X(A)\colon N\to G\) given by \(\Phi_X(A)=XAX^{-1}\) that Hall introduces earlier in the chapter, where we have \(X\in G\) and \(A\in N\). This map is obviously continuous in both \(X\) and \(A\) due to the nature of matrix multiplication. More motivation for thinking of this map comes from the fact that \(\Phi_X\) actually has its image as a subset of \(N\) since \(N\) is normal. I saw that if \(A\) was in the center, it must be true that \(\Phi_X(A)=XX^{-1}A=A\) for any choice of \(X\in G\). Moreover, the converse is true, since \(XAX^{-1}=A\) implies \(XA=AX\). So it sufficed to show that \(XAX^{-1}=A\). A common construction in path-connected abstract topological spaces is to draw paths connecting a point of interest to an "important point" in the space. I have seen this construction before in algebraic topology, where numerous proofs involving path-lifting requires paths to be drawn from a chosen point in the fiber (which determines the lift), to some other point of interest. In our case, \(G\) is a topological group, and an important element of any group is the identity. So we let \(\lambda_X\colon[0,1]\to G\) to be the path in \(G\) such that \(\lambda_X(0)=X\) and \(\lambda_X(1)=I\). Now define \(f\colon[0,1]\to G\) given by \[f(t)=\Phi_{\lambda_X(t)}(A).\] What I had to show was clear: the function \(\Phi\) evaluated at \(A\) must be constant along the path \(\lambda_X(t)\). In other words, \(f(t)\) needed to be the constant function with image \(A\). At this point, I began fiddling with a literal notion of "closeness". One may put the metric induced by the Hilbert-Schmidt (Frobenius) norm on \(G\) to turn it into a metric space, and do some \(\epsilon\)-\(\delta\) calculations to determine that in fact \(f(t)=A\) for all \(t\). But I later realized that this is not a nice solution, since we are imposing additional (unnecessary) structure on \(G\). It turns out that we can prove the claim entirely topologically. First note that as a composition of continuous functions, \(f\) is continuous. Since \(N\) is discrete, around each \(Y\in\mathrm{Im}\ f\), there exists an open neighborhood \(U_Y\subseteq G\) that intersects \(N\) only at \(Y\). By the continuity of \(f\), the sets of the form \(f^{-1}(U_Y)\) are open in \([0,1]\) (with the subspace topology inherited from \(\mathbb{R}\)) and form an open cover of \([0,1]\), which is compact by the Heine-Borel theorem. So we can extract a finite subcover. But the sets in the cover are preimages, so they are pairwise disjoint. The only way to write \([0,1]\) as a finite union of pairwise disjoint sets that are open in \([0,1]\) is if \([0,1]\) is the only (nonempty) set in the union. So one of the preimages is \([0,1]\), the entire domain of \(f\). That is, there is only one element in the image of \(f\). Since \(f(1)=IAI=A\), this element is \(A\). \(\square\) It is easy to see that the set of all \(3\times3\) real matrices, which I will denote by \(\mathbb{M}_{3\times3}\) is a vector space. Consider \(H\subset\mathbb{M}_{3\times3}\) defined by
\[H=\{A\in\mathbb{M}_{3\times3}\ \text{such that }\det{(A+A^T)}=0\}.\] Is \(H\) a subspace of \(\mathbb{M}_{3\times3}\)? It turns out, the answer is no. \(H\) has a zero element and is closed under scalar multiplication, but it is not closed under vector addition. It is sufficient to find \(A,B\in H\) with \(A+B\notin H\). Easy, right? Just take \[A=\begin{bmatrix} 1 & 2 & 1\\ 2 & 1 & 2\\ 1 & 2 & 1 \end{bmatrix}\] \[B=\begin{bmatrix} \sqrt{2} & \sqrt{3} & 2\sqrt{2}\\ \sqrt{3} & \sqrt{2} & \sqrt{3}\\ 2\sqrt{2} & \sqrt{3} & \sqrt{2} \end{bmatrix},\] and we are done. \(\square\) As satisfying it would be to just leave these matrices without any trace of where they come from, it contradicts the mantra: a slick write-up is not an instructional one. The first thing to observe is that a matrix plus its transpose is always a symmetric matrix. This is easy to show (addition on real numbers is commutative). So the condition \(\det{(A+A^T)}=0\) is really saying that a certain symmetric matrix related to \(A\) is singular. Moreover, if a matrix is already symmetric, then it is equal to its transpose. This means that the sum of itself with its transpose is equal to two times itself. That is, if \(A\) is a symmetric, \(A+A^T=2A\). But determinants are multilinear (linear on each column vector). So \(\det{2A}=2^n\det{A}\), if \(A\) is an \(n\times n\) matrix. This means that if we choose a symmetric \(A\in\mathbb{M}_{3\times3}\) that is singular, then \(A\in H\), since \(\det{(A+A^T)}=\det{2A}=8\det{A}=0\). In other words, if we choose a symmetric matrix from \(H\), instead of worrying about the sum of that matrix with its transpose, we can just focus on the matrix itself (that is, if \(A\) is symmetric, then \(\det{(A+A^T)}=0\) if and only if \(\det{A}=0\)). So let's investigate symmetric \(3\times3\) matrices. They look like this \[S=\begin{bmatrix} a & b & c\\ b & d & e\\ c & e & f \end{bmatrix}.\] Full disclosure: I stumbled upon this SE post which inspired me to consider the special case of a \(3\times3\) symmetric matrix where \(d=f=a\) and \(e=b\): \[S=\begin{bmatrix} a & b & c\\ b & a & b\\ c & b & a \end{bmatrix}.\] We compute the determinant of such a matrix. \[\begin{split} \det{S}&=a(a^2-b^2)-b(ab-bc)+c(b^2-ac)\\ &=a(a^2-b^2)-b^2(a-c)+b^2c-ac^2\\ &=a^3-b^2a+b^2c-ac^2-b^2(a-c)\\ &=a(a+c)(a-c)-b^2(a-c)-b^2(a-c)\\ &=(a-c)(a^2+ac-2b^2). \end{split}\] In essence, there are three ways that this special case symmetric matrix can be singular. Either only the first factor is equal to zero, only the second factor is equal to zero, or both factors are equal to zero. The sum of two symmetric matrices is also a symmetric matrix, so applying our previous reasoning, instead of worrying about the determinant \(\det{(A+B+(A+B)^T)}\), if we select \(A\) and \(B\) to be symmetric, we need to only worry about \(\det{(A+B)}\). Now watch what happens if we let \(A\) and \(B\) be special-case symmetric matrices in \(H\) with \(A\) having only the first factor in its determinant equal to zero and \(B\) having only the second factor in its determinant equal to zero. In other words, write \[A=\begin{bmatrix} a & b & c\\ b & a & b\\ c & b & a \end{bmatrix}\] \[B=\begin{bmatrix} x & y & z\\ y & x & y\\ z & y & x \end{bmatrix}\] with \(a-c=0\neq a^2+ac-2b^2\), and \(x^2-xz-2y^2=0\neq x-z\). In that case, \[A+B=\begin{bmatrix} p & q & r\\ q & p & q\\ r & q & p \end{bmatrix},\] where \(p=a+x\), \(q=b+y\), and \(r=c+z\). Clearly then it follows that \[p-r=a+x-c-z=x-z\neq0,\] \[p^2-pr-2q^2=(a+x)^2-(a+x)(c+z)-2(b+y)^2=a^2+ac-2b^2+2ax-az-cx-4by=\xi.\] So as long as we choose \(a,b,c,x,y,z\) such that \(\xi\) is nonzero, the determinant of \(A+B\) is guaranteed to be nonzero! This is easy to accomplish. Setting \(a=c=1\) and \(b=2\) satisfies our conditions for the matrix \(A\). Setting \(x=\sqrt{2}\), \(y=\sqrt{3}\), and \(z=2\sqrt{2}\) satisfies our conditions for matrix \(B\), and some quick algebra verifies that \(\xi\neq0\). The result follows! Definitely a fun problem. I reckon there is a simpler solution. My apartment-mate found a counterexample against the closure of \(H\) under addition by trying random things on Matlab. I haven't yet looked into why that solution works, and its corresponding generalization. Stick around for updates! Here is an interesting problem that has surprisingly far-reaching consequences.
Consider a permutation \(\sigma\colon S\to S'\) of some list, say \(S=\{1,2,\dots,n\}\) that maps \(S\mapsto\{k_1,k_2,\dots,k_n\}\), where \(k_i\in S\). Define a transposition to be a permutation that only swaps two elements of the list it acts on. It is clear that every permutation can be decomposed into transpositions. Seriously. It is intuitively obvious. But if we wanted to be a bit more rigorous, observe that we can give a state to every element \(j_i\in S\). Let \(j_i\) be correct if \(j_i=\sigma(j_i)\), that is, \(j_i\) is a fixed point under the permutation. Let \(j_i\) be incorrect otherwise. For every incorrect element in \(S\), there must exist another incorrect element in \(S\) such that the transposition of the two elements makes the former element correct. So, we can iteratively exclude correct elements, find the unique elements that are in the correct spot of each incorrect element, and perform the transpositions. Since \(S\) has finite cardinality, this algorithm terminates, and in fact leaves every element correct. So any permutation can be decomposed into transpositions. But this does not say anything about the decomposition itself. In fact, there are an infinite number of sequences of transpositions that when composed, yield a given permutation. This is trivial, since we don't have to necessarily stop after the aforementioned algorithm terminates and everything is in the correct spot. You could perform another transposition to "mess things up" and then reverse it, and then continue that an arbitrary number of times. However, a powerful invariant hides behind all of these decompositions. For a given permutation, the number of transpositions in any decomposition of that permutation always has the same parity. This intuitively makes a lot of sense, though it's not very clear how to prove it. I suppose it would perhaps be possible to do so by performing a case analysis with my states (by considering transpositions of the form "incorrect/correct to correct/incorrect", "correct/correct to incorrect/incorrect", etc.). But this is a bit ugly. There is a much more elegant way, though it is arguably more complicated if you are required to prove the preliminaries from scratch as well. This parity of a permutation is called its signature, and we are concerned with its existence and uniqueness. The signature of a permutation is \(1\) when a permutation can only be decomposed into an even number of transpositions and \(-1\) otherwise. But there is actually an equivalent definition of signature that we can give with which it is much easier to probe the questions of existence and uniqueness. Let \(P_n\) denote the set of all permutations that act on lists with cardinality \(n\). Then the signature is defined as the function \[\varsigma\colon P_n\to\{-1,1\}\] such that \(\varsigma(\sigma_1\circ\sigma_2)=\varsigma(\sigma_1)\varsigma(\sigma_2)\) for all \(\sigma_1,\sigma_2\in P_n\) and \(\varsigma(\tau)=-1\) if \(\tau\) is a transposition. The signature is usually denoted by "sgn", but due to MathJax limitations, I will denote it with \(\varsigma\). We need to show that this definition is equivalent to our parity definition, and also that the signature exists and is unique (using this definition). Showing the equivalence of the definitions is trivial. Since the signature of the composition is the product of the signatures, and the signature of a transposition is \(-1\), we must have that the signature is \(\varsigma(\sigma)=(-1)^{T(\sigma)}\), where \(T(\sigma)\) gives the number of transpositions in any one of the decompositions of \(\sigma\). Now we must prove that this function exists and is unique. Proving existence is the hard part. For any permutation \(\sigma\in P_n\), define the permutation matrix \(M_{\sigma}\) as the matrix that maps \[M_{\sigma}\vec{e}_i=\vec{e}_{\sigma(i)},\] where \(\vec{e}_i\) is just a standard basis vector in \(\mathbb{R}^n\). It follows that \(M_{\sigma}\) is some \(n\times n\) matrix which only 0's and 1's, which I won't compute explicitly (though it isn't hard to). It is easy to see for \(\sigma_1,\sigma_2\in P_n\), we have \(M_{\sigma_1\circ\sigma_2}=M_{\sigma_1}M_{\sigma_2}\). Here comes our leap of faith. Define \[\varsigma(\sigma)=\det{M_{\sigma}}.\] This fits our first property, since the determinant of a product is the product of the determinants! Furthermore, the second property is satisfied as well, since one can easily show that for some transposition \(\tau\), the matrix \(M_{\tau}\) is found by transposing two columns in the identity matrix (and so the determinant is negated, due to the antisymmetry of the determinant – this argument is sometimes phrased in terms of the determinant of an elementary matrix representing the transposition of two columns). Hence, the function \(\varsigma\) exists. But, if \(\varsigma\), a function, exists, then each element in the domain can only map to one element in the codomain, so it is also unique. By equivalence of definitions, this means that the number of transpositions of every decomposition of \(\sigma\) into transpositions always has the same parity! The signature ends up being quite important for various definitions in the theory of forms in vector calculus, such as the wedge product. All of this from just viewing a permutation as a series of transpositions. There is a notion of direct sums that I've known, but it has always bothered me that I never really had a feel for what it really is. Here I will attempt to construct an analogy between a sum of vector spaces and a union of sets, and in doing so, construct an analogy between a direct sum of vector spaces and a disjoint union of sets.
It is already well-known (to me) that the dimension of a vector space is in some sense a measure of how large that vector space is. This is made apparent by this incorrect proof (see the comment to see why the argument is flawed) of the fact that if a linear transformation is injective, then the dimension of the domain must be at most the dimension of the codomain, and that if the linear transformation is surjective, then the dimension of the domain must be at least the dimension of the codomain. While these statements are easily proven with the rank-nullity theorem, the approach in the incorrect proof imply that, for instance, if \(\dim{V}>\dim{W}\), then \(|V|\geq|W|\). So in some sense, the dimension represents how "large" a vector space is. This plays out in the following problem. Problem: Suppose \(U_1,\dotsc,U_m\) are finite-dimensional subspaces of \(V\). Prove that \(U_1+\dots+U_m\) is finite-dimensional and \[\dim{(U_1+\dots+U_m)}\leq\dim{U_1}+\dots+\dim{U_m}.\] Solution: Any vector in \(U_1+\dots+U_m\) can be written as a sum of vectors from each subspace, which themselves can be written as a linear combination of basis vectors of the respective subspaces. So vector can be written as a linear combination of all of these basis vectors. But across subspaces, the basis vectors may not be linearly independent, so we can extract a basis across all of the subspaces from the union of the sets of basis vectors from each subspace, since any spanning set contains a basis. The conclusion follows. \(\square\) We make a key observation in this solution. The strict inequality condition of the problem statement is that a basis vector in one subspace is not linearly independent to other basis vectors in the union of the sets of basis vectors from each subspace. That is, the equality condition is that all of the basis vectors across the subspaces are linearly independent. As it turns out, this is precisely what occurs only when the sum of the subspaces is a direct sum. That is, a direct sum of subspaces is analogous to disjoint unions of sets, in the sense that the union of the sets of basis vectors from the subspaces are still linearly independent. We can use this idea to prove one direction. Problem: Suppose \(U_1,\dotsc,U_m\) are finite-dimensional subspaces of \(V\) such that \(U_1+\dots+U_m\) is a direct sum. Prove that \(U_1\oplus\dots\oplus U_m\) is finite-dimensional and \[\dim{(U_1\oplus\dots\oplus U_m)}=\dim{U_1}+\dots+\dim{U_m}.\] Solution: Since the sum is direct, when we write a vector in the sum as a linear combination of all of the basis vectors across all subspaces, no basis vectors collapse (otherwise, we contradict the fact that the sum is direct). So all of these basis vectors are linearly independent, and they span the sum, so they form a basis for the sum and the conclusion follows. \(\square\) It turns out that the converse is also true. To prove this direction, we need to release that when we take the union of the sets of basis vectors for each subspace and we put them all together to form a vector in the sum of subspaces, we are essentially taking vectors in the Cartesian product of the subspaces and mapping them to the sum of the subspaces. There are few quick things that we can easily verify about the Cartesian product of vector spaces. The first, is that it itself is a vector space. Secondly, the dimension Cartesian product of the vector spaces is the sum of the dimensions of the vector spaces. Both of these are easy to check. Now, we can prove the other direction. In fact, we will prove both in a much cleaner manner. Theorem: Suppose \(V\) is finite-dimensional and \(U_1,\dotsc,U_m\) are subspaces of \(V\). Then \(U_1+\dots+U_m\) is a direct sum if and only if \[\dim{(U_1+\dots+U_m)}=\dim{U_1}+\dots+\dim{U_m}.\] Proof: Consider the linear transformation \(\Gamma\colon U_1\times\dots\times U_m\to U_1+\dots+U_m\) such that it maps \((u_1,\dotsc,u_m)\mapsto u_1+\dots+u_m\). \(\Gamma\) is injective if and only if the only vector in its kernel is the zero vector. So there is only one way to write the zero vector in the sum of subspaces, and that is by choosing the zero vector from each subspace. This is a well-known necessary and sufficient condition for a sum to be direct. So \(\Gamma\) is injective if and only if the sum of subspaces is direct. \(\Gamma\) is obviously surjective. By the theorem we discussed in the very beginning, regarding the dimensions of the domain and codomain of injective linear transformations and surjective linear transformations (or alternatively, by the rank-nullity theorem), we have that \(\Gamma\) is injective iff the dimension of the sum equals the dimension of the Cartesian product. But the dimension of the Cartesian product is the sum of the dimensions. The result follows. \(\square\) Here is something I never thought about.
Additivity and homogeneity are conditions for a linear transformation for a reason. That is, it's possible for a function to be homogeneous but not additive and to be additive but not homogeneous. For example, consider \(\varphi\colon\mathbb{R}^2\to\mathbb{R}\) defined by \(\varphi(x,y)=\sqrt{xy}\). This is homogeneous but not additive. Alternatively, for \(\varphi\colon\mathbb{C}\to\mathbb{C}\), we can consider the map \(a+bi\mapsto b+ai\). This is additive but not homogeneous! In fact, it can be shown that an additive but not homogeneous function exists \(\mathbb{R}\to\mathbb{R}\), though the proof is nonconstructive and uses the axiom of choice. Meanwhile, I learned the proof of the Fundamental Theorem of Algebra just a few weeks ago. Very tricky, but very interesting. If anything I think I can get a cool diagram out of it. Sorry for the inactivity. I have just started my studies at the University of California, San Diego.
Here is a method I've learned from Introductory Classical Mechanics to solve a certain system of ODEs. Consider a system of ODEs of the form: \[\left\{\begin{array}{ll}\ddot{x}+\omega^2(ax+by)=0\\\ddot{y}+\omega^2(cx+dy)=0\\\end{array}\right.\] Such a system often arises from coupled oscillations. To solve such equations, we suppose there exists solutions of the form \(x=Ae^{i\alpha t}\) and \(y=Be^{i\alpha t}\). Substituting these solutions into the the differential equations, we obtain the system of equations \[\left\{\begin{array}{ll}(a\omega^2-\alpha^2)A+b\omega^2B\\c\omega^2A+(d\omega^2-\alpha^2)B\\\end{array}\right.\] We wish to find nontrivial solutions \(A,B\) (which in this case means nonzero solutions). It is easier to see the condition for the independent nontrivial solution if we write the system in matrix form: \[\begin{bmatrix} a\omega^2-\alpha^2 & b\omega^2 \\ c\omega^2 & d\omega^2-\alpha^2 \end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \] Let \(\mathbf{M}\) be the first matrix and \(\mathbf{N}\) be the second matrix in the above equation. Then, if \(\mathbf{M}\) is invertible, then multiplying through with its inverse would yield \(\mathbf{N}=0\), the solutions that we wish to avoid. Hence, if nontrivial solutions are to exist, \(\mathbf{M}\) cannot be invertible. The inverse of \(\mathbf{M}\) is the adjugate matrix of \(\mathbf{M}\) divided by the determinant of \(\mathbf{M}\). So if we wish for this to not exist, the determinant of \(\mathbf{M}\) must be zero. That is, \[\begin{vmatrix} a\omega^2-\alpha^2 & b\omega^2 \\ c\omega^2 & d\omega^2-\alpha^2 \end{vmatrix}=\alpha^4-(a+d)\omega^2\alpha^2+(ad-bc)\omega^4=0.\] We can view this multivariate polynomial as biquadratic in \(\alpha\). We solve the biquadratic by letting \(\beta=\alpha^2\) and then using the quadratic formula. \[\beta=\frac{1}{2}\left[a+d\pm\sqrt{(a-d)^2+4bc}\right]\omega^2.\] Hence, we have four solutions for \(\alpha\): \[\alpha=\pm\frac{\omega}{\sqrt{2}}\sqrt{a+d\pm\sqrt{(a-d)^2+4bc}}.\] Suppose that these solutions are \(\pm R\omega\) and \(\pm S\omega\). We can then obtain \[A=-\frac{b}{a-R^2}B\] \[A=-\frac{b}{a-S^2}B,\] We can now use the principle of superposition to write the general solution as \[\begin{bmatrix} x \\ y \end{bmatrix}=A_1\begin{bmatrix} 1 \\ -\frac{b}{a-R^2} \end{bmatrix}e^{iR\omega t}+A_2\begin{bmatrix} 1 \\ -\frac{b}{a-R^2} \end{bmatrix}e^{-iR\omega t}+A_3\begin{bmatrix} 1 \\ -\frac{b}{a-S^2} \end{bmatrix}e^{iS\omega t}+A_4\begin{bmatrix} 1 \\ -\frac{b}{a-S^2} \end{bmatrix}e^{-iS\omega t}\] Where the coefficients \(A_i\) are determined by initial conditions. Depending on the \(A_i\), we can condense the exponentials into, for instance, sinusoidal functions. This sort of system of ODEs appears most often with coupled oscillation. This post was way overdue. Sorry. UC San Diego is keeping me busy. I'll grind this weekend to catch up on stuff on this website. Here are some problems from Multivariable and Vector Calculus by David Santos.
Varignon's Theorem: The quadrilateral formed by midpoints of sides of any quadrilateral is a parallelogram. Proof: We proceed by vectors. Let our arbitrary quadrilateral be \(ABCD\), and let \(W\) be the midpoint of \(\overline{AB}\), \(X\) be the midpoint of \(\overline{BC}\), etc. Observe that it suffices to show that \(\mathbf{ZW}+\mathbf{ZY}=\mathbf{ZX}\). We see that \(\mathbf{ZW}=\mathbf{ZA}+\mathbf{AW}\). Furthermore, \(\mathbf{ZY}=\mathbf{ZD}+\mathbf{DY}=-\mathbf{ZA}+\mathbf{DY}\). Adding these two equations yields: \[\mathbf{ZW}+\mathbf{ZY}=\mathbf{AW}+\mathbf{DY}\] We take a look at \(\mathbf{AW}\) and find \(\mathbf{AW}=\mathbf{WB}\) and \(\mathbf{WB}=\mathbf{WX}-\mathbf{BX}\), hence \(\mathbf{AW}=\mathbf{WX}-\mathbf{BX}\). In a similar manner we may show that \(\mathbf{DY}=\mathbf{YX}-\mathbf{CX}=\mathbf{YX}+\mathbf{BX}\). Therefore: \[\mathbf{AW}+\mathbf{DY}=\mathbf{YX}+\mathbf{WX}\] And so: \[\mathbf{ZW}+\mathbf{ZY}=\mathbf{YX}+\mathbf{WX}\] We're almost done. We add \(\mathbf{YX}+\mathbf{WX}\) to both sides to obtain: \[\mathbf{ZW}+\mathbf{WX}+\mathbf{ZY}+\mathbf{YX}=2\left(\mathbf{YX}+\mathbf{WX}\right)\] This simplifies to: \[2\mathbf{ZX}=2\left(\mathbf{YX}+\mathbf{WX}\right)=2\left(\mathbf{ZW}+\mathbf{ZY}\right)\] And dividing by \(2\), we obtain the desired result. \(\square\) Problem: Let \(X\), \(Y\), and \(Z\) be points on the plane with \(X\neq Y\). Demonstrate that the point \(A\) belongs to \(\overleftrightarrow{XY}\) if and only if there exists scalars \(\alpha,\beta\) with \(\alpha+\beta=1\) such that: \[\mathbf{ZA}=\alpha\mathbf{ZX}+\beta\mathbf{ZY}\] Solution: We define \(\overleftrightarrow{XY}\) to be the standard horizontal axis. Then, the component of any vector from \(Z\) to \(\overleftrightarrow{XY}\) orthogonal to \(\overleftrightarrow{XY}\) must be a constant. Hence we may write: \[\mathbf{ZX}=\left<a,k\right>\] \[\mathbf{ZY}=\left<b,k\right>\] Suppose \(\exists\alpha,\beta\) such that \(\mathbf{ZA}=\alpha\mathbf{ZX}+\beta\mathbf{ZY}\) and \(\alpha+\beta=1\). Then it follows that we may write: \[\mathbf{ZA}=\left<\alpha a+\beta b,(\alpha+\beta)k\right>\] But since \(\alpha+\beta=1\), this becomes: \[\mathbf{ZA}=\left<\alpha a+\beta b,k\right>\] Since the component orthogonal to \(\overleftrightarrow{XY}\) of \(\mathbf{ZA}\) is equal to the constant \(k\), \(A\) must lie on \(\overleftrightarrow{XY}\). The converse of the above is also true. Suppose that \(A\) does lie on \(\overleftrightarrow{XY}\). Then: \[\mathbf{ZA}=\left<c,k\right>\] It is possible to uniquely determine \(\alpha,\beta\) that satisfy: \[\left\{\begin{array}{ll}\alpha a+\beta b=c\\\ \alpha+\beta=1\\\end{array}\right.\] By solving the system of equations above. A unique solution exists because we are given \(X\neq Y\) and thus \(a\neq b\), so the system is independent. Those values of \(\alpha\) and \(\beta\) permit the relation: \[\mathbf{ZA}=\alpha\mathbf{ZX}+\beta\mathbf{ZY}\] Hence, \(A\) lies on \(\overleftrightarrow{XY}\) iff \(\alpha\) and \(\beta\) exist as described. \(\square\) |
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