Andrew Walker Problems - Revision history

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Created page with "'''Exercise''' Show that <math>\{1+i,1-i \}</math> form a linearly independent set of vectors in <math>\mathbb{C}</math>, viewed as a vector space over <math>\mathbb{R}</math>..."

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'''Exercise'''
Show that <math>\{1+i,1-i \}</math> form a linearly independent set of vectors in <math>\mathbb{C}</math>, viewed as a vector space over <math>\mathbb{R}</math>.

''Proof''
Recall that the set of vectors <math>\{v_{1},\ldots, v_{n} \}</math> in a vector space <math>V</math> (over a field <math>\mathbb{F}</math>) are said to be '''linearly independent''' if whenever <math>c_{1},\ldots,c_{n}</math> are scalars in <math>\mathbb{F}</math> such that <math>c_{1}v_{1} + \cdots + c_{n}v_{n} = 0,</math> then <math>c_{1} = \cdots = c_{n} = 0</math>. So for this problem, since we’re considering the complex numbers <math>\mathbb{C}</math> as a vector space over <math>\mathbb{R}</math>, we must show that whenever <math>c_{1},c_{2} \in \mathbb{R}</math> and <math>c_{1}(1+i) + c_{2}(1-i) = 0,</math> then <math>c_{1} = c_{2} = 0</math>. Rearranging the above equation, we obtain <math>(c_{1} + c_{2}) + (c_{1} - c_{2}) i = 0.</math> Now, a complex number is equal to <math>0</math> if and only if its real and imaginary parts are both <math>0</math>. So in this case, we conclude that <math>c_{1} + c_{2} = 0 \text{ and } c_{1} - c_{2} = 0.</math> This implies <math>c_{1} = c_{2}</math>, so that <math>c_{1} + c_{2} = 2c_{2} = 0</math>, which yields <math>c_{1} = c_{2} = 0</math>. Thus we conclude the vectors <math>1+i,1-i</math> are linearly independent in <math>\mathbb{C}</math> (over <math>\mathbb{R}</math>).

'''Exercise'''
Show that <math>\{1+i,1-i \}</math> form a linearly independent set of vectors in <math>\mathbb{C}</math>, viewed as a vector space over <math>\mathbb{R}</math>.

''Proof''
Recall that a set of vectors <math>\{v_{1},\ldots,v_{n}\}</math> in a vector space <math>V</math> (over a field <math>\mathbb{F}</math>) is said to be <math>\textbf{linearly dependent}</math> if they are not linearly independent. More concretely, these vectors are linearly dependent if we can find scalars <math>c_{1},\ldots, c_{n} \in \mathbb{F}</math> ''not all equal to zero'' such that <math>c_{1}v_{1} + \cdots + c_{n}v_{n} = 0.</math>

So for this problem, to show that <math>1+i</math> and <math>1-i</math> are not linearly dependent over <math>\mathbb{C}</math>, all we need to do is exhibit two complex scalars <math>c_{1}</math> and <math>c_{2}</math> that are not ''both'' zero such that <math>c_{1}(1+i) + c_{2}(1-i) = 0.</math> There are many choices for <math>c_{1}</math> and <math>c_{2}</math>, but one such example is <math>c_{1} = i</math> and <math>c_{2} = 1</math>.

'''Exercise'''
Let <math>V</math> be a vector space over a field <math>\mathbb{F}</math>. If <math>\{v_{1},v_{2},v_{3},v_{4}\} \subseteq V</math> are a linearly independent set of vectors, then show that <math>\{v_{1} - v_{2}, v_{2} - v_{3}, v_{3} - v_{4},v_{4}\}</math> also form a linearly independent set of vectors in <math>V</math>.

''Proof''
Recall that the set of vectors <math>\{w_{1},\ldots, w_{n} \}</math> in a vector space <math>V</math> (over a field <math>\mathbb{F}</math>) are said to be '''linearly independent''' if whenever <math>c_{1},\ldots,c_{n}</math> are scalars in <math>\mathbb{F}</math> such that <math>c_{1}w_{1} + \cdots + c_{n}w_{n} = 0,</math> then <math>c_{1} = \cdots = c_{n} = 0</math>.

'''Exercise'''
So for this problem, we must show that whenever <math>c_{1},c_{2},c_{3},c_{4} \in \mathbb{F}</math> and <math>c_{1}(v_{1} - v_{2}) + c_{2}(v_{2} - v_{3}) + c_{3}(v_{3} - v_{4}) + c_{4}v_{4} = 0,</math> we have that <math>c_{1} = c_{2} = c_{3} = c_{4} = 0.</math> After rearranging terms in the above equation, we have that <math>c_{1}v_{1} + (c_{2} - c_{1})v_{2} + (c_{3} - c_{2})v_{3} + (c_{4} - c_{3})v_{4} = 0.</math> Now since the vectors <math>\{v_{1},v_{2},v_{3},v_{4}\}</math> are linearly independent in <math>V</math> by assumption, we have that <math>
c_{1} = 0 \\
c_{2} - c_{1} = 0 \\
c_{3} - c_{2} = 0 \\
c_{4} - c_{3} = 0.
</math> In other words, <math>c_{1} = c_{2} = c_{3} = c_{4} = 0</math>, so that <math>\{v_{1} - v_{2}, v_{2} - v_{3}, v_{3} - v_{4},v_{4}\}</math> form a linearly independent set as desired.

@@ Line 74: / Line 74: @@
 |}
 Before we begin, we will need the following notation: for an arbitrary non-empty set <math>X</math>, let <math>\mathbb{R}^{X}</math> denote the set of all functions <math>f \colon X \to \mathbb{R}</math>. Then <math>\mathbb{R}^{X}</math> is always a vector space, with addition and scalar multiplication defined pointwise.
 '''Exercise'''

@@ Line 94: / Line 94: @@
 <math>U_{1} = \{ (0,y) \in \mathbb{R}^{2} \colon y \in \mathbb{R} \} </math>
-<math>U_{2} = \{ (z,z) \in \mathbb{R}^{2} \colon z \in \mathbb{R}  \} c
+<math>U_{2} = \{ (z,z) \in \mathbb{R}^{2} \colon z \in \mathbb{R}  \} </math>
 Then <math>U_{1}</math> and <math>U_{2}</math> are not the same subspaces of <math>V</math>, so that all we need to check is <math>U_{1} \oplus W = V = U_{2} \oplus W.</math> Suppose that <math>(x,y) \in V</math>. Then <math>(x,y) = (x,0) + (0,y)</math>, where <math>(x,0) \in W</math> and <math>(0,y) \in U_{1}</math>, so that <math>V = W + U_{1}</math>. Now if <math>(a,b) \in W \cap U_{1}</math>, then <math>(a,b) = (x,0) \in W</math> for some <math>x \in \mathbb{R}</math>, and thus <math>b = 0</math>. Likewise, <math>(a,b) = (0,y) \in W</math> for some <math>y \in \mathbb{R}</math>, so that <math>a = 0</math>. This shows <math>W \cap U_{1} = \{(0,0)\}</math>, and hence <math>V = U_{1} \oplus W</math>.<br />
 Now again say <math>(x,y) \in V</math>. Then <math>(x,y) = (x-y,0) + (y,y)</math>, where <math>(x-y,0) \in W</math> and <math>(y,y) \in U_{2}</math> so that <math>V = W + U_{2}</math>. Now suppose <math>(a,b) \in W \cap U_{2}</math>. Then <math>(a,b) = (x,0) \in W</math> for some <math>x \in \mathbb{R}</math>, so that <math>b = 0</math>. Likewise, <math>(a,b) = (a,0) = (z,z) \in U_{2}</math> for some <math>z \in \mathbb{R}</math>, thus <math>z = 0 = a</math>, so that we conclude <math>W \cap U_{2} = \{(0,0)\}</math>, and thus <math>V = U_{2} \oplus W</math>.

@@ Line 89: / Line 89: @@
 ''Proof''
 Let’s think about what it means for two subspaces <math>A,B</math> of a vector space <math>C</math> to satisfy <math>C = A \oplus B</math>. This means that <math>A \cap B = \{0_{C}\}</math> and that <math>C = A + B</math>. In other words, for any <math>c \in C</math>, we may write <math>c</math> uniquely in the form <math>c = a + b</math>, where <math>a \in A</math> and <math>b \in B</math>.<br />
-It turns out the statement of the problem is '''false''', so that we must provide a counterexample to this statement: Let <math>V = \mathbb{R}^{2}</math> and consider its subspaces (one should check that they actually form subspaces first): <math>
+It turns out the statement of the problem is '''false''', so that we must provide a counterexample to this statement: Let <math>V = \mathbb{R}^{2}</math> and consider its subspaces (one should check that they actually form subspaces first):
-W = \{ (x,0) \in \mathbb{R}^{2} \colon x \in \mathbb{R}  \} <br />
+<math>W = \{ (x,0) \in \mathbb{R}^{2} \colon x \in \mathbb{R}  \} </math>
-U_{1} = \{ (0,y) \in \mathbb{R}^{2} \colon y \in \mathbb{R} \} <br />
-U_{2} = \{ (z,z) \in \mathbb{R}^{2} \colon z \in \mathbb{R}  \}
+<math>U_{1} = \{ (0,y) \in \mathbb{R}^{2} \colon y \in \mathbb{R} \} </math>
-</math>
 Then <math>U_{1}</math> and <math>U_{2}</math> are not the same subspaces of <math>V</math>, so that all we need to check is <math>U_{1} \oplus W = V = U_{2} \oplus W.</math> Suppose that <math>(x,y) \in V</math>. Then <math>(x,y) = (x,0) + (0,y)</math>, where <math>(x,0) \in W</math> and <math>(0,y) \in U_{1}</math>, so that <math>V = W + U_{1}</math>. Now if <math>(a,b) \in W \cap U_{1}</math>, then <math>(a,b) = (x,0) \in W</math> for some <math>x \in \mathbb{R}</math>, and thus <math>b = 0</math>. Likewise, <math>(a,b) = (0,y) \in W</math> for some <math>y \in \mathbb{R}</math>, so that <math>a = 0</math>. This shows <math>W \cap U_{1} = \{(0,0)\}</math>, and hence <math>V = U_{1} \oplus W</math>.<br />
 Now again say <math>(x,y) \in V</math>. Then <math>(x,y) = (x-y,0) + (y,y)</math>, where <math>(x-y,0) \in W</math> and <math>(y,y) \in U_{2}</math> so that <math>V = W + U_{2}</math>. Now suppose <math>(a,b) \in W \cap U_{2}</math>. Then <math>(a,b) = (x,0) \in W</math> for some <math>x \in \mathbb{R}</math>, so that <math>b = 0</math>. Likewise, <math>(a,b) = (a,0) = (z,z) \in U_{2}</math> for some <math>z \in \mathbb{R}</math>, thus <math>z = 0 = a</math>, so that we conclude <math>W \cap U_{2} = \{(0,0)\}</math>, and thus <math>V = U_{2} \oplus W</math>.

← Older revision		Revision as of 19:28, 9 November 2015
Line 35:		Line 35:

	In other words, <math>c_{1} = c_{2} = c_{3} = c_{4} = 0</math>, so that <math>\{v_{1} - v_{2}, v_{2} - v_{3}, v_{3} - v_{4},v_{4}\}</math> form a linearly independent set as desired.		In other words, <math>c_{1} = c_{2} = c_{3} = c_{4} = 0</math>, so that <math>\{v_{1} - v_{2}, v_{2} - v_{3}, v_{3} - v_{4},v_{4}\}</math> form a linearly independent set as desired.
		+
		+
		+	'''Exercise'''
		+	Prove that a vector space <math>V</math> over a field <math>\mathbb{F}</math> is infinite-dimensional if and only if there is a sequence <math>v_{1},v_{2},\ldots</math> in <math>V</math> such that <math>v_{1},\ldots,v_{m}</math> is linearly independent for every <math>m \in \mathbb{N}</math>.
		+
		+	''Proof''
		+	Recall that a vector space <math>V</math> is said to be '''finite dimensional''' if it is spanned by a finite list of vectors <math>w_{1},\ldots,w_{m} \in V.</math> In other words, <math>V</math> has finite dimension if every vector in <math>V</math> may be written as a linear combination of some list of vectors <math>w_{1},\ldots, w_{m} \in V</math>. On the other hand, a vector space <math>V</math> is '''infinite dimensional''' if it is not finite dimensional, i.e., <math>V</math> cannot be spanned by a finite list of vectors. Now before we proceed in the proof, we will need the following fact:
		+
		+	''Lemma''
		+	Suppose <math>V</math> is a vector space over a field <math>\mathbb{F}</math>, and <math>v_{1},\ldots,v_{n}</math> are vectors that span <math>V</math>. If <math>w_{1},\ldots,w_{m}</math> in <math>V</math> are linearly independent, then <math>m \leq n</math>.
		+
		+	We are ready now to proceed with the proof:<br />
		+	’<math>\Rightarrow</math>’: Suppose that <math>V</math> is an infinite dimensional vector space. Then, in particular, <math>V \neq 0</math>, so that there is some <math>v_{1} \neq 0</math> in <math>V</math>. Then <math>v_{1}</math> is a linearly independent vector in <math>V</math>. By way of induction now, suppose that for some <math>k \geq 1</math>, we have produced vectors <math>v_{1},\ldots, v_{k} \in V</math> such that <math>v_{1},\ldots,v_{k}</math> are linearly independent. Since <math>V</math> is infinite-dimensional, it cannot be spanned by the (finite!) list of vectors <math>v_{1},\ldots,v_{k}</math>. Thus we have that there is some <math>v_{k+1} \in V</math> such that <math>v_{k+1} \neq c_{1}v_{1} + \cdots + c_{k}v_{k}, \text{ for any }c_{1},\ldots, c_{k} \in \mathbb{F}.</math> We claim that now that <math>v_{1},\ldots,v_{k},v_{k+1}</math> form a linearly independent set in <math>V</math>. To see this, suppose that <math>a_{1}v_{1} + \cdots + a_{k}v_{k} + a_{k+1}v_{k+1} = 0 \text{ for some } a_{1},\ldots,a_{k},a_{k+1} \in \mathbb{F}.</math> Now if <math>a_{k+1} \neq 0</math>, then we may re-write the above equation as <math>v_{k+1} = \Big( \frac{-a_{1}}{a_{k+1}} \Big)v_{1} + \cdots + \Big( \frac{-a_{k}}{a_{k+1}} \Big)v_{k},</math> contradicting the fact that <math>v_{k+1}</math> is not in the span of <math>v_{1},\ldots,v_{k}</math>. So we conclude <math>a_{k+1} = 0</math>, and thus we have that <math>a_{1}v_{1} + \cdots + a_{k}v_{k} + a_{k+1}v_{k+1}
		+	= a_{1}v_{1} + \cdots + a_{k}v_{k} + (0)v_{k+1}</math> <math>= a_{1}v_{1} + \cdots + a_{k}v_{k} = 0.</math> Now by induction hypothesis, since <math>v_{1},\ldots,v_{k}</math> are linearly independent, we must have <math>a_{1},\ldots,a_{k}</math> are all zero. Thus we’ve shown that <math>v_{1},\ldots, v_{k},v_{k+1}</math> also form a linearly independent set, completing the induction. Thus we have constructed a sequence of vectors <math>\{v_{k}\}^{\infty}_{k =1}</math> in <math>V</math> so that <math>v_{1},\ldots, v_{m}</math> is linearly independent for each <math>m \in \mathbb{N}</math>.
		+
		+	’<math>\Leftarrow</math>’: On the other hand, suppose that <math>V</math> contains a sequence of vectors <math>\{v_{k}\}^{\infty}_{k =1}</math> so that <math>v_{1},\ldots, v_{m}</math> is linearly independent for each <math>m \in \mathbb{N}</math>. By way of contradiction, let’s suppose <math>V</math> is not infinite dimensional, i.e. is finite dimensional. Then <math>V</math> can be spanned by a finite list of vectors <math>w_{1},\ldots, w_{n} \in V</math>.<br />
		+	Now, since <math>V</math> contains a linearly independent set of <math>V</math>.
		+
		+
		+	'''Exercise'''
		+	Suppose that <math>U</math> and <math>W</math> are subspaces of a vector space <math>V</math>. Prove that <math>U \cup W</math> is a subspace of <math>V</math> if and only if <math>U \subseteq W</math> or <math>W \subseteq U</math>.
		+
		+	''Proof''
		+	Recall that a subset <math>A</math> of a vector space <math>V</math> is a '''subspace''' of <math>V</math> if <math>A</math> itself is a vector space with the same addition and scalar multiplication operations as <math>V</math>.<br />
		+	’<math>\Rightarrow</math>’: Instead of proving that <math>U \cup W</math> is a subspace of <math>V</math> implies <math>U \subseteq W</math> or <math>W \subseteq U</math>, we’ll show the ''contrapositive'' of this statement. That is, if <math>U \not\subseteq W</math> and <math>W \not\subseteq U</math>, then <math>U \cup W</math> is not a subspace of <math>V</math>. So suppose there is some <math>x \in U</math> that is not in <math>W</math>, and likewise that there is some <math>y \in W</math> that is not in <math>U</math>. We claim that <math>x + y \notin U \cup W</math>. For if it were, then <math>x+y</math> would lie in either <math>U</math> or <math>W</math>. If <math>x + y \in U</math>, then since <math>U</math> is a subspace, this would imply <math>y = (x+y) - x \in U,</math> contradicting our choice of <math>y</math>. Likewise, if <math>x + y \in W</math>, this would yield <math>x \in W</math>, which is again a contradiction. So we conclude that <math>x + y \notin U \cup W</math>, and thus <math>U \cup W</math> fails to be closed under addition, so cannot be a subspace of <math>V</math>.<br />
		+	’<math>\Leftarrow</math>’: Suppose now that <math>U \subseteq W</math> or <math>W \subseteq U</math>. Then <math>U \cup W</math> is equal to either <math>W</math> or <math>U</math> respectively, which, by assumption are subspaces of <math>V</math>.
		+
		+	---
		+
		+	Before we begin, we will need the following notation: for an arbitrary non-empty set <math>X</math>, let <math>\mathbb{R}^{X}</math> denote the set of all functions <math>f \colon X \to \mathbb{R}</math>. Then <math>\mathbb{R}^{X}</math> is always a vector space, with addition and scalar multiplication defined pointwise.
		+
		+	'''Exercise'''
		+	Let <math>b \in \mathbb{R}</math> and consider the set <math>W = \Big\{ f \in \mathbb{R}^{[0,1]} \colon f \text{ is continuous and} \int^{1}_{0} f(x) dx = b \Big\}.</math> Show that <math>W</math> is a subspace of <math>\mathbb{R}^{[0,1]}</math> if and only if <math>b = 0</math>.
		+
		+	''Proof''
		+	Recall that a subset <math>A</math> of a vector space <math>V</math> is a '''subspace''' of <math>V</math> if <math>A</math> itself is a vector space with the same addition and scalar multiplication operations as <math>V</math>. There is a very convenient test that determines if <math>A</math> is a subspace of <math>V</math>, sometimes called the ''subspace test''. It says the following:
		+
		+	('''Subspace Test''') Suppose that <math>A \subseteq V</math>, where <math>V</math> is a vector space over a field <math>\mathbb{F}</math>. Then <math>A</math> is a subspace of <math>V</math> if and only if the following conditions are met:
		+
		+	# <math>0_{V} \in A</math>,
		+	# <math>c \in \mathbb{F}, v \in A \Rightarrow cv \in A</math>
		+	# <math>v,w \in A \Rightarrow v + w \in A</math>.
		+
		+	We are now ready to proceed with the proof:
		+
		+	’<math>\Rightarrow</math>’: Suppose <math>W</math> is a subspace of <math>\mathbb{R}^{[0,1]}</math>. Then by condition <math>(1)</math> of the subspace test, <math>W</math> contains the zero vector of <math>\mathbb{R}^{[0,1]}</math>, which is just the function that maps <math>x</math> to <math>0</math> for all <math>x \in [0,1]</math>. We will write this zero vector as <math>\textbf{0}</math>. Now since <math>\textbf{0} \in W</math>, by definition of being in <math>W</math>, we must have that <math>\int^{1}_{0} \textbf{0}(x)dx = b.</math> On the other hand, when we actually integrate <math>\textbf{0}</math>, we find the integral must be zero. Thus <math>b = 0</math> as desired.<br />
		+	’<math>\Leftarrow</math>’: Say <math>b = 0</math>. We will show that <math>W</math> is a subspace of <math>V</math> by showing that it passes all three conditions of the subspace test above. For condition (1), just note that by our previous remark, <math>\int^{1}_{0} \textbf{0}(x)dx = 0 = b</math>, and since <math>\textbf{0}</math> is continuous, we have that <math>\textbf{0} \in W</math>. For condition (2), suppose that <math>c \in \mathbb{R}</math> and <math>f \in W</math>. We must show that <math>cf \in W</math>. Since a continuous function multiplied by a constant is still continuous, <math>cf</math> is still a continuous function. Now, <math>\int^{1}_{0}(cf)(x)dx = \int^{1}_{0}c[f(x)]dx = c\int^{1}_{0} f(x)dx = c(0) = 0,</math> so that we conclude <math>cf \in W</math>. Lastly, for condition (3), we must show that if <math>f,g \in W</math>, then <math>f + g \in W</math>. The addition of two continuous functions is always continuous, so that <math>f + g</math> is continuous. Now since <math>f,g \in W</math>, we have that <math>\int^{1}_{0} (f + g)(x)dx = \int^{1}_{0}[f(x) + g(x)]dx = \int^{1}_{0}f(x)dx + \int^{1}_{0}g(x)dx = 0 + 0 = 0,</math> so that <math>f + g \in W</math>, and thus <math>W</math> satisfies all three conditions of the subspace test.
		+
		+
		+	'''Exercise'''
		+	Prove or give a counterexample to the following statement: If <math>U_{1},U_{2},W</math> are subspaces of a vector space <math>V</math> with <math>V = U_{1} \oplus W \text { and } V = U_{2} \oplus W,</math> then <math>U_{1} = U_{2}</math>.
		+
		+	''Proof''
		+	Let’s think about what it means for two subspaces <math>A,B</math> of a vector space <math>C</math> to satisfy <math>C = A \oplus B</math>. This means that <math>A \cap B = \{0_{C}\}</math> and that <math>C = A + B</math>. In other words, for any <math>c \in C</math>, we may write <math>c</math> uniquely in the form <math>c = a + b</math>, where <math>a \in A</math> and <math>b \in B</math>.<br />
		+	It turns out the statement of the problem is '''false''', so that we must provide a counterexample to this statement: Let <math>V = \mathbb{R}^{2}</math> and consider its subspaces (one should check that they actually form subspaces first): <math>
		+	W = \{ (x,0) \in \mathbb{R}^{2} \colon x \in \mathbb{R} \} <br />
		+	U_{1} = \{ (0,y) \in \mathbb{R}^{2} \colon y \in \mathbb{R} \} <br />
		+	U_{2} = \{ (z,z) \in \mathbb{R}^{2} \colon z \in \mathbb{R} \}
		+	</math>
		+
		+	Then <math>U_{1}</math> and <math>U_{2}</math> are not the same subspaces of <math>V</math>, so that all we need to check is <math>U_{1} \oplus W = V = U_{2} \oplus W.</math> Suppose that <math>(x,y) \in V</math>. Then <math>(x,y) = (x,0) + (0,y)</math>, where <math>(x,0) \in W</math> and <math>(0,y) \in U_{1}</math>, so that <math>V = W + U_{1}</math>. Now if <math>(a,b) \in W \cap U_{1}</math>, then <math>(a,b) = (x,0) \in W</math> for some <math>x \in \mathbb{R}</math>, and thus <math>b = 0</math>. Likewise, <math>(a,b) = (0,y) \in W</math> for some <math>y \in \mathbb{R}</math>, so that <math>a = 0</math>. This shows <math>W \cap U_{1} = \{(0,0)\}</math>, and hence <math>V = U_{1} \oplus W</math>.<br />
		+	Now again say <math>(x,y) \in V</math>. Then <math>(x,y) = (x-y,0) + (y,y)</math>, where <math>(x-y,0) \in W</math> and <math>(y,y) \in U_{2}</math> so that <math>V = W + U_{2}</math>. Now suppose <math>(a,b) \in W \cap U_{2}</math>. Then <math>(a,b) = (x,0) \in W</math> for some <math>x \in \mathbb{R}</math>, so that <math>b = 0</math>. Likewise, <math>(a,b) = (a,0) = (z,z) \in U_{2}</math> for some <math>z \in \mathbb{R}</math>, thus <math>z = 0 = a</math>, so that we conclude <math>W \cap U_{2} = \{(0,0)\}</math>, and thus <math>V = U_{2} \oplus W</math>.

@@ Line 10: / Line 10: @@
 ''Proof''
-Recall that a set of vectors <math>\{v_{1},\ldots,v_{n}\}</math> in a vector space <math>V</math> (over a field <math>\mathbb{F}</math>) is said to be <math>\textbf{linearly dependent}</math> if they are not linearly independent. More concretely, these vectors are linearly dependent if we can find scalars <math>c_{1},\ldots, c_{n} \in \mathbb{F}</math> ''not all equal to zero'' such that <math>c_{1}v_{1} + \cdots + c_{n}v_{n} = 0.</math>
+Recall that a set of vectors <math>\{v_{1},\ldots,v_{n}\}</math> in a vector space <math>V</math> (over a field <math>\mathbb{F}</math>) is said to be ''linearly dependent'' if they are not linearly independent. More concretely, these vectors are linearly dependent if we can find scalars <math>c_{1},\ldots, c_{n} \in \mathbb{F}</math> ''not all equal to zero'' such that <math>c_{1}v_{1} + \cdots + c_{n}v_{n} = 0.</math>
 So for this problem, to show that <math>1+i</math> and <math>1-i</math> are not linearly dependent over <math>\mathbb{C}</math>, all we need to do is exhibit two complex scalars <math>c_{1}</math> and <math>c_{2}</math> that are not ''both'' zero such that <math>c_{1}(1+i) + c_{2}(1-i) = 0.</math> There are many choices for <math>c_{1}</math> and <math>c_{2}</math>, but one such example is <math>c_{1} = i</math> and <math>c_{2} = 1</math>.
@@ Line 21: / Line 21: @@
 Recall that the set of vectors <math>\{w_{1},\ldots, w_{n} \}</math> in a vector space <math>V</math> (over a field <math>\mathbb{F}</math>) are said to be '''linearly independent''' if whenever <math>c_{1},\ldots,c_{n}</math> are scalars in <math>\mathbb{F}</math> such that <math>c_{1}w_{1} + \cdots + c_{n}w_{n} = 0,</math> then <math>c_{1} = \cdots = c_{n} = 0</math>.
 So for this problem, we must show that whenever <math>c_{1},c_{2},c_{3},c_{4} \in \mathbb{F}</math> and <math>c_{1}(v_{1} - v_{2})  + c_{2}(v_{2} - v_{3}) + c_{3}(v_{3} - v_{4}) + c_{4}v_{4} = 0,</math> we have that <math>c_{1} = c_{2} = c_{3} = c_{4} = 0.</math> After rearranging terms in the above equation, we have that <math>c_{1}v_{1} + (c_{2} - c_{1})v_{2} + (c_{3} - c_{2})v_{3} + (c_{4} - c_{3})v_{4} = 0.</math> Now since the vectors <math>\{v_{1},v_{2},v_{3},v_{4}\}</math> are linearly independent in <math>V</math> by assumption, we have that

@@ Line 24: / Line 24: @@
 '''Exercise'''
-So for this problem, we must show that whenever <math>c_{1},c_{2},c_{3},c_{4} \in \mathbb{F}</math> and <math>c_{1}(v_{1} - v_{2})  + c_{2}(v_{2} - v_{3}) + c_{3}(v_{3} - v_{4}) + c_{4}v_{4} = 0,</math> we have that <math>c_{1} = c_{2} = c_{3} = c_{4} = 0.</math> After rearranging terms in the above equation, we have that <math>c_{1}v_{1} + (c_{2} - c_{1})v_{2} + (c_{3} - c_{2})v_{3} + (c_{4} - c_{3})v_{4} = 0.</math> Now since the vectors <math>\{v_{1},v_{2},v_{3},v_{4}\}</math> are linearly independent in <math>V</math> by assumption, we have that <math>
+So for this problem, we must show that whenever <math>c_{1},c_{2},c_{3},c_{4} \in \mathbb{F}</math> and <math>c_{1}(v_{1} - v_{2})  + c_{2}(v_{2} - v_{3}) + c_{3}(v_{3} - v_{4}) + c_{4}v_{4} = 0,</math> we have that <math>c_{1} = c_{2} = c_{3} = c_{4} = 0.</math> After rearranging terms in the above equation, we have that <math>c_{1}v_{1} + (c_{2} - c_{1})v_{2} + (c_{3} - c_{2})v_{3} + (c_{4} - c_{3})v_{4} = 0.</math> Now since the vectors <math>\{v_{1},v_{2},v_{3},v_{4}\}</math> are linearly independent in <math>V</math> by assumption, we have that
-c_{1} = 0 \\
-c_{2} - c_{1} = 0 \\
+<math>c_{1} = 0 </math>
-c_{3} - c_{2} = 0 \\
-c_{4} - c_{3} = 0.
+<math>c_{2} - c_{1} = 0 </math>
 </math> In other words, <math>c_{1} = c_{2} = c_{3} = c_{4} = 0</math>, so that <math>\{v_{1} - v_{2}, v_{2} - v_{3}, v_{3} - v_{4},v_{4}\}</math> form a linearly independent set as desired.