Subspaces of Rn and Linear Independence - Revision history

Khanh at 02:15, 15 November 2021

2021-11-15T02:15:43Z

Lila: Lila moved page Subspaces of Rⁿ and Linear Independence to Subspaces of Rn and Linear Independence

2021-11-03T13:23:04Z

Lila moved page Subspaces of Rⁿ and Linear Independence to Subspaces of Rn and Linear Independence

Lila at 22:17, 29 September 2021

2021-09-29T22:17:55Z

Lila at 22:12, 29 September 2021

2021-09-29T22:12:27Z

Lila at 22:05, 29 September 2021

2021-09-29T22:05:02Z

Lila at 21:19, 29 September 2021

2021-09-29T21:19:17Z

Lila at 21:00, 29 September 2021

2021-09-29T21:00:11Z

Lila at 20:58, 29 September 2021

2021-09-29T20:58:44Z

Lila: /* Subspaces */

2021-09-29T20:39:37Z

Subspaces

Lila at 20:38, 29 September 2021

2021-09-29T20:38:19Z

@@ Line 319: / Line 319: @@
 where not all of the scalars are zero (the fact that some of the scalars are zero doesn't matter).
 ==Resources==
 * [https://textbooks.math.gatech.edu/ila/subspaces.html Subspaces], Interactive Linear Algebra from Georgia Tech

← Older revision		Revision as of 22:17, 29 September 2021
Line 319:		Line 319:

	where not all of the scalars are zero (the fact that some of the scalars are zero doesn't matter).		where not all of the scalars are zero (the fact that some of the scalars are zero doesn't matter).
		+
		+
		+
		+
		+	==Resources==
		+	* [https://en.wikibooks.org/wiki/Linear_Algebra/Definition_and_Examples_of_Linear_Independence Definition and Examples of Linear Independence], WikiBooks Linear Algebra
		+	* [https://en.wikibooks.org/wiki/Linear_Algebra/Subspaces_and_Spanning_sets Subspaces and Spanning Sets], WikiBooks Linear Algebra
		+	* [https://en.wikibooks.org/wiki/Linear_Algebra/Subspaces Subspaces], WikiBooks Linear Algebra
		+	* [https://textbooks.math.gatech.edu/ila/subspaces.html Subspaces], Interactive Linear Algebra from Georgia Tech

@@ Line 141: / Line 141: @@
 : (<math> p </math>, <math> r </math> scalars) is a linear combination of elements of <math> S </math> and so is in <math> [S] </math> (possibly some of the <math>\vec{s}_i</math>'s forming <math>\vec{v}</math> equal some of the <math>\vec{s}_j</math>'s from <math>\vec{w}</math>, but it does not matter).
-===Example===
+===Example 7===
 The span of this set
 is all of <math>\mathbb{R}^2</math>.
@@ Line 236: / Line 236: @@
 is the trivial one: <math> c_1=0,\dots,\,c_n=0 </math>.
-}}
+Proof: This is a direct consequence of the observation above.
-{{TextBox|1=
+the set <math> S=\{\vec{v}_1,\vec{v}_2,\vec{v}_3\} </math> is linearly dependent because this is a relationship
-If the set <math> S </math> is linearly independent then no vector <math>\vec{s}_i</math> can be written as a linear combination of the other vectors from <math>S</math> so there is no linear relationship where some of the <math>\vec{s}\,</math>'s have nonzero coefficients. If <math> S </math> is not linearly independent then some <math> \vec{s}_i </math> is a linear combination <math>\vec{s}_i=c_1\vec{s}_1+\dots+c_{i-1}\vec{s}_{i-1} +c_{i+1}\vec{s}_{i+1}+\dots+c_n\vec{s}_n</math> of other vectors from <math> S </math>, and subtracting <math>\vec{s}_i</math> from both sides of that equation gives a linear relationship involving a nonzero coefficient, namely the <math> -1 </math> in front of <math> \vec{s}_i </math>.
+:<math>
-}}
+\cdot\vec{v}_1

@@ Line 2: / Line 2: @@
 For any vector space, a '''subspace''' is a subset that is itself a vector space, under the inherited operations.
-===Important Lemma on Subspaces===
+===Lemma 1===
 For a nonempty subset <math> S </math> of a vector space, under the inherited
 operations, the following are equivalent statements.
@@ Line 126: / Line 126: @@
 The span of the empty subset of a vector space is the trivial subspace. No notation for the span is completely standard. The square brackets used here are common, but so are "<math>\mbox{span}(S)</math>" and "<math>\mbox{sp}(S)</math>".
-===Lemma===
+===Lemma 2===
 In a vector space, the span of any subset is a subspace.
@@ Line 178: / Line 178: @@
 We first characterize when a vector can be removed from a set without changing the span of that set.
-===Lemma===
+===Lemma 3===
 Where <math> S </math> is a subset of a vector space <math>V</math>,
 :<math>
@@ Line 199: / Line 199: @@
 : the spans <math> [\{\vec{v}_1,\vec{v}_2\}] </math> and <math> [\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}] </math> are equal since <math> \vec{v}_3 </math> is in the span <math> [\{\vec{v}_1,\vec{v}_2\}] </math>.
-The lemma says that if we have a spanning set then we can remove a <math>\vec{v}</math> to get a new set <math>S</math> with the same span if and only if <math>\vec{v}</math> is a linear combination of vectors from <math>S</math>. Thus, under the second sense described above, a spanning set is minimal if and only if it contains no vectors that are linear combinations of the others in that set. We have a term for this important property.
+Lemma 2 says that if we have a spanning set then we can remove a <math>\vec{v}</math> to get a new set <math>S</math> with the same span if and only if <math>\vec{v}</math> is a linear combination of vectors from <math>S</math>. Thus, under the second sense described above, a spanning set is minimal if and only if it contains no vectors that are linear combinations of the others in that set. We have a term for this important property.
 ===Definition of Linear Independence===
@@ Line 226: / Line 226: @@
 what is usually the easiest way to compute whether a finite set is
 dependent or independent.

@@ Line 141: / Line 141: @@
 : (<math> p </math>, <math> r </math> scalars) is a linear combination of elements of <math> S </math> and so is in <math> [S] </math> (possibly some of the <math>\vec{s}_i</math>'s forming <math>\vec{v}</math> equal some of the <math>\vec{s}_j</math>'s from <math>\vec{w}</math>, but it does not matter).
-===Example 1===
+===Example===
 The span of this set
 is all of <math>\mathbb{R}^2</math>.
@@ Line 174: / Line 174: @@
 </math>
 with back substitution gives <math>c_2=(x-y)/2</math> and <math>c_1=(x+y)/2</math>. These two equations show that for any <math>x</math> and <math>y</math> that we start with, there are appropriate coefficients <math>c_1</math> and <math>c_2</math> making the above vector equation true. For instance, for <math>x=1</math> and <math>y=2</math> the coefficients <math>c_2=-1/2</math> and <math>c_1=3/2</math> will do. That is, any vector in <math>\mathbb{R}^2</math> can be written as a linear combination of the two given vectors.

← Older revision	Revision as of 13:23, 3 November 2021
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← Older revision		Revision as of 21:00, 29 September 2021
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	</math>		</math>
	: (<math> p </math>, <math> r </math> scalars) is a linear combination of elements of <math> S </math> and so is in <math> [S] </math> (possibly some of the <math>\vec{s}_i</math>'s forming <math>\vec{v}</math> equal some of the <math>\vec{s}_j</math>'s from <math>\vec{w}</math>, but it does not matter).		: (<math> p </math>, <math> r </math> scalars) is a linear combination of elements of <math> S </math> and so is in <math> [S] </math> (possibly some of the <math>\vec{s}_i</math>'s forming <math>\vec{v}</math> equal some of the <math>\vec{s}_j</math>'s from <math>\vec{w}</math>, but it does not matter).
		+
		+	===Example 1===
		+	The span of this set
		+	is all of <math>\mathbb{R}^2</math>.
		+
		+	:<math>
		+	\{\begin{pmatrix} 1 \\ 1 \end{pmatrix},\begin{pmatrix} 1 \\ -1 \end{pmatrix}\}
		+	</math>
		+
		+	To check this we must show that any member of <math>\mathbb{R}^2</math> is a linear combination
		+	of these two vectors.
		+	So we ask: for which
		+	vectors (with real components <math>x</math> and <math>y</math>)
		+	are there scalars <math>c_1</math> and <math>c_2</math> such that this holds?
		+
		+	:<math>
		+	c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix}+c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix}=\begin{pmatrix} x \\ y \end{pmatrix}
		+	</math>
		+
		+	Gauss' method
		+
		+	:<math>\begin{array}{rcl}
		+	\begin{array}{*{2}{rc}r}
		+	c_1 &+ &c_2 &= &x \\
		+	c_1 &- &c_2 &= &y
		+	\end{array}
		+	&\xrightarrow[]{-\rho_1+\rho_2}
		+	&\begin{array}{*{2}{rc}r}
		+	c_1 &+ &c_2 &= &x \\
		+	& &-2c_2 &= &-x+y
		+	\end{array}
		+	\end{array}
		+	</math>
		+	with back substitution gives <math>c_2=(x-y)/2</math> and <math>c_1=(x+y)/2</math>. These two equations show that for any <math>x</math> and <math>y</math> that we start with, there are appropriate coefficients <math>c_1</math> and <math>c_2</math> making the above vector equation true. For instance, for <math>x=1</math> and <math>y=2</math> the coefficients <math>c_2=-1/2</math> and <math>c_1=3/2</math> will do. That is, any vector in <math>\mathbb{R}^2</math> can be written as a linear combination of the two given vectors.

@@ Line 36: / Line 36: @@
 We usually show that a subset is a subspace with <math> (2)\implies (1) </math>.
-====Example 1====
+===Example 1===
 : The plane <math> P=\{\begin{pmatrix} x \\ y \\ z \end{pmatrix}\,\big|\, x+y+z=0\} </math> is a subspace of <math> \mathbb{R}^3 </math>. As specified in the definition, the operations are the ones inherited from the larger space, that is, vectors add in <math>P</math> as they add in <math>\mathbb{R}^3</math>
@@ Line 46: / Line 46: @@
 : and scalar multiplication is also the same as it is in <math>\mathbb{R}^3</math>. To show that <math>P</math> is a subspace, we need only note that it is a subset and then verify that it is a space. Checking that <math>P</math> satisfies the conditions in the definition of a vector space is routine. For instance, for closure under addition, just note that if the summands satisfy that <math>x_1+y_1+z_1=0</math> and <math>x_2+y_2+z_2=0</math> then the sum satisfies that <math>(x_1+x_2)+(y_1+y_2)+(z_1+z_2)=(x_1+y_1+z_1)+(x_2+y_2+z_2)=0</math>.
-====Example 2====
+===Example 2===
 : The <math> x </math>-axis in <math> \mathbb{R}^2 </math> is a subspace where the addition and scalar multiplication operations are the inherited ones.
@@ Line 62: / Line 62: @@
 : As above, to verify that this is a subspace, we simply note that it is a subset and then check that it satisfies the conditions in definition of a vector space. For instance, the two closure conditions are satisfied: (1) adding two vectors with a second component of zero results in a vector with a second component of zero, and (2) multiplying a scalar times a vector with a second component of zero results in a vector with a second component of zero.
-====Example 3====
+===Example 3===
 : Another subspace of <math>\mathbb{R}^2</math> is
@@ Line 76: / Line 76: @@
 {{anchor|proper}}Other subspaces are '''proper'''.
-====Example 4====
+===Example 4===
 The condition in the definition requiring that the addition and scalar multiplication operations must be the ones inherited from the larger space is important. Consider the subset <math> \{1\} </math> of the vector space <math> \mathbb{R}^1 </math>. Under the operations <math>1+1=1</math> and  <math>r\cdot 1=1</math> that set is a vector space, specifically, a trivial space. But it is not a subspace of <math> \mathbb{R}^1 </math> because those aren't the inherited operations, since of course <math> \mathbb{R}^1 </math> has <math> 1+1=2 </math>.
-====Example 5====
+===Example 5===
 : All kinds of vector spaces, not just <math>\mathbb{R}^n</math>'s, have subspaces. The vector space of cubic polynomials <math> \{a+bx+cx^2+dx^3\,\big|\, a,b,c,d\in\mathbb{R}\} </math> has a subspace comprised of all linear polynomials <math> \{m+nx\,\big|\, m,n\in\mathbb{R}\} </math>.
-====Example 6====
+===Example 6===
 : This is a subspace of the <math> 2 \! \times \! 2 </math> matrices
@@ Line 115: / Line 115: @@
 : As above, we've described the subspace as a collection of unrestricted linear combinations (by coincidence, also of two elements).