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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 16415, 472]*) (*NotebookOutlinePosition[ 17269, 502]*) (* CellTagsIndexPosition[ 17225, 498]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Subspaces", "Title", ImageRegion->{{-0, 1}, {0, 1}}], Cell[BoxData[{ \(Author : \ Thomas\ Shores\), "\[IndentingNewLine]", \(University\ of\ Nebraska\), "\[IndentingNewLine]", \(Send\ comments\ \(to : \ tshores@math . unl . edu\)\)}], "Input", FontFamily->"Helvetica", FontWeight->"Plain", FontVariations->{"CompatibilityType"->0}], Cell[CellGroupData[{ Cell["Review", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ Let us recall the definition of an (abstract) vector space: Definition: A VECTOR SPACE (v.s.) over a field of scalars is a set V of \ elements called vectors, together with two binary operations, scalar \ multiplication * (usually suppressed) and vector addition + , subject to the \ following vector space laws: For all u, v, w \[Element] V and scalars \ a and b, (1) (Closure of vector additon) u + v \[Element] V. (2) (Commutativity of addition) u + v = v + u. (3) (Associativity of addition) u + (v + w) = (u + v) + w. (4) (Additivive Identity) There exists an element 0 \[Element] V such that \ u + 0 = u = 0 + u. (5) (Additive Inverse) There exists, for each u \[Element] V, an element -u \ \[Element] in V such that \tu + (-u) = 0 = (-u) + u. (6) (Closure of scalar multiplication) a*v \[Element] V. (7) (Distributive Law) a*(u+v) = a*u + a*v. (8) (Distributive Law) (a + b)*u = a*u + b*u. (9) (Associative Law) (a b) * u = a*(b*u). (10) (Monoidal Law) 1*u = u. NOTATION: u + (-v) = u - v. ALSO: we often refer to the \[OpenCurlyDoubleQuote]vector space V\ \[CloseCurlyDoubleQuote], without identifying explicitly the binary \ operations. They are taken for granted. \t SIMPLE FACTS: (deducible from the vector space properties alone): \tFor all vectors u, 0*u = 0, where the latter 0 means the zero vector. \tFor all vectors u, (-1)*u = -u.\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ EXAMPLES OF VECTOR SPACES: (1) The spaces Rn and Cn, n >=1, over the fields R and C, \ respectively. Also, C is a vector space over R. (2) The set M(2,2) of all 2x2 real matrices, together with the usual \ matrix addition and scalar multiplication. (3) The set \tV = C[0,1] = {f(x) | f(x) is a continuous function on the interval [0,1] }, \ together with the usual function addition and scalar multiplication: \t\t(f+g)(x) = f(x) + g(x) \t\t(a*f)(x) = a f(x). \tThis is a very non-standard vector space. It seems kind of odd, but we are \ really thinking of functions f(x), 0 <= x <= 1, as \ \[OpenCurlyDoubleQuote]vectors\[CloseCurlyDoubleQuote]. Of course, one has \ to check that all the laws are satisfied. This is routine: they are. What \ plays the part of the zero vector is the zero constant function. \t \tNow, what is a subspace of a given vector space V? DEFINITION: A subspace of the vector space V is a subset W of V such \ that W, together with the binary operations it inherits from V , forms a \ vector space (over the same field of scalars) in its own right. EXAMPLES OF SUBSPACES: \t(1) Let V = R3 and W = {[x,y,x] | x, y \:02db R}. \t(2) Let V = M(2,2) and W = {{{a,b},{0,a}} | a , b \:02db R } \t(3) Let V = C[0,1] and W = P_2 = {f(x) = a + b x + c x2 | a, \ b, c in R} In every case we have to verify that laws (1) - (10) are valid for each \ subset W of the v.s. V. This is unpleasant!! Do we have to do all this \ work?? No, and here is why: SUBSPACE TEST: Let W be a nonempty subset of the v.s. V. Then W is \ a subspace of V if and only if \t(1) For all u, v \:02db W, u + v \:02db W (this is really closure of \ addition on elements of W). \t(2) For all u \:02db W and scalars a, a*u \:02db W (this is really \ closure of scalar multiplication on elements of W). \t \tIt isn't too hard to see what the Subspace Test works: certainly, any \ subspace W of V must satisfy these conditions since it is a vector space \ in its own right with the operations inherited from V. Conversely, if the \ nonempty subset W of V satisfies these conditions, then vector space \ properties (1) and (6) follow. Properties (2)-(3) and (7)-(10) follow \ simply because elements of W are elements of V and they hold for elements of \ V. The only properties to be checked are (4) and (5). Since W is \ nonempty, there is an element v \:02db W. Now one of the Simple Facts above \ is that (-1)v = -v, the additive inverse of v as an element of V. By \ condition (2) of the Subspace Test, (-1)v \:02db W. Therefore -v \:02db W. \ By condition (1) of the Subspace Test, v + (-v) = 0 (the zero vector of \ the vector space V) also belongs to W. It follows that W contains an \ additive identity, namely, the very same one as V has, and that every \ element of W has an additive inverse in W. This disposes of vector space \ laws (4) and (5) and shows that W is a vector space in its own right, hence \ a subspace of V. \t APPLICATION: A very common occurrence is a linear combination (l.c.) of \ vectors v1, v2, ... , vn, by which we mean an expression of the form, for \ suitable scalars a1, a2, ...an, \t \t\ta1 v1 + a2 v2 + ... + an vn. \t\t The set of all possible linear combinations of the vectors v1, v2, ... ,vn \ has a name: the linear span of v1, v2, ..., vn. DEFINITION: The linear span of vectors v1, v2, ...., vn is the set of all \ vectors \tspan{v1, v2, ..., vn} = { a1 v1 + a2 v2 + ... + an vn | a1, a2, ..., an \ are scalars}. \t Note that the span depends on the field of scalars in use. Here is the \ application we referred to: FACT: If v1, v2, ..., vn are vector in the v.s. V then lin{v1, v2, ..., \ vn} is a subspace of V. This is a fact that you should be able to prove. It is also in your text. ANOTHER IDEA: if V is a vector space such that V = lin{v1, v2, ..., vn}, \ then the vectors v1, v2, ..., vn are said to SPAN the vector space V. EXAMPLE: {1,0,0}, {0,1,0}, and {0,0,1} span the vector space R3. As \ a matter of fact, so does the set of vectors {1,0,0}, {0, 1,0},{0,0,1}, \ {1,1,2}. Why do you think that one would prefer the first spanning set? DEFINITION: A set of vectors v1, v2, ... , vn is a BASIS of the vector \ space V if this set spans V and no subset of v1, v2, ... vn spans \ V.\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Subspaces and Matrices", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ Suppose we are given a matrix A. For the sake of definiteness, we \ will work with the following matrix A:\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["A={{1,-2,0,0,3},{2,-5,-3,-2,6},{0,5,15,10,0},{2,6,18,8,6}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Here are three very important vector spaces that are associated with the \ matrix A. Let us say that A = [C", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", C", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ..., C", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], "] where the A", StyleBox["i", FontVariations->{"CompatibilityType"->"Subscript"}], "'s are the columns of A, and that \nA = [R", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", R", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ..., R", StyleBox["m", FontVariations->{"CompatibilityType"->"Subscript"}], "]", StyleBox["T", FontVariations->{"CompatibilityType"->"Superscript"}], " where the R", StyleBox["i", FontVariations->{"CompatibilityType"->"Subscript"}], "'s are the rows of A. Then:\n\n\tThe column space of A: C(A) = \ lin{C", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", C", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ..., C", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], "}\n\t\n\tThe row space of A: R(A) = lin{R", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", R", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ..., R", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], "}\n\t\n\tThe null space of A: N(A) = { v ", "\[Element]", " R", StyleBox["n", FontVariations->{"CompatibilityType"->"Superscript"}], " ", "\:f2bd", " A.v = 0} \n(Note: replace R by C in the case of complex scalars in all \ that follows. We have to use one symbol, so we default to R.)\n\t\nR(A) and \ C(A) are definitely vector spaces. As a matter of fact, R(A) is a subspace \ of R", StyleBox["m", FontVariations->{"CompatibilityType"->"Superscript"}], " and C(A) is a subspace of R", StyleBox["n", FontVariations->{"CompatibilityType"->"Superscript"}], ". (How come?) \n\nEXAMPLE: Describe the row space, column space and null \ space of the simple matrix of the following cell in simple geometrical terms. \ Where do these subspaces live? For the null space, by the way, it might \ help to click on the cells after the definition of" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["B = {1, 3, 1}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["v = {x,y,z}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["B.v", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ Now let's take a look at a harder example, namely, the matrix A \ we defined above. How would you define the row space and column space of A? \ Write this out on a separate piece of paper. It might help to use the \ following comman\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["MatrixForm[A]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["NullSpace[A]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ " Now for another idea about spanning sets. Remember the examples of the \ last section (look them up). There was something undesirable about one \ spanning set: it had unnecessarily many vectors in the list. A spanning set \ which does not have too many vectors is called a basis of the vector space \ it spans. Thus:\n\nDEFINITION: A ", StyleBox["basis", FontSlant->"Italic"], " for the vector space V is a minimal spanning set, that is, a spanning \ set which has the property that no smaller set will span the space.\n\nGive \ an example of a basis for R", StyleBox["3", FontVariations->{"CompatibilityType"->"Superscript"}], ". Now for a key question: how can we spot redundancies in the list of \ rows or columns of a given matrix like our A? I claim the answer lies in \ the following cell:" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["RowReduce[A]", "Input", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Exercises", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ Find the row space, the column space, and the null space of the \ following matrices. Are there any relationships between the dimensions of \ (the number of independent vectors in) these spaces ?\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ StyleBox[" ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e6", FontWeight->"Plain"], StyleBox["1 0 2 7 5", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f6", FontWeight->"Plain"], StyleBox["\n1) A = ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f7 ", FontWeight->"Plain"], StyleBox["3 3 -1 -3 0", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f7", FontWeight->"Plain"], StyleBox["\n ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e8", FontWeight->"Plain"], StyleBox["2 1 -2 0 5", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f8", FontWeight->"Plain"] }], "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ StyleBox[" ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e6", FontWeight->"Plain"], StyleBox["3 2 10 5", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f6", FontWeight->"Plain"], StyleBox["\n2) B = ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e7", FontWeight->"Plain"], StyleBox["1 1 0 2", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f7", FontWeight->"Plain"], StyleBox["\n ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e7", FontWeight->"Plain"], StyleBox["4 -2 8 0", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f7", FontWeight->"Plain"], StyleBox["\n ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e8", FontWeight->"Plain"], StyleBox["-3 5 -2 5", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f8", FontWeight->"Plain"] }], "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ StyleBox["\n ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e6", FontWeight->"Plain"], StyleBox[" 2 2 1 0", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f6", FontWeight->"Plain"], StyleBox["\n ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e7", FontWeight->"Plain"], StyleBox[" 1 0 2 1", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f7", FontWeight->"Plain"], StyleBox["\n3) C = ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e7", FontWeight->"Plain"], StyleBox[" 3 -1 -1 0", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f7", FontWeight->"Plain"], StyleBox["\n ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e7", FontWeight->"Plain"], StyleBox[" -2 0 0 2", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f7", FontWeight->"Plain"], StyleBox["\n ", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2e8", FontWeight->"Plain"], StyleBox[" 4 -4 2 1", FontFamily->"Helvetica", FontWeight->"Plain"], StyleBox["\:f2f8", FontWeight->"Plain"] }], "Input", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]] }, Closed]] }, FrontEndVersion->"4.0 for X", ScreenRectangle->{{0, 1152}, {0, 864}}, WindowToolbars->{}, CellGrouping->Manual, WindowSize->{520, 600}, WindowMargins->{{Automatic, 308}, {104, Automatic}}, PrivateNotebookOptions->{"ColorPalette"->{RGBColor, -1}}, ShowCellLabel->True, ShowCellTags->False, RenderingOptions->{"ObjectDithering"->True, "RasterDithering"->False} ] (*********************************************************************** Cached data follows. 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