Title Basis and Dimension of vector space_01.pdf ✅

Basis and Dimension of a Vector Space Page | 26 Definition 11.1.1. A set of vectors SS = {vv1, vv2, ... , vvnn} in a vector space VV is called a basis for VV if the following conditions hold true. 1. SS spans VV. 2. SS is linearly independent. Remark. This definition tells us that a basis has two features: It • must have enough vectors to span the space; • should not have so many vectors that one of them could be written as a linear combination of the other vectors in it. Example 11.1.1. Show that SS = ��1 0 0 �, � 0 1 0 �, � 0 0 1 �� is a basis for R3. Solution. In Example 10.2.3 and Example 10.2.6, we proved that span(SS) = R3 and SS is linearly independent, respectively. Thus SS is a basis for R3. This basis is called the standard basis for R3. Similarly, the set SS = {ee1, ee2, ... , eenn} where ee1 = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ 1 0 ⋮ ⋮ ⋮ 0⎦ ⎥ ⎥ ⎥ ⎥ ⎤ , ee2 = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ 0 1 ⋮ ⋮ ⋮ 0⎦ ⎥ ⎥ ⎥ ⎥ ⎤ , ⋯ eekk = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ 0 0 ⋮ 1 ⋮ 0⎦ ⎥ ⎥ ⎥ ⎥ ⎤ , ⋯ eenn = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ 0 0 ⋮ ⋮ ⋮ 1⎦ ⎥ ⎥ ⎥ ⎥ ⎤ . eeii has 1 in the iith position and 0’s elsewhere. The set SS is called the standard basis for Rnn. Example 11.1.2. Show that SS = ⎩ ⎪ ⎨ ⎪ ⎧ � 1 2 3 � �vv1 , � 0 1 2 � �vv2 , � −2 0 1 � �vv3 ⎭ ⎪ ⎬ ⎪ ⎫ is a basis for R3. Solution. In Example 10.2.4 and Example 10.2.7, we obtained that span(SS) = R3 and SS is linearly independent, respectively. Hence SS is a basis for R3. Example 11.1.1 and Example 11.1.2 show that basis for a vector space is not unique.
Basis and Dimension of a Vector Space Page | 27 Example 11.1.3. Show that the set SS = �� 1 0 0 0�,� 0 1 0 0�,� 0 0 1 0�,� 0 0 0 1�� is a basis for MM2,2. This is called the standard basis for MM2,2. Solution. Take an arbitrary matrix � aa11 aa12 aa21 aa22� ∈ MM2,2. Clearly, � aa11 aa12 aa21 aa22� = aa11 � 1 0 0 0� + aa12 � 0 1 0 0� + aa21 � 0 0 1 0� + aa22 � 0 0 0 1� Therefore, span(SS) = MM2,2. To show the linear independence, we solve the vector equation for cc1, cc2, cc3, and cc4 cc1vv1 + cc2vv2 + cc3vv3 + cc4vv4 = 00 where vv1 = � 1 0 0 0�, vv2 = � 0 1 0 0�, vv3 = � 0 0 1 0�, vv4 = � 0 0 0 1� So, cc1 � 1 0 0 0� + cc2 � 0 1 0 0� + cc3 � 0 0 1 0� + cc4 � 0 0 0 1� = � 0 0 0 0� � cc1 cc2 cc3 cc4 � = � 0 0 0 0� This equation is true if and only if cc1 = cc2 = cc3 = cc4 = 0. This shows that SS is linearly independent. Thus SS is basis for MM2,2.
Basis and Dimension of a Vector Space Page | 28 Example 11.1.4. Show that the set SS = {1, xx, xx2} is a basis for PP2(xx) where PP2(xx) = {aa2xx2 + aa1xx + aa0| aa2, aa1, aa0 ∈ R}. Solution. Let us denote vv1 = 1, vv2 = xx, vv3 = xx2 Take an arbitrary polynomial aa2xx2 + aa1xx + aa0 from PP2(xx). Then obviously, aa2xx2 + aa1xx + aa0 = aa0vv1 + aa1vv2 + aa2vv3 This shows that span(SS) = PP2(xx). Now we will show the linear independence of SS. cc1vv1 + cc2vv2 + cc3vv3 = 00 cc1 + cc2xx + cc3xx2 = 0 By differentiating both sides with respect to xx, we obtain cc2 + 2cc3xx = 0 Again differentiating, we get 2cc3 = 0 gives cc1 = cc2 = cc3 = 0. ∎ In Example 10.2.1, we observed that vv = � 1 1 1 � ∈ R3 can be written as a linear combination of vectors in the set SS = ⎩ ⎪ ⎨ ⎪ ⎧ � 1 2 3 � �vv1 , � 0 1 2 � �vv2 , � −1 0 1 � �vv3 ⎭ ⎪ ⎬ ⎪ ⎫ in various ways. For example, vv = vv1 − vv2 or vv = 2vv1 − 3vv2 + vv3. The next theorem tells us that under what condition this representation is unique.
Basis and Dimension of a Vector Space Page | 29 Theorem 11.1.1. (Uniqueness of Basis Representation) If SS = {vv1, vv2, ... , vvnn} is a basis for a vector space VV, then every vector in VV can be written in one and only one way as a linear combination of vectors in S. Proof. The existence portion of the proof is straightforward. Since SS is a basis for VV, SS spans VV, thus an arbitrary vector vv in VV can be expressed as vv = cc1vv1 + cc2vv2 + ⋯ + ccnnvvnn (1) for some scalars cc1, cc2, ... , ccnn. To prove uniqueness (that a vector can be represented in only one way), assume vv has another representation vv = dd1vv1 + dd2vv2 + ⋯ + ddnnvvnn (2) Subtracting eq. (2) from eq. (1), we obtain (cc1 − dd1)vv1 + (cc2 − dd2)vv2 + ⋯ + (ccnn − ddnn)vvnn = 00. SS is linearly independent, however, so the only solution to this equation is the trivial solution cc1 − dd1 = 0, cc2 − dd2 = 0, ⋯ , ccnn − ddnn = 0, which means that ccii = ddii for all ii = 1, 2, ... , nn and vv has only one representation for the basis SS. ∎ Theorem 11.1.2. If SS = {vv1, vv2, ... , vvnn} is a basis for a vector space VV, then every set containing more than nn vectors in VV is linearly dependent, hence cannot be a basis. Proof. Consult the textbook. Question. Is SS = ⎩ ⎪ ⎨ ⎪ ⎧ � 1 2 3 � �vv1 , � 0 1 2 � �vv2 , � −2 0 1 � �vv3 , � 3 1 −2 � �vv4 ⎭ ⎪ ⎬ ⎪ ⎫ a basis for R3? Answer.NO. Because in Example 11.1.1 we found that one basis of R3 contains three vectors, since SS has four vectors, it must be linearly dependent, hence SS is not a basis for R3.