basis — Jemoka Knowledge Base

A basis is a list of vectors in V that spans V and is linearly independent constituents a LIST! of vectors in vector space V requirements the list is… linear independent spans V additional information criteria for basis A list v_1, \dots v_{n} of vectors in V is a basis of V IFF every v \in V can be written uniquely as:

\begin{equation} v = a_1v_1+ \dots + a_{n}v_{n} \end{equation}

where a_1, \dots, a_{n} \in \mathbb{F}. forward direction Suppose we have v_1, \dots, v_{n} as the basis in V. We desire that v_1, \dots v_{n} uniquely constructs each v \in V. By definition, they span V and are linear independent in V. Because of the spanning quality, there exists at least one set of a_1, \dots, a_{n} \in \mathbb{F} such that we can write:

\begin{equation} v \in V = a_1v_1+ \dots + a_{n}v_{n} \end{equation}

Suppose now that we have another representation of v via scalars c_1, \dots, c_{n} and our same list of vectors:

\begin{equation} v \in V =^{?} c_1v_1+ \dots + c_{n}v_{n} \end{equation}

Subtracting the two expressions, we have that:

\begin{equation} 0 = (a_1-c_1)v_1 + \dots +(a_{n}-c_{n}) v_{n} \end{equation}

By definition that v_1 \dots v_{n} is linearly independent, we have that a_j-c_j=0 \implies a_{j}=c_{j}. Therefore, there is only one unique representation for v as a linear combination of vectors v_1, \dots v_{n}. (to be honest, we could have just applied that as the definition of linear independence that the scalars in a linear combo of linearly independent list is unique but this is the more careful definition.) backward direction Suppose we have a list v_1, \dots v_{n} which uniquely constructs each v \in V. We desire that v_1, \dots v_{n} is a basis in V. Given a linear combination thereof can construct all v \in V, we can say that v_1, \dots v_{n} spans V. As V is a vector space, we have 0 \in V. Therefore, there exists some scalars a_1, \dots a_{n} for which:

\begin{equation} 0 = a_1v_1 + \dots +a_{n}v_{n} \end{equation}

(as we already established v_1, \dots, v_{n} spans V and 0 \in V) Of course, we are given that v_1, \dots v_{n} uniquely constructs each v \in V. As the trivial solution does exist: that a_1 = \dots = a_{n} = 0, it is the only solution. By definition of linear independence, then, v_1, \dots v_{n} is linearly independent. Having constructed that v_1, \dots v_{n} is both a spanning set in V and are linearly independent, we have that they are a basis of V. \blacksquare Dualing Basis Construction These are two results that says: “you can build up a linearly independent list to a basis or you can pluck away a spanning list to a basis”. all spanning lists contains a basis of which you are spanning Every spanning list in V contains the basis (and possibly some more) in V. Read: “apply Linear Dependence Lemma your way to success”. Begin with a spanning list v_1, \dots v_{m} of V. We run a for loop for the list. Step 0: If v_1=0 (i.e. v_1 \in span(\{\})), delete v_1. Otherwise, do nothing. Step j: If v_{j} is in span(v_1, \dots v_{j-1}), v_{j} satisfies the Linear Dependence Lemma’s first condition, and therefore naturally satisfies the second condition (removal from list keeps the same span because v_{j} can just be rewritten from v_1, \dots v_{j-1}). So we remove v_{j} if it is indeed in the span of the previous vectors. By the Linear Dependence Lemma, the new list spans the same space the old list. Conclusion By the end of this process, no vectors left in the list will satisfy the Linear Dependence Lemma (read: we got rid of all of them.) Therefore, the list is linearly independent. However, every step of the way the Linear Dependence Lemma ensures that the new list spans the same space; therefore, the new list still spans V. Having constructed a linearly independent list that spans V, we declare the new list as a basis of V. As all we did was pluck vectors out of the old list, the new list is a sublist of the old list. This means that the spanning list (old list) contains the new list, which is a basis. \blacksquare a linearly independent list expends to a basis Every linearly independent list of vectors in finite-dimensional vector spaces can be extended to a basis. Recall first that every finite-dimensional vector space has a basis. Let’s begin with a linearly independent list in V u_1, \dots u_{m}. Let’s recruit also a basis of V: w_{1}, \dots w_{m}. Naturally: u_1, \dots u_{m}, w_1, \dots w_{m} spans V (as the w vectors already span V). We will now apply the fact that all spanning lists contains a basis of which you are spanning (the order of u vectors first and w vectors second ensuring that you try to remove the w, and, as u are linearly independent, none of them will be removed) to get back a basis in V consisting of all u and some w. \blacksquare