Basis and Dimension In our previous discussion, we introduced the concepts of span and linear independence. In a way a set of vectors S = {v_{1}, ... , v_{k}} span a vector space V if there are enough of the right vectors in S, while they are linearly independent if there are no redundancies. We now combine the two concepts.
Definition of Basis Let V be a vector space and S = {v_{1}, v_{2}, ... , v_{k}} be a subset of V. Then S is a basis for V if the following two statements are true.
Let V = R^{n} and let S = {e_{1}, e_{2}, ... ,e_{n}} where e_{i} has i^{th} component equal to 1 and the rest 0. For example e_{2} = (0,1,0,0,...,0) Then S is a basis for V called the standard basis. Example Let V = P_{3} and let S = {1, t, t^{2}, t^{3}}. Show that S is a basis for V.
Solution We must show both linear independence and span.
Linear Independence: Let c_{1}(1) + c_{2}(t) + c_{3}(t^{2}) + c_{4}(t^{3}) = 0 Then since a polynomial is zero if and only if its coefficients are all zero we have c_{1} = c_{2} = c_{3} = c_{4} = 0 Hence S is a linearly independent set of vectors in V.
Span A general vector in P_{3} is given by a + bt + ct^{2} + dt^{3} We need to find constants c_{1}, c_{2}, c_{3}, c_{4} such that c_{1}(1) + c_{2}(t) + c_{3}(t^{2}) + c_{4}(t^{3}) = a + bt + ct^{2} + dt^{3} We just let c_{1} = a, c_{2} = b, c_{3} = c, c_{4} = d Hence S spans V. We can conclude that S is a basis for V.
In general the basis {1, t, t^{2}, ... , t^{n}} is called the standard basis for P_{n}. Example Show that S = {v_{1}, v_{2}, v_{3}, v_{4}} where
is a basis for V = M^{2x2}.
Solution We need to show that S spans V and is linearly independent.
Linear Independence We write
this gives the four equations
c_{1} + c_{2} + c_{3} + 2c_{4} = 0 Which has the corresponding homogeneous matrix equation Ac = 0 with
We have det A = 12 Since the determinant is nonzero, we can conclude that only the trivial solution exists. That is c_{1} = c_{2} = c_{3} = c_{4} = 0
Span We set
which gives the equations
c_{1} + c_{2} + c_{3} + 2c_{4} = x_{1} The corresponding matrix equation is Ac = x Since A is nonsingular, this has a unique solution, namely c = A^{1}x Hence S spans V. We conclude that S is a basis for V. If S spans V then we know that every vector in V can be written as a linear combination of vectors in S. If S is a basis, even more is true. Theorem Let S = {v_{1}, v_{2}, ... , v_{k}} be a basis for V. Then every vector in V can be written uniquely as a linear combination of vectors in S. Remark: What is new here is the word uniquely.
Proof Suppose that v = a_{1}v_{1} + ... + a_{n}v_{n} = b_{1}v_{1} + ... + b_{n}v_{n} then (a_{1}  b_{1})v_{1} + ... + (a_{n}  b_{n})v_{n} = 0 Since S is a basis for V, it is linearly independent and the above equation implies that all the coefficients are zero. That is a_{1}  b_{1} = ... = a_{n}  b_{n} = 0 We can conclude that a_{1} = b_{1}, ... , a_{n} = b_{n} Since linear independence is all about having no redundant vectors, it should be no surprise that if S = {v_{1}, v_{2}, ... , v_{k}}spans V, then if S is not linearly independent then we can remove the redundant vectors until we arrive at a basis. That is if S is not linearly independent, then one of the vectors is a linear combination of the rest. Without loss of generality, we can assume that this is the vector v_{k}. We have that v_{k} = a_{1}v_{1} + ... + a_{k1}v_{k1} If v is any vector in S we can write v = c_{1}v_{1} + ... + c_{k}v_{k} = c_{1}v_{1} + ... + c_{k1}v_{k1} + c_{k}(a_{1}v_{1} + ... + a_{k1}v_{k1}) which is a linear combination of the smaller set S' = {v_{1}, v_{2}, ... , v_{k1}} If S' is not a basis, as above we can get rid of another vector. We can continue this process until the vectors are finally linear independent. We have proved the following theorem.
Theorem Let S span a vector space V, then there is a subset of S that is a basis for V.
Dimension We have seen that any vector space that contains at least two vectors contains infinitely many. It is uninteresting to ask how many vectors there are in a vector space. However there is still a way to measure the size of a vector space. For example, R^{3} should be larger than R^{2}. We call this size the dimension of the vector space and define it as the number of vectors that are needed to form a basis. Tow show that the dimensions is well defined, we need the following theorem. Theorem If S = {v_{1}, v_{2}, ... , v_{n}} is a basis for a vector space V and T = {w_{1}, w_{2}, ... , w_{k}} is a linearly independent set of vectors in V, then k < n. Remark: If S and T are both bases for V then k = n. This says that every basis has the same number of vectors. Hence the dimension is will defined. The dimension of a vector space V is the number of vectors in a basis. If there is no finite basis we call V an infinite dimensional vector space. Otherwise, we call V a finite dimensional vector space.
Proof If k > n, then we consider the set R_{1} = {w_{1},v_{1}, v_{2}, ... , v_{n}} Since S spans V, w_{1} can be written as a linear combination of the v_{i}'s. w_{1} = c_{1}v_{1} + ... + c_{n}v_{n} Since T is linearly independent, w_{1} is nonzero and at least one of the coefficients c_{i} is nonzero. Without loss of generality assume it is c_{1}. We can solve for v_{1} and write v_{1} as a linear combination of w_{1}, v_{2}, ... v_{n}. Hence T_{1} = {w_{1}, v_{2}, ... , v_{n}} is a basis for V. Now let R_{2} = {w_{1}, w_{2}, v_{2}, ... , v_{n}} Similarly, w_{2} can be written as a linear combination of the rest and one of the coefficients is non zero. Note that since w_{1} and w_{2} are linearly independent, at least one of the v_{i} coefficients must be nonzero. We can assume that this nonzero coefficient is v_{2} and as before we see that T_{2} = {w_{1}, w_{2},v_{3,} ... , v_{n}} is a basis for V. Continuing this process we see that T_{n} = {w_{1}, w_{2}, ... , w_{n}} is a basis for V. But then T_{n} spans V and hence w_{n+1} is a linear combination of vectors in T_{n}. This is a contradiction since the w's are linearly independent. Hence n > k.
Example Since E = {e_{1}, e_{2}, ... , e_{n}} is a basis for R^{n} then dim R^{n} = n. Example dim P_{n} = n + 1 since E = {1, t, t^{2}, ... , t^{n}} is a basis for P_{n}. Example dim M^{mxn} = mn We will leave it to you to find a basis containing mn vectors.
If we have a set of linearly independent vectors S = {v_{1}, v_{2}, ... , v_{k}} with k < n, then S is not a basis. From the definition of basis, S does not span V, hence there is a v_{k+1} such that v_{k+1} is not in the span of S. Let S_{1} = {v_{1}, v_{2}, ... ,v_{k },v_{k+1}} S_{1} is linearly independent. We can continue this until we get a basis.
Theorem Let S = {v_{1}, v_{2}, ... , v_{k}} be a linearly independent set of vectors in a vector space V. Then there are vectors v_{k+1}, ... , v_{n} such that {v_{1}, v_{2}, ... , v_{k}, v_{k+1}, ... , v_{n}} is a basis for V. We finish this discussion with some very good news. We have seen that to find out if a set is a basis for a vector space, we need to check for both linear independence and span. We know that if there are not the right number of vectors in a set, then the set cannot form a basis. If the number is the right number we have the following theorem. Theorem Let V be an n dimensional vector space and let S be a set with n vectors. Then the following are equivalent.
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