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These notes closely follow the presentation of the material given in David C. Lay’s textbook Linear Algebra and its Applications (3rd edition). These notes are intended primarily for in-class presentation and should not be regarded as a substitute for thoroughly reading the textbook itself and working through the exercises therein. The Dimension of a Vector Space Recall that a basis for a vector space, V, is a set of vectors in V that is linearly independent and spans V. Theorem If a vector space, V, has a basis that contains exactly n vectors, then every basis for V contains exactly n vectors. Proof Suppose that V is a vector space and that B v 1 , v 2 , , v n is a basis for V (containing exactly n vectors). We will first prove that any set of vectors in V that contains more than n vectors cannot be a basis for V. Let C w 1 , w 2 , , w n , , w m be a set of vectors in V that contains more than n vectors. (We are assuming that C contains exactly m vectors where m n.). Since B is a basis for V, every vector in C can be written (in a unique way) as a linear combination of the vectors in B. That is, w 1 a 11 v 1 a 12 v 2 a 1n v n w 2 a 21 v 1 a 22 v 2 a 2n v n w n a n1 v 1 a n2 v 2 a nn v n w m a m1 v 1 a m2 v 2 a mn v n . Now consider the equation c 1w1 c 2w2 c nwn c mwm 0V. This equation can be written as c 1 a 11 v 1 a 12 v 2 a 1n v n c 2 a 21 v 1 a 22 v 2 a 2n v n c n a n1 v 1 a n2 v 2 a nn v n c m a m1 v 1 a m2 v 2 a mn v n 0V or as 1 c 1 a 11 c 2 a 21 c n a n1 c m a m1 v 1 c 1 a 12 c 2 a 22 c n a n2 c m a m2 v 2 c 1 a 1n c 2 a 2n c n a nn c m a mn v n 0V. Since the set B is linearly independent, all of the coefficients (weights) in the above equation must be zero. Thus, a 11 c 1 a 21 c 2 a n1 c n a m1 c m 0 a 12 c 1 a 22 c 2 a n2 c n a m2 c m 0 a 1n c 1 a 2n c 2 a nn c n a mn c m 0. The above system is a homogeneous linear system with more unknowns (m) than equations (n). Such a system must have non–trivial solutions. Thus, there exist scalars c 1 , c 2 , c n , , c m , not all zero, that satisfy the above system. These same scalars also satisfy the vector equation c 1w1 c 2w2 c nwn c mwm 0V, which shows that the set C is linearly dependent. Thus, C cannot be a basis for V. We have proved that if V has a basis containing exactly n vectors, then no basis for V can contain more than n vectors – but, if we think about it, we have actually also proved that no basis for V can contain fewer than n vectors. The reason is as follows: If we have a basis for V that contains exactly m vectors where m n, then the argument given above shows that no basis for V can contain more than m vectors. In particular, no basis for V can contain n vectors, but this contradicts our assumption that B is a basis for V. Definition The dimension of a vector space is defined to be the number of vectors in any basis of V. If each basis of V contains only a finite number, n, of vectors, then we say that V is an n–dimensional vector space. If V does not have a basis consisting of a finite number of vectors, then we say that V is an infinite–dimensional vector space. The trivial vector space, V 0, is said to have dimension zero (even though it actually has no basis). Example For each positive integer n, the vector space n has dimension n. Example For each pair of positive integers m and n, the vector space M mn , consisting of all m n matrices, has dimension m n. Example For each positive integer n, the vector space P n , which consists of all polynomial functions p : that have degree less than or equal to n, has dimension n 1. 2 Example The vector space P, which consists of all polynomial functions p : , is infinite–dimensional. Example The vector space C 0 , which consists of all continuous functions f : , is infinite dimensional. Example If V is any non–trivial vector space and S v 1 ,v 2 ,,v n is any linearly independent set of vectors in V, then the vector space SpanS (which is a subspace of V) has dimension n. Example If A is an m n matrix, then nulA, the nullspace of A, has dimension equal to the number of non–pivot columns of A. Also, colA, the column space of A has dimension equal to the number of pivot columns of A. This tells us that dimnulA dimcolA n. As a specific example of this, we can immediately observe for the matrix 1 A 2 3 4 4 3 0 6 0 , 1 0 3 that dimnulA dimcolA 4. To verify that this is correct, find bases for nulA and colA. 3