By Kallenrode

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Remark: This particular basis S = {ep e2, ... , en} is called the standard basis for Vn• Since the number of element in S is n, and hence dimension of Vn is n, and we write dim Vn =n. Example 28. Consider the set S = {t, x, x\ ... ,xn} in Pn. 1 + (lIX + (l2X2 Then + ... + (l"X" = 0, a zero pOlynomial for some scalars (li' (lo =O=(ll =(l2 = .. ·=(In' 46 Introductory Linear Algebra Further, each polynomial of degree less than or equal to n is just the linear combination of the polynomials in S. Hence S spans Pw This implies that S is a basis for P n and dim p" = n + 1 = the number of elements in S.

B = [A 1]. 3. Perform elementary row operations on B. 4. If [A I] is transformed into [I C], then C = A-I; otherwise A does not have an inverse. Remark 1. Note that by performing elementary row operations on B = [A 1] if it is possible to find row-reduced echelon form of A as I, then row-rank of A = n and so A is non singular. If row-reduced echelon form of A I by performing elementary row operations on B = [A 1], then row-rank of A < n, and so A is singular. * Remark 2. If we wish to check whether the given n x n matrix is singular or non-singular, then perform elementary row operations only on A.

An as S is LI Thus, a ll = 0 = a 2 2 = ... = an = a => SI is LI, which is a contradiction. 4) can be rewritten as Hence, Cl * v = (-~}I+(-C;:}2+ ... +(-:}n => S spans V Hence the theorem follows. 1. Let S = {vI' v2' ... , v,,} be LI in a vector space V. Suppose v E V be such that v is not in [S]. Then S u {v} is LI. Proof: Let al' a 2 ••• ,, a" and ex Cl be scalars such that 2 ,, ... alvli +a 2v2+ .... +a"v" +av =Ov. 5) for if a:,t: 0, then And it would imply that v E [S], a contradiction. 5) reduces to alvli +a 2v2+....