By Bobylev N. A., Bulatov V.

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Proof: if M is a subspace, and if x is an arbitrary element of M, then 0 · x belongs to M (scalar multiples), and since it is already known that 0 · x = 0, it follows that 0 ∈ M for all M. The definition of subspaces could have been formulated this way: a subset M of V is a subspace if M itself is a vector space with respect to the same linear operations (vector addition and scalar multiplication, or, in one phrase, linear combination) as are given in V. Since every vector space contains its zero vector, the presence of 0 in M should not come as a surprise.

In view of the definition, the question is this: which vectors can be obtained as linear combinations of no vectors at all? This formulation calls attention to a blemish of the definition; it doesn't apply to the conceptually trivial but technically very important empty set. The cure is to rephrase the definition: the span of VECTORS 31 a set of vectors is the intersection of all the subspaces that include the set. It is a non-profound exercise to show that for non-empty sets the rephrasing is equivalent to the original definition; its virtue is that it applies with no change to every set|including, in particular, the empty set.

In Z8 , 2 × 4 = 0; in Z9 , 3 × 3 = 0; etc. ) General fact (not hard to prove): Zn is a field if and only if the modulus n is a prime. The fact that Z 4 is not a field shows that a certain way of defining addition and multiplication for four elements does not result in a field. Is it possible that different definitions would lead to a different result? Problem 19. Is there a field with four elements? CHAPTER 2 VECTORS 20. Vector spaces 20 2 Real numbers can be added, and so can pairs of real numbers.

### A Bound on the Real Stability Radius of Continuous-Time Linear Infinite-Dimensional Systems by Bobylev N. A., Bulatov V.

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