{"id":8,"date":"2020-11-29T15:02:42","date_gmt":"2020-11-29T15:02:42","guid":{"rendered":"http:\/\/blog.metu.edu.tr\/e235505\/?p=8"},"modified":"2020-11-29T16:58:24","modified_gmt":"2020-11-29T16:58:24","slug":"fifth-axiom-of-vector-spaces-can-be-slightly-changed","status":"publish","type":"post","link":"https:\/\/blog.metu.edu.tr\/e235505\/2020\/11\/29\/fifth-axiom-of-vector-spaces-can-be-slightly-changed\/","title":{"rendered":"Fifth axiom of vector spaces can be slightly changed"},"content":{"rendered":"\n

Let S be a Vector Space which consists of V (set of vectors) and F (field). Fifth axiom of vector spaces <\/strong>states that<\/p>\n\n\n\n

1<\/strong>\u2217\u03b1=\u03b1<\/p>\n\n\n\n

where 1<\/strong> \u2208 F and \u03b1 \u2208 V. <\/p>\n\n\n\n

We can change this axiom as follows; there is an element, b, in F such that<\/p>\n\n\n\n

b*\u03b1=\u03b1<\/p>\n\n\n\n

Now, we can prove that for any vector space (with slightly different axiom), b can be replaced by 1<\/strong>. <\/p>\n\n\n\n

Proposition: b can be replaced by 1<\/strong>. In fact, if V has some non-zero vectors and S is a vector space, then b must be 1<\/strong>.<\/p>\n\n\n\n

Proof: If V has only zero vector, then <\/p>\n\n\n\n

b*0=0<\/p>\n\n\n\n

for every b \u2208 F, particularly for b=1<\/strong><\/p>\n\n\n\n

If V has some non-zero vectors and for every \u03b1 \u2208 V <\/p>\n\n\n\n

b*\u03b1=\u03b1<\/p>\n\n\n\n

then b must be equal to 1<\/strong>. Since there is at least one non-zero vector, <\/p>\n\n\n\n

b\u22600<\/p>\n\n\n\n

From another axiom of vector spaces, we know that<\/p>\n\n\n\n

c1<\/sub>\u2217c2<\/sub>(\u03b1)=c1<\/sub>(c2<\/sub>\u03b1)<\/p>\n\n\n\n

Since b\u22600, below equalities are valid<\/p>\n\n\n\n

(b-1<\/sup>\u2217b)\u03b1=1<\/strong>\u03b1<\/p>\n\n\n\n

b-1<\/sup>(b\u03b1)=b-1<\/sup>\u03b1<\/p>\n\n\n\n

then,<\/p>\n\n\n\n

1<\/strong>\u03b1=b-1<\/sup>\u03b1<\/p>\n\n\n\n

(1<\/strong>-b-1<\/sup>)\u03b1=0<\/p>\n\n\n\n

since this is true for even non-zero vectors in S,<\/p>\n\n\n\n

1<\/strong>-b-1<\/sup>=0<\/p>\n\n\n\n

1<\/strong>=b-1<\/sup><\/p>\n\n\n\n

b=1<\/strong><\/p>\n\n\n\n

<\/p>\n\n\n\n

What does this proposition mean? It means that every vector space in which fifth axiom is slightly different is “normal” vector space.<\/p>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

Let S be a Vector Space which consists of V (set of vectors) and F (field). Fifth axiom of vector spaces states that 1\u2217\u03b1=\u03b1 where 1 \u2208 F and \u03b1 \u2208 V. We can change this axiom as follows; there is an element, b, in F such that b*\u03b1=\u03b1 Now, we can prove that for … <\/p>\n

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