Question

6. Show that an isomorphism of vector spaces must preserve dimension. That is, an isomorphism sends a basis of $n$ elements to a basis of $n$ elements.

          6. Show that an isomorphism of vector spaces must preserve dimension. That is, an isomorphism sends a basis of $n$ elements to a basis of $n$ elements.
        
6. Show that an isomorphism of vector spaces must preserve dimension. That is, an isomorphism sends a basis of n elements to a basis of n elements.

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Geometry A Common Core Curriculum
Geometry A Common Core Curriculum
Ron Larson, Laurie Boswell 1st Edition
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6. Show that an isomorphism of vector spaces must preserve dimension. That is, an isomorphism sends a basis of n elements to a basis of n elements
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Transcript

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00:01 To discuss in this question, firstly we write first part there is let d imv is equal to dimv that is n.
00:10 Then here in this question let from alpha 1 dot dot a dot alpha and that is b of basis or v.
00:23 Then here is in this question this is new t v that is w is is is is is is iso.
00:34 Then in this question we can take this as here we knew show that we now to show that this is v now to show that that that is our that is alpha on dot dot dot dot here is t that is alpha n this is alpha n this is alpha and this is t b this is the linearly then after us in this question this is independent let let let this is c1 t that is alpha 1 plus dot dot dot c n t alpha n that is equal to 0 then here is t c1 alpha 1 alpha 1 plus cn alpha and alpha and c1 alpha and dot dot dot here that is belongs to k and this is then here is since t is 1 1 1 since t is 1 write that since t is 1 1 so here as we have to put in this question here that t is equal to 0 then after that hence hence we have to put in this question that have to firstly take next step hence there is in this question firstly we have to take that hence t to write that t that is how is write that t firstly, we have to write that t there is 0.
02:25 Then therefore, this is therefore t alpha 1 dot dot t alpha n is linearly then independent.
02:43 Then after that in this question, now let this is now let this is since the t is into then x is belong to v.
02:57 Then in this question here is t.
03:00 X is equal to w then this is equal to t summation that is from n i is equal to 1 from that phi that is w that is summation here is n i is equal to 1 this is lambda t that is alpha i that is equal to omega then in this question there is t alpha plus t alpha n is and also this is an span so that t alpha 1 to dot dot alpha n be the basis for w this is basis for w thus then in this question we get next step that take we have to take here is beta is equal to t alpha 1 so this is beta 1 .0 .beta n.
04:02 Then this is basis.
04:07 Okay, then for j .w.
04:09 Then after that, hence t maps a basis v into the basis of javlou.
04:15 If the t is is is isomorphism...
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