Constructing new vector spaces out of old ones
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
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Video
\(\LaTeX\)
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Exercise:
Let VW be two vector spaces over K. Then textHomVW has the structure of a vector space over K if we ow it with the following operations: forall T_ T_in textHomVW T_+T_v&:T_v+T_v forall vin V forall Tin textHomVW alpha in K alpha Tv&:alpha Tv forall vin V
Solution:
Proof. We claim that T_+T_in textHomVW i.e. T_+T_ is a linear map. Indeed T_+T_av T_av+T_avquad textlin. aT_v+aT_v aT_v+T_c aT_+T_vquad textdef. T_+T_v_+v_quad textdef. T_v_+v_+T_v_+v_quad textlin. T_v_+T_v_+T_v_+T_v_quad textdef. T_+T_v_+T_+T_v_ This shows T_+T_ is a linear map. The proof that alpha T is linear is similar. So far we have defined +: textHomVWtimes textHomVW longrightarrow textHomVW : Ktimes textHomVW longrightarrow textHomVW. bf Neutral element. The map :Vlongrightarrow W vmapsto forall vin V is linear. We claim that +TT. Indeed +Tvv+Tv+TvTv forall vin V. Similarly one shows that T+T. Now it remains to be proven that the eight axioms that define a vector space hold. Example distributivity: forall alpha in K T_ T_ in textHomVW: alpha T_+T_alpha T_+alpha T_ Indeed forall vin V: alphaT_+T_valpha T_+T_v alphaT_v+T_v alpha T_v+alpha T_v alpha T_v+alpha T_v alpha T_+alpha T_v &Longrightarrow alpha T_+T_alpha T_+alpha T_
Let VW be two vector spaces over K. Then textHomVW has the structure of a vector space over K if we ow it with the following operations: forall T_ T_in textHomVW T_+T_v&:T_v+T_v forall vin V forall Tin textHomVW alpha in K alpha Tv&:alpha Tv forall vin V
Solution:
Proof. We claim that T_+T_in textHomVW i.e. T_+T_ is a linear map. Indeed T_+T_av T_av+T_avquad textlin. aT_v+aT_v aT_v+T_c aT_+T_vquad textdef. T_+T_v_+v_quad textdef. T_v_+v_+T_v_+v_quad textlin. T_v_+T_v_+T_v_+T_v_quad textdef. T_+T_v_+T_+T_v_ This shows T_+T_ is a linear map. The proof that alpha T is linear is similar. So far we have defined +: textHomVWtimes textHomVW longrightarrow textHomVW : Ktimes textHomVW longrightarrow textHomVW. bf Neutral element. The map :Vlongrightarrow W vmapsto forall vin V is linear. We claim that +TT. Indeed +Tvv+Tv+TvTv forall vin V. Similarly one shows that T+T. Now it remains to be proven that the eight axioms that define a vector space hold. Example distributivity: forall alpha in K T_ T_ in textHomVW: alpha T_+T_alpha T_+alpha T_ Indeed forall vin V: alphaT_+T_valpha T_+T_v alphaT_v+T_v alpha T_v+alpha T_v alpha T_v+alpha T_v alpha T_+alpha T_v &Longrightarrow alpha T_+T_alpha T_+alpha T_
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Exercise:
Let VW be two vector spaces over K. Then textHomVW has the structure of a vector space over K if we ow it with the following operations: forall T_ T_in textHomVW T_+T_v&:T_v+T_v forall vin V forall Tin textHomVW alpha in K alpha Tv&:alpha Tv forall vin V
Solution:
Proof. We claim that T_+T_in textHomVW i.e. T_+T_ is a linear map. Indeed T_+T_av T_av+T_avquad textlin. aT_v+aT_v aT_v+T_c aT_+T_vquad textdef. T_+T_v_+v_quad textdef. T_v_+v_+T_v_+v_quad textlin. T_v_+T_v_+T_v_+T_v_quad textdef. T_+T_v_+T_+T_v_ This shows T_+T_ is a linear map. The proof that alpha T is linear is similar. So far we have defined +: textHomVWtimes textHomVW longrightarrow textHomVW : Ktimes textHomVW longrightarrow textHomVW. bf Neutral element. The map :Vlongrightarrow W vmapsto forall vin V is linear. We claim that +TT. Indeed +Tvv+Tv+TvTv forall vin V. Similarly one shows that T+T. Now it remains to be proven that the eight axioms that define a vector space hold. Example distributivity: forall alpha in K T_ T_ in textHomVW: alpha T_+T_alpha T_+alpha T_ Indeed forall vin V: alphaT_+T_valpha T_+T_v alphaT_v+T_v alpha T_v+alpha T_v alpha T_v+alpha T_v alpha T_+alpha T_v &Longrightarrow alpha T_+T_alpha T_+alpha T_
Let VW be two vector spaces over K. Then textHomVW has the structure of a vector space over K if we ow it with the following operations: forall T_ T_in textHomVW T_+T_v&:T_v+T_v forall vin V forall Tin textHomVW alpha in K alpha Tv&:alpha Tv forall vin V
Solution:
Proof. We claim that T_+T_in textHomVW i.e. T_+T_ is a linear map. Indeed T_+T_av T_av+T_avquad textlin. aT_v+aT_v aT_v+T_c aT_+T_vquad textdef. T_+T_v_+v_quad textdef. T_v_+v_+T_v_+v_quad textlin. T_v_+T_v_+T_v_+T_v_quad textdef. T_+T_v_+T_+T_v_ This shows T_+T_ is a linear map. The proof that alpha T is linear is similar. So far we have defined +: textHomVWtimes textHomVW longrightarrow textHomVW : Ktimes textHomVW longrightarrow textHomVW. bf Neutral element. The map :Vlongrightarrow W vmapsto forall vin V is linear. We claim that +TT. Indeed +Tvv+Tv+TvTv forall vin V. Similarly one shows that T+T. Now it remains to be proven that the eight axioms that define a vector space hold. Example distributivity: forall alpha in K T_ T_ in textHomVW: alpha T_+T_alpha T_+alpha T_ Indeed forall vin V: alphaT_+T_valpha T_+T_v alphaT_v+T_v alpha T_v+alpha T_v alpha T_v+alpha T_v alpha T_+alpha T_v &Longrightarrow alpha T_+T_alpha T_+alpha T_
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