Linear maps characteristics II
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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Exercise:
Let V be a finite dimensional vector space and v_...v_nin V be a basis for V. Let w_...w_nin W be arbitrary vectors. Then exists ! a linear map T:Vlongrightarrow W s.t. Tv_iw_iquad forall leq ileq n.
Solution:
Proof of the existence of T Let vin V. We want to define Tv. Since v_...v_n is a basis exists a_...a_nin K s.t. va_v_+...+a_nv_n. We define Tv:a_w_+...+a_nw_n. Note that T is well defined because forall v the coeffs. a_...a_n of the linear combination a_v_+...+a_nv_n that gives v are unique. bf We claim that T is linear. Indeed let vv'in V. Longrightarrow a_...a_nin K s.t. va_v_+...+a_nv_n and exists a_'...a_n'in K s.t. v'a_'v_+...+a_n'v_n Longrightarrow v+v'a_+a_'v_+...+a_n+a_n'v_n Longrightarrow by our definition of T Tv+v'a_+a_'w_+...+a_n+a_n'w_n a_w_+...+a_nw_n+a_'w_+...+a_n'w_n Tv+Tv'. Also if alphain K then alpha valpha a_v_+...+alpha a_nv_n Longrightarrow Talpha valpha a_w_+...+alpha a_nw_n alphaa_w_+...+a_nw_nalpha Tv. This proves that T:Vlongrightarrow W is a linear map. bf We claim that Tv_iw_iquad forall leq ileq n. Indeed v_i v_+...+ v_i+...+ v_n Longrightarrow Tv_i w_+...+ w_i+...+ w_nw_i. This concludes the proof of the existence of T. Proof of the uniqueness of T Suppose TS:Vlongrightarrow W are two linear maps with Tv_iSv_iw_iquad forall leq ileq n. Let vin V. We have to prove that TvSv. Write va_v_+...+a_nv_n. TvTa_v_+...+a_nv_na_Tv_+...+a_nTv_n a_Sv_+...+a_nSv_n Sa_v_+...+a_nv_n Sv
Let V be a finite dimensional vector space and v_...v_nin V be a basis for V. Let w_...w_nin W be arbitrary vectors. Then exists ! a linear map T:Vlongrightarrow W s.t. Tv_iw_iquad forall leq ileq n.
Solution:
Proof of the existence of T Let vin V. We want to define Tv. Since v_...v_n is a basis exists a_...a_nin K s.t. va_v_+...+a_nv_n. We define Tv:a_w_+...+a_nw_n. Note that T is well defined because forall v the coeffs. a_...a_n of the linear combination a_v_+...+a_nv_n that gives v are unique. bf We claim that T is linear. Indeed let vv'in V. Longrightarrow a_...a_nin K s.t. va_v_+...+a_nv_n and exists a_'...a_n'in K s.t. v'a_'v_+...+a_n'v_n Longrightarrow v+v'a_+a_'v_+...+a_n+a_n'v_n Longrightarrow by our definition of T Tv+v'a_+a_'w_+...+a_n+a_n'w_n a_w_+...+a_nw_n+a_'w_+...+a_n'w_n Tv+Tv'. Also if alphain K then alpha valpha a_v_+...+alpha a_nv_n Longrightarrow Talpha valpha a_w_+...+alpha a_nw_n alphaa_w_+...+a_nw_nalpha Tv. This proves that T:Vlongrightarrow W is a linear map. bf We claim that Tv_iw_iquad forall leq ileq n. Indeed v_i v_+...+ v_i+...+ v_n Longrightarrow Tv_i w_+...+ w_i+...+ w_nw_i. This concludes the proof of the existence of T. Proof of the uniqueness of T Suppose TS:Vlongrightarrow W are two linear maps with Tv_iSv_iw_iquad forall leq ileq n. Let vin V. We have to prove that TvSv. Write va_v_+...+a_nv_n. TvTa_v_+...+a_nv_na_Tv_+...+a_nTv_n a_Sv_+...+a_nSv_n Sa_v_+...+a_nv_n Sv
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Exercise:
Let V be a finite dimensional vector space and v_...v_nin V be a basis for V. Let w_...w_nin W be arbitrary vectors. Then exists ! a linear map T:Vlongrightarrow W s.t. Tv_iw_iquad forall leq ileq n.
Solution:
Proof of the existence of T Let vin V. We want to define Tv. Since v_...v_n is a basis exists a_...a_nin K s.t. va_v_+...+a_nv_n. We define Tv:a_w_+...+a_nw_n. Note that T is well defined because forall v the coeffs. a_...a_n of the linear combination a_v_+...+a_nv_n that gives v are unique. bf We claim that T is linear. Indeed let vv'in V. Longrightarrow a_...a_nin K s.t. va_v_+...+a_nv_n and exists a_'...a_n'in K s.t. v'a_'v_+...+a_n'v_n Longrightarrow v+v'a_+a_'v_+...+a_n+a_n'v_n Longrightarrow by our definition of T Tv+v'a_+a_'w_+...+a_n+a_n'w_n a_w_+...+a_nw_n+a_'w_+...+a_n'w_n Tv+Tv'. Also if alphain K then alpha valpha a_v_+...+alpha a_nv_n Longrightarrow Talpha valpha a_w_+...+alpha a_nw_n alphaa_w_+...+a_nw_nalpha Tv. This proves that T:Vlongrightarrow W is a linear map. bf We claim that Tv_iw_iquad forall leq ileq n. Indeed v_i v_+...+ v_i+...+ v_n Longrightarrow Tv_i w_+...+ w_i+...+ w_nw_i. This concludes the proof of the existence of T. Proof of the uniqueness of T Suppose TS:Vlongrightarrow W are two linear maps with Tv_iSv_iw_iquad forall leq ileq n. Let vin V. We have to prove that TvSv. Write va_v_+...+a_nv_n. TvTa_v_+...+a_nv_na_Tv_+...+a_nTv_n a_Sv_+...+a_nSv_n Sa_v_+...+a_nv_n Sv
Let V be a finite dimensional vector space and v_...v_nin V be a basis for V. Let w_...w_nin W be arbitrary vectors. Then exists ! a linear map T:Vlongrightarrow W s.t. Tv_iw_iquad forall leq ileq n.
Solution:
Proof of the existence of T Let vin V. We want to define Tv. Since v_...v_n is a basis exists a_...a_nin K s.t. va_v_+...+a_nv_n. We define Tv:a_w_+...+a_nw_n. Note that T is well defined because forall v the coeffs. a_...a_n of the linear combination a_v_+...+a_nv_n that gives v are unique. bf We claim that T is linear. Indeed let vv'in V. Longrightarrow a_...a_nin K s.t. va_v_+...+a_nv_n and exists a_'...a_n'in K s.t. v'a_'v_+...+a_n'v_n Longrightarrow v+v'a_+a_'v_+...+a_n+a_n'v_n Longrightarrow by our definition of T Tv+v'a_+a_'w_+...+a_n+a_n'w_n a_w_+...+a_nw_n+a_'w_+...+a_n'w_n Tv+Tv'. Also if alphain K then alpha valpha a_v_+...+alpha a_nv_n Longrightarrow Talpha valpha a_w_+...+alpha a_nw_n alphaa_w_+...+a_nw_nalpha Tv. This proves that T:Vlongrightarrow W is a linear map. bf We claim that Tv_iw_iquad forall leq ileq n. Indeed v_i v_+...+ v_i+...+ v_n Longrightarrow Tv_i w_+...+ w_i+...+ w_nw_i. This concludes the proof of the existence of T. Proof of the uniqueness of T Suppose TS:Vlongrightarrow W are two linear maps with Tv_iSv_iw_iquad forall leq ileq n. Let vin V. We have to prove that TvSv. Write va_v_+...+a_nv_n. TvTa_v_+...+a_nv_na_Tv_+...+a_nTv_n a_Sv_+...+a_nSv_n Sa_v_+...+a_nv_n Sv
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