Span and linear combinations
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 varnothing neq S subseteq V. Then textSpSa_v_+...+a_nv_n|nin mathbbN a_iin K v_iin S forall leq ileq n textsubset of W obtained by taking all possible linear combinations of vectors from S Reminder: The set S can be finite or infinite. Regardless of that each linear combination uses only a finite number of elements from S.
Solution:
Proof. Denote by overlinetextSpSa_v_+...+a_nv_n|nin mathbbN a_iin K v_iin S forall leq i leq n. We'll first show that: abcliste abc S subseteq overlinetextSpS. abc overlinetextSpS subseteq V is a linear subspace. abc forall subspaces Wsubseteq V with Ssubseteq W we have overlinetextSpSsubseteq W. abcliste Note that a+bLongrightarrow textSpSsubseteq overlinetextSpS by the previous Lemma and cLongrightarrow overlinetextSpS subseteq bigcap_W in mathcalN W textSpS with mathcalNW|Ssubseteq W Wsubseteq V is a subspace. So together we have overlinetextSpStextSpS which is what we wanted. Now we continue by proving a-c. abcliste abc If vin S Longrightarrow v v in overlinetextSpS. abc We need to show that in overlinetextSpS and that forall vwin overlinetextSpS forall alpha beta in K: alpha v+beta w in overlinetextSpS. Indeed Sneq varnothing by asption. Pick any u in S. Then uin overlinetextSpS. Let vwin overlinetextSpS alpha betain K. Write va_v_+...+a_nv_n wb_w_+...+b_mw_m where nmin mathbbN a_ib_j in K v_iw_j in S forall leq i leq n leq jleq m. Now alpha v+beta w alpha a_v_+...+alpha a_nv_n+beta b_w_+...+beta b_mw_m which is a linear combination of the vectors v_...v_nw_...w_min S with coeffs alpha a_...alpha a_nbeta b_...beta b_m in K. This proves b. abc Let Wsubseteq V be a linear subspace s.t. S subseteq W. Let vin overlinetextSpS. We need to show that vin W. Indeed write va_v_+...+a_nv_n for some nin mathbbN v_iin S a_i in K. By asption S subseteq W hence v_iin W forall i. Since W is a linear subspace we have va_v_+...+a_nv_n in W. abcliste
Let varnothing neq S subseteq V. Then textSpSa_v_+...+a_nv_n|nin mathbbN a_iin K v_iin S forall leq ileq n textsubset of W obtained by taking all possible linear combinations of vectors from S Reminder: The set S can be finite or infinite. Regardless of that each linear combination uses only a finite number of elements from S.
Solution:
Proof. Denote by overlinetextSpSa_v_+...+a_nv_n|nin mathbbN a_iin K v_iin S forall leq i leq n. We'll first show that: abcliste abc S subseteq overlinetextSpS. abc overlinetextSpS subseteq V is a linear subspace. abc forall subspaces Wsubseteq V with Ssubseteq W we have overlinetextSpSsubseteq W. abcliste Note that a+bLongrightarrow textSpSsubseteq overlinetextSpS by the previous Lemma and cLongrightarrow overlinetextSpS subseteq bigcap_W in mathcalN W textSpS with mathcalNW|Ssubseteq W Wsubseteq V is a subspace. So together we have overlinetextSpStextSpS which is what we wanted. Now we continue by proving a-c. abcliste abc If vin S Longrightarrow v v in overlinetextSpS. abc We need to show that in overlinetextSpS and that forall vwin overlinetextSpS forall alpha beta in K: alpha v+beta w in overlinetextSpS. Indeed Sneq varnothing by asption. Pick any u in S. Then uin overlinetextSpS. Let vwin overlinetextSpS alpha betain K. Write va_v_+...+a_nv_n wb_w_+...+b_mw_m where nmin mathbbN a_ib_j in K v_iw_j in S forall leq i leq n leq jleq m. Now alpha v+beta w alpha a_v_+...+alpha a_nv_n+beta b_w_+...+beta b_mw_m which is a linear combination of the vectors v_...v_nw_...w_min S with coeffs alpha a_...alpha a_nbeta b_...beta b_m in K. This proves b. abc Let Wsubseteq V be a linear subspace s.t. S subseteq W. Let vin overlinetextSpS. We need to show that vin W. Indeed write va_v_+...+a_nv_n for some nin mathbbN v_iin S a_i in K. By asption S subseteq W hence v_iin W forall i. Since W is a linear subspace we have va_v_+...+a_nv_n in W. abcliste
Meta Information
Exercise:
Let varnothing neq S subseteq V. Then textSpSa_v_+...+a_nv_n|nin mathbbN a_iin K v_iin S forall leq ileq n textsubset of W obtained by taking all possible linear combinations of vectors from S Reminder: The set S can be finite or infinite. Regardless of that each linear combination uses only a finite number of elements from S.
Solution:
Proof. Denote by overlinetextSpSa_v_+...+a_nv_n|nin mathbbN a_iin K v_iin S forall leq i leq n. We'll first show that: abcliste abc S subseteq overlinetextSpS. abc overlinetextSpS subseteq V is a linear subspace. abc forall subspaces Wsubseteq V with Ssubseteq W we have overlinetextSpSsubseteq W. abcliste Note that a+bLongrightarrow textSpSsubseteq overlinetextSpS by the previous Lemma and cLongrightarrow overlinetextSpS subseteq bigcap_W in mathcalN W textSpS with mathcalNW|Ssubseteq W Wsubseteq V is a subspace. So together we have overlinetextSpStextSpS which is what we wanted. Now we continue by proving a-c. abcliste abc If vin S Longrightarrow v v in overlinetextSpS. abc We need to show that in overlinetextSpS and that forall vwin overlinetextSpS forall alpha beta in K: alpha v+beta w in overlinetextSpS. Indeed Sneq varnothing by asption. Pick any u in S. Then uin overlinetextSpS. Let vwin overlinetextSpS alpha betain K. Write va_v_+...+a_nv_n wb_w_+...+b_mw_m where nmin mathbbN a_ib_j in K v_iw_j in S forall leq i leq n leq jleq m. Now alpha v+beta w alpha a_v_+...+alpha a_nv_n+beta b_w_+...+beta b_mw_m which is a linear combination of the vectors v_...v_nw_...w_min S with coeffs alpha a_...alpha a_nbeta b_...beta b_m in K. This proves b. abc Let Wsubseteq V be a linear subspace s.t. S subseteq W. Let vin overlinetextSpS. We need to show that vin W. Indeed write va_v_+...+a_nv_n for some nin mathbbN v_iin S a_i in K. By asption S subseteq W hence v_iin W forall i. Since W is a linear subspace we have va_v_+...+a_nv_n in W. abcliste
Let varnothing neq S subseteq V. Then textSpSa_v_+...+a_nv_n|nin mathbbN a_iin K v_iin S forall leq ileq n textsubset of W obtained by taking all possible linear combinations of vectors from S Reminder: The set S can be finite or infinite. Regardless of that each linear combination uses only a finite number of elements from S.
Solution:
Proof. Denote by overlinetextSpSa_v_+...+a_nv_n|nin mathbbN a_iin K v_iin S forall leq i leq n. We'll first show that: abcliste abc S subseteq overlinetextSpS. abc overlinetextSpS subseteq V is a linear subspace. abc forall subspaces Wsubseteq V with Ssubseteq W we have overlinetextSpSsubseteq W. abcliste Note that a+bLongrightarrow textSpSsubseteq overlinetextSpS by the previous Lemma and cLongrightarrow overlinetextSpS subseteq bigcap_W in mathcalN W textSpS with mathcalNW|Ssubseteq W Wsubseteq V is a subspace. So together we have overlinetextSpStextSpS which is what we wanted. Now we continue by proving a-c. abcliste abc If vin S Longrightarrow v v in overlinetextSpS. abc We need to show that in overlinetextSpS and that forall vwin overlinetextSpS forall alpha beta in K: alpha v+beta w in overlinetextSpS. Indeed Sneq varnothing by asption. Pick any u in S. Then uin overlinetextSpS. Let vwin overlinetextSpS alpha betain K. Write va_v_+...+a_nv_n wb_w_+...+b_mw_m where nmin mathbbN a_ib_j in K v_iw_j in S forall leq i leq n leq jleq m. Now alpha v+beta w alpha a_v_+...+alpha a_nv_n+beta b_w_+...+beta b_mw_m which is a linear combination of the vectors v_...v_nw_...w_min S with coeffs alpha a_...alpha a_nbeta b_...beta b_m in K. This proves b. abc Let Wsubseteq V be a linear subspace s.t. S subseteq W. Let vin overlinetextSpS. We need to show that vin W. Indeed write va_v_+...+a_nv_n for some nin mathbbN v_iin S a_i in K. By asption S subseteq W hence v_iin W forall i. Since W is a linear subspace we have va_v_+...+a_nv_n in W. abcliste
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