Bases and identity
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
Short
Video
\(\LaTeX\)
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Exercise:
Let V and W be finite dimensional spaces over K. Let ntextdimV mtextdimW. Let mathcalBB' be two bases for V and mathcalCC' two bases for W. Then textid_V_mathcalB^mathcalB'in textGL_nKtextid_Win textGL_mK and textid_V_mathcalB^mathcalB'lefttextid_V_mathcalB'^mathcalBright^- textid_W_mathcalC^mathcalC'lefttextid_W_mathcalC'^mathcalCright^-. Let T:Vlongrightarrow W be a linear map. Then: T_mathcalC'^mathcalB'textid_W_mathcalC'^mathcalC T_mathcalC^mathcalB textid_W_mathcalB^mathcalB'.
Solution:
Proof. Ttextid_Wcirc Tcirc textid_V textid_Wcirc Tcirc textid_V. &Longrightarrow T_mathcalC'^mathcalB' textid_W_mathcalC'^mathcalC Tcirc textid_V_mathcalC^mathcalB' textid_W_mathcalC'^mathcalC T_mathcalC^mathcalB textid_V_mathcalB^mathcalB'. Also we have: textid_V textid_Vcirc textid_W Longrightarrow I_n textid_V_mathcalB^mathcalBtextid_V_mathcalB^mathcalB' textid_V_mathcalB'^mathcalB and Longrightarrow I_n textid_V_mathcalB'^mathcalB'textid_V_mathcalB'^mathcalB textid_V_mathcalB^mathcalB' which results in Longrightarrow textid_V_mathcalB^mathcalB'lefttextid_V_mathcalB'^mathcalBright^- And similarly for textid_W_mathcalC^mathcalC'.
Let V and W be finite dimensional spaces over K. Let ntextdimV mtextdimW. Let mathcalBB' be two bases for V and mathcalCC' two bases for W. Then textid_V_mathcalB^mathcalB'in textGL_nKtextid_Win textGL_mK and textid_V_mathcalB^mathcalB'lefttextid_V_mathcalB'^mathcalBright^- textid_W_mathcalC^mathcalC'lefttextid_W_mathcalC'^mathcalCright^-. Let T:Vlongrightarrow W be a linear map. Then: T_mathcalC'^mathcalB'textid_W_mathcalC'^mathcalC T_mathcalC^mathcalB textid_W_mathcalB^mathcalB'.
Solution:
Proof. Ttextid_Wcirc Tcirc textid_V textid_Wcirc Tcirc textid_V. &Longrightarrow T_mathcalC'^mathcalB' textid_W_mathcalC'^mathcalC Tcirc textid_V_mathcalC^mathcalB' textid_W_mathcalC'^mathcalC T_mathcalC^mathcalB textid_V_mathcalB^mathcalB'. Also we have: textid_V textid_Vcirc textid_W Longrightarrow I_n textid_V_mathcalB^mathcalBtextid_V_mathcalB^mathcalB' textid_V_mathcalB'^mathcalB and Longrightarrow I_n textid_V_mathcalB'^mathcalB'textid_V_mathcalB'^mathcalB textid_V_mathcalB^mathcalB' which results in Longrightarrow textid_V_mathcalB^mathcalB'lefttextid_V_mathcalB'^mathcalBright^- And similarly for textid_W_mathcalC^mathcalC'.
Meta Information
Exercise:
Let V and W be finite dimensional spaces over K. Let ntextdimV mtextdimW. Let mathcalBB' be two bases for V and mathcalCC' two bases for W. Then textid_V_mathcalB^mathcalB'in textGL_nKtextid_Win textGL_mK and textid_V_mathcalB^mathcalB'lefttextid_V_mathcalB'^mathcalBright^- textid_W_mathcalC^mathcalC'lefttextid_W_mathcalC'^mathcalCright^-. Let T:Vlongrightarrow W be a linear map. Then: T_mathcalC'^mathcalB'textid_W_mathcalC'^mathcalC T_mathcalC^mathcalB textid_W_mathcalB^mathcalB'.
Solution:
Proof. Ttextid_Wcirc Tcirc textid_V textid_Wcirc Tcirc textid_V. &Longrightarrow T_mathcalC'^mathcalB' textid_W_mathcalC'^mathcalC Tcirc textid_V_mathcalC^mathcalB' textid_W_mathcalC'^mathcalC T_mathcalC^mathcalB textid_V_mathcalB^mathcalB'. Also we have: textid_V textid_Vcirc textid_W Longrightarrow I_n textid_V_mathcalB^mathcalBtextid_V_mathcalB^mathcalB' textid_V_mathcalB'^mathcalB and Longrightarrow I_n textid_V_mathcalB'^mathcalB'textid_V_mathcalB'^mathcalB textid_V_mathcalB^mathcalB' which results in Longrightarrow textid_V_mathcalB^mathcalB'lefttextid_V_mathcalB'^mathcalBright^- And similarly for textid_W_mathcalC^mathcalC'.
Let V and W be finite dimensional spaces over K. Let ntextdimV mtextdimW. Let mathcalBB' be two bases for V and mathcalCC' two bases for W. Then textid_V_mathcalB^mathcalB'in textGL_nKtextid_Win textGL_mK and textid_V_mathcalB^mathcalB'lefttextid_V_mathcalB'^mathcalBright^- textid_W_mathcalC^mathcalC'lefttextid_W_mathcalC'^mathcalCright^-. Let T:Vlongrightarrow W be a linear map. Then: T_mathcalC'^mathcalB'textid_W_mathcalC'^mathcalC T_mathcalC^mathcalB textid_W_mathcalB^mathcalB'.
Solution:
Proof. Ttextid_Wcirc Tcirc textid_V textid_Wcirc Tcirc textid_V. &Longrightarrow T_mathcalC'^mathcalB' textid_W_mathcalC'^mathcalC Tcirc textid_V_mathcalC^mathcalB' textid_W_mathcalC'^mathcalC T_mathcalC^mathcalB textid_V_mathcalB^mathcalB'. Also we have: textid_V textid_Vcirc textid_W Longrightarrow I_n textid_V_mathcalB^mathcalBtextid_V_mathcalB^mathcalB' textid_V_mathcalB'^mathcalB and Longrightarrow I_n textid_V_mathcalB'^mathcalB'textid_V_mathcalB'^mathcalB textid_V_mathcalB^mathcalB' which results in Longrightarrow textid_V_mathcalB^mathcalB'lefttextid_V_mathcalB'^mathcalBright^- And similarly for textid_W_mathcalC^mathcalC'.
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