Constructing new vector spaces out of old ones II
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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 -
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Exercise:
Let VW be finite dimensional vector spaces over K mathcalBv_...v_n a basis for V and mathcalCw_...w_n a basis for W. Then Psi_mathcalC^mathcalB:textHomVWlongrightarrow M_mtimes nK defined by Tlongmapsto T_mathcalC^mathcalB:pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & & vdots pmatrix. is a linear map and moreover an isomorphism.
Solution:
Proof. Recall that T_T_mathcalC^mathcalBPsi_mathcalCcirc Tcirc Psi_mathcalB^-. Recall also that Psi_mathcalBvv_mathcalBin K^n forall vin V Psi_mathcalCww_mathcalCin K^m forall win W Psi_mathcalB^-leftarrayc a_ vdots a_n arrayright a_v_+...+a_nv_n Psi_mathcalC^-leftarrayc b_ vdots b_n arrayright b_w_+...+b_mw_m We claim that the map Psi_mathcalC^mathcalB is bijective. Indeed define mathcalH:M_mtimes nKlongrightarrow textHomVW forall Ain M_mtimes nK mathcalHA:Psi_mathcalC^-circ T_Acirc Psi_mathcalB in textHomVW. We have Psi_mathcalC^mathcalBcirc mathcalHA Psi_mathcalC^mathcalBPsi_mathcalC^-circ T_A circ Psi_mathcalB pmatrix vdots & & vdots Psi_mathcalC^-circ T_A circ Psi_mathcalBv__mathcalC & hdots & Psi_mathcalC^-circ T_A circ Psi_mathcalBv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots T_A circ Psi_mathcalBv_ & hdots & T_A circ Psi_mathcalBv_n vdots & &vdots pmatrix pmatrix vdots & & vdots T_Ae_ & hdots & T_Ae_n vdots & &vdots pmatrix A. This proves that Psi_mathcalC^mathcalBcirc mathcalHtextid_M_mtimes nK. Now mathcalHcirc Psi_mathcalC^mathcalBPhi_mathcalC^-circ T_T_mathcalC^mathcalBcircPhi_mathcalB T This shows that mathcalHcirc Psi_mathcalC^mathcalBtextid_textHomVW. This proves that Phi_mathcalC^mathcalB is bijective. It remains to prove that Psi_mathcalC^mathcalB is linear. Indeed let alphain K Tin textHomVW. Then Psi_mathcalC^mathcalBalpha T pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix alpha pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & &vdots pmatrix alpha Psi_mathcalC^mathcalBT Similarly one shows that Psi_mathcalC^mathcalBT_+T_Psi_mathcalC^mathcalBT_+Psi_mathcalC^mathcalBT_.
Let VW be finite dimensional vector spaces over K mathcalBv_...v_n a basis for V and mathcalCw_...w_n a basis for W. Then Psi_mathcalC^mathcalB:textHomVWlongrightarrow M_mtimes nK defined by Tlongmapsto T_mathcalC^mathcalB:pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & & vdots pmatrix. is a linear map and moreover an isomorphism.
Solution:
Proof. Recall that T_T_mathcalC^mathcalBPsi_mathcalCcirc Tcirc Psi_mathcalB^-. Recall also that Psi_mathcalBvv_mathcalBin K^n forall vin V Psi_mathcalCww_mathcalCin K^m forall win W Psi_mathcalB^-leftarrayc a_ vdots a_n arrayright a_v_+...+a_nv_n Psi_mathcalC^-leftarrayc b_ vdots b_n arrayright b_w_+...+b_mw_m We claim that the map Psi_mathcalC^mathcalB is bijective. Indeed define mathcalH:M_mtimes nKlongrightarrow textHomVW forall Ain M_mtimes nK mathcalHA:Psi_mathcalC^-circ T_Acirc Psi_mathcalB in textHomVW. We have Psi_mathcalC^mathcalBcirc mathcalHA Psi_mathcalC^mathcalBPsi_mathcalC^-circ T_A circ Psi_mathcalB pmatrix vdots & & vdots Psi_mathcalC^-circ T_A circ Psi_mathcalBv__mathcalC & hdots & Psi_mathcalC^-circ T_A circ Psi_mathcalBv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots T_A circ Psi_mathcalBv_ & hdots & T_A circ Psi_mathcalBv_n vdots & &vdots pmatrix pmatrix vdots & & vdots T_Ae_ & hdots & T_Ae_n vdots & &vdots pmatrix A. This proves that Psi_mathcalC^mathcalBcirc mathcalHtextid_M_mtimes nK. Now mathcalHcirc Psi_mathcalC^mathcalBPhi_mathcalC^-circ T_T_mathcalC^mathcalBcircPhi_mathcalB T This shows that mathcalHcirc Psi_mathcalC^mathcalBtextid_textHomVW. This proves that Phi_mathcalC^mathcalB is bijective. It remains to prove that Psi_mathcalC^mathcalB is linear. Indeed let alphain K Tin textHomVW. Then Psi_mathcalC^mathcalBalpha T pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix alpha pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & &vdots pmatrix alpha Psi_mathcalC^mathcalBT Similarly one shows that Psi_mathcalC^mathcalBT_+T_Psi_mathcalC^mathcalBT_+Psi_mathcalC^mathcalBT_.
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Exercise:
Let VW be finite dimensional vector spaces over K mathcalBv_...v_n a basis for V and mathcalCw_...w_n a basis for W. Then Psi_mathcalC^mathcalB:textHomVWlongrightarrow M_mtimes nK defined by Tlongmapsto T_mathcalC^mathcalB:pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & & vdots pmatrix. is a linear map and moreover an isomorphism.
Solution:
Proof. Recall that T_T_mathcalC^mathcalBPsi_mathcalCcirc Tcirc Psi_mathcalB^-. Recall also that Psi_mathcalBvv_mathcalBin K^n forall vin V Psi_mathcalCww_mathcalCin K^m forall win W Psi_mathcalB^-leftarrayc a_ vdots a_n arrayright a_v_+...+a_nv_n Psi_mathcalC^-leftarrayc b_ vdots b_n arrayright b_w_+...+b_mw_m We claim that the map Psi_mathcalC^mathcalB is bijective. Indeed define mathcalH:M_mtimes nKlongrightarrow textHomVW forall Ain M_mtimes nK mathcalHA:Psi_mathcalC^-circ T_Acirc Psi_mathcalB in textHomVW. We have Psi_mathcalC^mathcalBcirc mathcalHA Psi_mathcalC^mathcalBPsi_mathcalC^-circ T_A circ Psi_mathcalB pmatrix vdots & & vdots Psi_mathcalC^-circ T_A circ Psi_mathcalBv__mathcalC & hdots & Psi_mathcalC^-circ T_A circ Psi_mathcalBv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots T_A circ Psi_mathcalBv_ & hdots & T_A circ Psi_mathcalBv_n vdots & &vdots pmatrix pmatrix vdots & & vdots T_Ae_ & hdots & T_Ae_n vdots & &vdots pmatrix A. This proves that Psi_mathcalC^mathcalBcirc mathcalHtextid_M_mtimes nK. Now mathcalHcirc Psi_mathcalC^mathcalBPhi_mathcalC^-circ T_T_mathcalC^mathcalBcircPhi_mathcalB T This shows that mathcalHcirc Psi_mathcalC^mathcalBtextid_textHomVW. This proves that Phi_mathcalC^mathcalB is bijective. It remains to prove that Psi_mathcalC^mathcalB is linear. Indeed let alphain K Tin textHomVW. Then Psi_mathcalC^mathcalBalpha T pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix alpha pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & &vdots pmatrix alpha Psi_mathcalC^mathcalBT Similarly one shows that Psi_mathcalC^mathcalBT_+T_Psi_mathcalC^mathcalBT_+Psi_mathcalC^mathcalBT_.
Let VW be finite dimensional vector spaces over K mathcalBv_...v_n a basis for V and mathcalCw_...w_n a basis for W. Then Psi_mathcalC^mathcalB:textHomVWlongrightarrow M_mtimes nK defined by Tlongmapsto T_mathcalC^mathcalB:pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & & vdots pmatrix. is a linear map and moreover an isomorphism.
Solution:
Proof. Recall that T_T_mathcalC^mathcalBPsi_mathcalCcirc Tcirc Psi_mathcalB^-. Recall also that Psi_mathcalBvv_mathcalBin K^n forall vin V Psi_mathcalCww_mathcalCin K^m forall win W Psi_mathcalB^-leftarrayc a_ vdots a_n arrayright a_v_+...+a_nv_n Psi_mathcalC^-leftarrayc b_ vdots b_n arrayright b_w_+...+b_mw_m We claim that the map Psi_mathcalC^mathcalB is bijective. Indeed define mathcalH:M_mtimes nKlongrightarrow textHomVW forall Ain M_mtimes nK mathcalHA:Psi_mathcalC^-circ T_Acirc Psi_mathcalB in textHomVW. We have Psi_mathcalC^mathcalBcirc mathcalHA Psi_mathcalC^mathcalBPsi_mathcalC^-circ T_A circ Psi_mathcalB pmatrix vdots & & vdots Psi_mathcalC^-circ T_A circ Psi_mathcalBv__mathcalC & hdots & Psi_mathcalC^-circ T_A circ Psi_mathcalBv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots T_A circ Psi_mathcalBv_ & hdots & T_A circ Psi_mathcalBv_n vdots & &vdots pmatrix pmatrix vdots & & vdots T_Ae_ & hdots & T_Ae_n vdots & &vdots pmatrix A. This proves that Psi_mathcalC^mathcalBcirc mathcalHtextid_M_mtimes nK. Now mathcalHcirc Psi_mathcalC^mathcalBPhi_mathcalC^-circ T_T_mathcalC^mathcalBcircPhi_mathcalB T This shows that mathcalHcirc Psi_mathcalC^mathcalBtextid_textHomVW. This proves that Phi_mathcalC^mathcalB is bijective. It remains to prove that Psi_mathcalC^mathcalB is linear. Indeed let alphain K Tin textHomVW. Then Psi_mathcalC^mathcalBalpha T pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix pmatrix vdots & & vdots alpha Tv__mathcalC & hdots & alpha Tv_n_mathcalC vdots & &vdots pmatrix alpha pmatrix vdots & & vdots Tv__mathcalC & hdots & Tv_n_mathcalC vdots & &vdots pmatrix alpha Psi_mathcalC^mathcalBT Similarly one shows that Psi_mathcalC^mathcalBT_+T_Psi_mathcalC^mathcalBT_+Psi_mathcalC^mathcalBT_.
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