Matrix multiplication characteristics
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:
abcliste abc forall Ain M_mtimes nK Bin M_ntimes pK Cin M_ptimes qK we have: ABC ABC in M_mtimes qK. abc forall Ain M_mtimes nK Cin M_ntimes pK C'in M_qtimes mK we have: A+BC AC+BC in M_mtimes pK C'A+B C'A+C'B in M_qtimes nK. abc forall Ain M_mtimes nK we have I_mAA AI_nA abc forall alpha Ain M_mtimes nK Bin M_ntimes pK we have: alpha ABalpha ABAalpha B abcliste
Solution:
abcliste abc Let T_A:K^nlongrightarrow K^m T_B:K^plongrightarrow K^n T_C:K^qlongrightarrow K^p be the linear maps defined by ABC. i.e. T_AvAvquad forall vin K^n T_BwBwquad forall win K^p T_CuCuquad forall uin K^q forall r denote by epsilon_r the standard basis of K^r. We have seen that T_A_epsilon_m^epsilon_nAT_B_epsilon_n^epsilon_pB T_C_epsilon_p^epsilon_qC. Now T_Acirc T_Bcirc T_C T_Acirc T_Bcirc T_C K^prightarrow K^mK^qrightarrow K^mK^nrightarrow K^mK^qrightarrow K^n We now pass to the matrix representation of the above leftT_Acirc T_Bcirc T_Cright_epsilon_m^epsilon_q leftT_Acirc T_Bcirc T_Cright_epsilon_p^epsilon_q &Longrightarrow T_Acirc T_B_epsilon_m^epsilon_p T_C_epsilon_p^epsilon_q T_A_epsilon_m^epsilon_n T_Bcirc T_C_epsilon_n^epsilon_q &Longrightarrow leftT_A_epsilon_m^epsilon_n T_B_epsilon_n^epsilon_pright T_C_epsilon_p^epsilon_qT_A_epsilon_m^epsilon_n leftT_B_epsilon_n^epsilon_p T_C_epsilon_p^epsilon_qright &Longrightarrow A B C A B C. abc T_A+T_Bcirc T_C T_Acirc T_C +T_B circ T_C &Longrightarrow leftT_Acirc T_Bcirc T_Cright_epsilon_m^epsilon_p T_Acirc T_C + T_Bcirc T_C_epsilon_m^epsilon_p &Longrightarrow T_A+ T_B_epsilon_m^epsilon_n T_C_epsilon_n^epsilon_p T_Acirc T_C_epsilon_m^epsilon_p+T_Bcirc T_C_epsilon_m^epsilon_p &Longrightarrow A+B C A C + B C. The other identity in b is similar. abc I_mid_K^m_epsilon_m^epsilon_mquad id_K^mcirc T_A T_A &Longrightarrow id_K^mcirc T_A_epsilon_m^epsilon_nT_A_epsilon_m^epsilon_n A &Longrightarrow I_m AA. The other identity is similar. abc bf claim. Let V be a finite dimensional VS over K alpha in K. Let B be a basis for V. Consider the map Q:Vlongrightarrow V Qv: alpha v. Then V is linear and moreover we have Q_mathcalB^mathcalBalpha I_n where ntextdimV. to be proven. abcliste
abcliste abc forall Ain M_mtimes nK Bin M_ntimes pK Cin M_ptimes qK we have: ABC ABC in M_mtimes qK. abc forall Ain M_mtimes nK Cin M_ntimes pK C'in M_qtimes mK we have: A+BC AC+BC in M_mtimes pK C'A+B C'A+C'B in M_qtimes nK. abc forall Ain M_mtimes nK we have I_mAA AI_nA abc forall alpha Ain M_mtimes nK Bin M_ntimes pK we have: alpha ABalpha ABAalpha B abcliste
Solution:
abcliste abc Let T_A:K^nlongrightarrow K^m T_B:K^plongrightarrow K^n T_C:K^qlongrightarrow K^p be the linear maps defined by ABC. i.e. T_AvAvquad forall vin K^n T_BwBwquad forall win K^p T_CuCuquad forall uin K^q forall r denote by epsilon_r the standard basis of K^r. We have seen that T_A_epsilon_m^epsilon_nAT_B_epsilon_n^epsilon_pB T_C_epsilon_p^epsilon_qC. Now T_Acirc T_Bcirc T_C T_Acirc T_Bcirc T_C K^prightarrow K^mK^qrightarrow K^mK^nrightarrow K^mK^qrightarrow K^n We now pass to the matrix representation of the above leftT_Acirc T_Bcirc T_Cright_epsilon_m^epsilon_q leftT_Acirc T_Bcirc T_Cright_epsilon_p^epsilon_q &Longrightarrow T_Acirc T_B_epsilon_m^epsilon_p T_C_epsilon_p^epsilon_q T_A_epsilon_m^epsilon_n T_Bcirc T_C_epsilon_n^epsilon_q &Longrightarrow leftT_A_epsilon_m^epsilon_n T_B_epsilon_n^epsilon_pright T_C_epsilon_p^epsilon_qT_A_epsilon_m^epsilon_n leftT_B_epsilon_n^epsilon_p T_C_epsilon_p^epsilon_qright &Longrightarrow A B C A B C. abc T_A+T_Bcirc T_C T_Acirc T_C +T_B circ T_C &Longrightarrow leftT_Acirc T_Bcirc T_Cright_epsilon_m^epsilon_p T_Acirc T_C + T_Bcirc T_C_epsilon_m^epsilon_p &Longrightarrow T_A+ T_B_epsilon_m^epsilon_n T_C_epsilon_n^epsilon_p T_Acirc T_C_epsilon_m^epsilon_p+T_Bcirc T_C_epsilon_m^epsilon_p &Longrightarrow A+B C A C + B C. The other identity in b is similar. abc I_mid_K^m_epsilon_m^epsilon_mquad id_K^mcirc T_A T_A &Longrightarrow id_K^mcirc T_A_epsilon_m^epsilon_nT_A_epsilon_m^epsilon_n A &Longrightarrow I_m AA. The other identity is similar. abc bf claim. Let V be a finite dimensional VS over K alpha in K. Let B be a basis for V. Consider the map Q:Vlongrightarrow V Qv: alpha v. Then V is linear and moreover we have Q_mathcalB^mathcalBalpha I_n where ntextdimV. to be proven. abcliste
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Exercise:
abcliste abc forall Ain M_mtimes nK Bin M_ntimes pK Cin M_ptimes qK we have: ABC ABC in M_mtimes qK. abc forall Ain M_mtimes nK Cin M_ntimes pK C'in M_qtimes mK we have: A+BC AC+BC in M_mtimes pK C'A+B C'A+C'B in M_qtimes nK. abc forall Ain M_mtimes nK we have I_mAA AI_nA abc forall alpha Ain M_mtimes nK Bin M_ntimes pK we have: alpha ABalpha ABAalpha B abcliste
Solution:
abcliste abc Let T_A:K^nlongrightarrow K^m T_B:K^plongrightarrow K^n T_C:K^qlongrightarrow K^p be the linear maps defined by ABC. i.e. T_AvAvquad forall vin K^n T_BwBwquad forall win K^p T_CuCuquad forall uin K^q forall r denote by epsilon_r the standard basis of K^r. We have seen that T_A_epsilon_m^epsilon_nAT_B_epsilon_n^epsilon_pB T_C_epsilon_p^epsilon_qC. Now T_Acirc T_Bcirc T_C T_Acirc T_Bcirc T_C K^prightarrow K^mK^qrightarrow K^mK^nrightarrow K^mK^qrightarrow K^n We now pass to the matrix representation of the above leftT_Acirc T_Bcirc T_Cright_epsilon_m^epsilon_q leftT_Acirc T_Bcirc T_Cright_epsilon_p^epsilon_q &Longrightarrow T_Acirc T_B_epsilon_m^epsilon_p T_C_epsilon_p^epsilon_q T_A_epsilon_m^epsilon_n T_Bcirc T_C_epsilon_n^epsilon_q &Longrightarrow leftT_A_epsilon_m^epsilon_n T_B_epsilon_n^epsilon_pright T_C_epsilon_p^epsilon_qT_A_epsilon_m^epsilon_n leftT_B_epsilon_n^epsilon_p T_C_epsilon_p^epsilon_qright &Longrightarrow A B C A B C. abc T_A+T_Bcirc T_C T_Acirc T_C +T_B circ T_C &Longrightarrow leftT_Acirc T_Bcirc T_Cright_epsilon_m^epsilon_p T_Acirc T_C + T_Bcirc T_C_epsilon_m^epsilon_p &Longrightarrow T_A+ T_B_epsilon_m^epsilon_n T_C_epsilon_n^epsilon_p T_Acirc T_C_epsilon_m^epsilon_p+T_Bcirc T_C_epsilon_m^epsilon_p &Longrightarrow A+B C A C + B C. The other identity in b is similar. abc I_mid_K^m_epsilon_m^epsilon_mquad id_K^mcirc T_A T_A &Longrightarrow id_K^mcirc T_A_epsilon_m^epsilon_nT_A_epsilon_m^epsilon_n A &Longrightarrow I_m AA. The other identity is similar. abc bf claim. Let V be a finite dimensional VS over K alpha in K. Let B be a basis for V. Consider the map Q:Vlongrightarrow V Qv: alpha v. Then V is linear and moreover we have Q_mathcalB^mathcalBalpha I_n where ntextdimV. to be proven. abcliste
abcliste abc forall Ain M_mtimes nK Bin M_ntimes pK Cin M_ptimes qK we have: ABC ABC in M_mtimes qK. abc forall Ain M_mtimes nK Cin M_ntimes pK C'in M_qtimes mK we have: A+BC AC+BC in M_mtimes pK C'A+B C'A+C'B in M_qtimes nK. abc forall Ain M_mtimes nK we have I_mAA AI_nA abc forall alpha Ain M_mtimes nK Bin M_ntimes pK we have: alpha ABalpha ABAalpha B abcliste
Solution:
abcliste abc Let T_A:K^nlongrightarrow K^m T_B:K^plongrightarrow K^n T_C:K^qlongrightarrow K^p be the linear maps defined by ABC. i.e. T_AvAvquad forall vin K^n T_BwBwquad forall win K^p T_CuCuquad forall uin K^q forall r denote by epsilon_r the standard basis of K^r. We have seen that T_A_epsilon_m^epsilon_nAT_B_epsilon_n^epsilon_pB T_C_epsilon_p^epsilon_qC. Now T_Acirc T_Bcirc T_C T_Acirc T_Bcirc T_C K^prightarrow K^mK^qrightarrow K^mK^nrightarrow K^mK^qrightarrow K^n We now pass to the matrix representation of the above leftT_Acirc T_Bcirc T_Cright_epsilon_m^epsilon_q leftT_Acirc T_Bcirc T_Cright_epsilon_p^epsilon_q &Longrightarrow T_Acirc T_B_epsilon_m^epsilon_p T_C_epsilon_p^epsilon_q T_A_epsilon_m^epsilon_n T_Bcirc T_C_epsilon_n^epsilon_q &Longrightarrow leftT_A_epsilon_m^epsilon_n T_B_epsilon_n^epsilon_pright T_C_epsilon_p^epsilon_qT_A_epsilon_m^epsilon_n leftT_B_epsilon_n^epsilon_p T_C_epsilon_p^epsilon_qright &Longrightarrow A B C A B C. abc T_A+T_Bcirc T_C T_Acirc T_C +T_B circ T_C &Longrightarrow leftT_Acirc T_Bcirc T_Cright_epsilon_m^epsilon_p T_Acirc T_C + T_Bcirc T_C_epsilon_m^epsilon_p &Longrightarrow T_A+ T_B_epsilon_m^epsilon_n T_C_epsilon_n^epsilon_p T_Acirc T_C_epsilon_m^epsilon_p+T_Bcirc T_C_epsilon_m^epsilon_p &Longrightarrow A+B C A C + B C. The other identity in b is similar. abc I_mid_K^m_epsilon_m^epsilon_mquad id_K^mcirc T_A T_A &Longrightarrow id_K^mcirc T_A_epsilon_m^epsilon_nT_A_epsilon_m^epsilon_n A &Longrightarrow I_m AA. The other identity is similar. abc bf claim. Let V be a finite dimensional VS over K alpha in K. Let B be a basis for V. Consider the map Q:Vlongrightarrow V Qv: alpha v. Then V is linear and moreover we have Q_mathcalB^mathcalBalpha I_n where ntextdimV. to be proven. abcliste
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