Row-equivalence and row space
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:
If AB in M_mtimes nK are row-equivalent matrices then Row SA Row SB. The statement does NOT hold for cols!
Solution:
Proof. Let Min M_mtimes nK. When we exchange two rows of M the span of the rows is not affected. So Row SM doesn't change under such an operation. The operation R_i longrightarrow lambda R_i where lambdain Kbackslash does NOT change the span of the rows of M. So Row SM is left unchanged after such an operation. The operation R_i+lamba R_j longrightarrow R_i where ineq j lambdain K. We claim that Spw_...w_mSpw_...w_i+lambda w_j...w_m. Indeed Spw_...w_i+lambda w_j...w_m subseteq Spw_...w_m because every linear combination of the left side is obviously also a linear combination of the right side. But we also have Spw_...w_msubseteq Spw_...w_i+lambda w_j...w_m because w_iw_i+lambda w_j-lambda w_j and since ineq j when we write a linear combination of w_...w_n we will get a linear combination of w_...w_i+lambda w_j...w_m. This proves the claim. From the claim it follows that the operation R_i+lambda R_j longrightarrow R_i ineq j lambda in K does not change Row SM. Now if A B are row equivalent then by definition exists afinite sequence of matrices A_AA_...A_r- A_rB s.t. A_i+ is obtained from A_i by an elementary row operation forall leq i leq r-. AA_longrightarrow A_ longrightarrow ... longrightarrow A_rB. By what we proved earlier we have Row SA Row SA_ Row SA_ ... Row SA_r Row SB
If AB in M_mtimes nK are row-equivalent matrices then Row SA Row SB. The statement does NOT hold for cols!
Solution:
Proof. Let Min M_mtimes nK. When we exchange two rows of M the span of the rows is not affected. So Row SM doesn't change under such an operation. The operation R_i longrightarrow lambda R_i where lambdain Kbackslash does NOT change the span of the rows of M. So Row SM is left unchanged after such an operation. The operation R_i+lamba R_j longrightarrow R_i where ineq j lambdain K. We claim that Spw_...w_mSpw_...w_i+lambda w_j...w_m. Indeed Spw_...w_i+lambda w_j...w_m subseteq Spw_...w_m because every linear combination of the left side is obviously also a linear combination of the right side. But we also have Spw_...w_msubseteq Spw_...w_i+lambda w_j...w_m because w_iw_i+lambda w_j-lambda w_j and since ineq j when we write a linear combination of w_...w_n we will get a linear combination of w_...w_i+lambda w_j...w_m. This proves the claim. From the claim it follows that the operation R_i+lambda R_j longrightarrow R_i ineq j lambda in K does not change Row SM. Now if A B are row equivalent then by definition exists afinite sequence of matrices A_AA_...A_r- A_rB s.t. A_i+ is obtained from A_i by an elementary row operation forall leq i leq r-. AA_longrightarrow A_ longrightarrow ... longrightarrow A_rB. By what we proved earlier we have Row SA Row SA_ Row SA_ ... Row SA_r Row SB
Meta Information
Exercise:
If AB in M_mtimes nK are row-equivalent matrices then Row SA Row SB. The statement does NOT hold for cols!
Solution:
Proof. Let Min M_mtimes nK. When we exchange two rows of M the span of the rows is not affected. So Row SM doesn't change under such an operation. The operation R_i longrightarrow lambda R_i where lambdain Kbackslash does NOT change the span of the rows of M. So Row SM is left unchanged after such an operation. The operation R_i+lamba R_j longrightarrow R_i where ineq j lambdain K. We claim that Spw_...w_mSpw_...w_i+lambda w_j...w_m. Indeed Spw_...w_i+lambda w_j...w_m subseteq Spw_...w_m because every linear combination of the left side is obviously also a linear combination of the right side. But we also have Spw_...w_msubseteq Spw_...w_i+lambda w_j...w_m because w_iw_i+lambda w_j-lambda w_j and since ineq j when we write a linear combination of w_...w_n we will get a linear combination of w_...w_i+lambda w_j...w_m. This proves the claim. From the claim it follows that the operation R_i+lambda R_j longrightarrow R_i ineq j lambda in K does not change Row SM. Now if A B are row equivalent then by definition exists afinite sequence of matrices A_AA_...A_r- A_rB s.t. A_i+ is obtained from A_i by an elementary row operation forall leq i leq r-. AA_longrightarrow A_ longrightarrow ... longrightarrow A_rB. By what we proved earlier we have Row SA Row SA_ Row SA_ ... Row SA_r Row SB
If AB in M_mtimes nK are row-equivalent matrices then Row SA Row SB. The statement does NOT hold for cols!
Solution:
Proof. Let Min M_mtimes nK. When we exchange two rows of M the span of the rows is not affected. So Row SM doesn't change under such an operation. The operation R_i longrightarrow lambda R_i where lambdain Kbackslash does NOT change the span of the rows of M. So Row SM is left unchanged after such an operation. The operation R_i+lamba R_j longrightarrow R_i where ineq j lambdain K. We claim that Spw_...w_mSpw_...w_i+lambda w_j...w_m. Indeed Spw_...w_i+lambda w_j...w_m subseteq Spw_...w_m because every linear combination of the left side is obviously also a linear combination of the right side. But we also have Spw_...w_msubseteq Spw_...w_i+lambda w_j...w_m because w_iw_i+lambda w_j-lambda w_j and since ineq j when we write a linear combination of w_...w_n we will get a linear combination of w_...w_i+lambda w_j...w_m. This proves the claim. From the claim it follows that the operation R_i+lambda R_j longrightarrow R_i ineq j lambda in K does not change Row SM. Now if A B are row equivalent then by definition exists afinite sequence of matrices A_AA_...A_r- A_rB s.t. A_i+ is obtained from A_i by an elementary row operation forall leq i leq r-. AA_longrightarrow A_ longrightarrow ... longrightarrow A_rB. By what we proved earlier we have Row SA Row SA_ Row SA_ ... Row SA_r Row SB
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