Algorithm to find a basis
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
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Let mathcalUW subseteq V subspaces. Then: abcliste abc mathcalU+mathcalWtextSpmathcalUcapmathcalW. In particular mathcalU+mathcalWsubseteq V is a subspace and mathcalUmathcalWsubseteq mathcalU+mathcalW are subspaces of mathcalU+mathcalW. abc If mathcalUmathcalW are finite dimensional then so is mathcalU+mathcalW. Moreover one can find a basis for mathcalU+mathcalW as follows: itemize item choose a basis p_dotsp_k for mathcalUcapmathcalW where ktextdimmathcalUcapmathcalW. item Ext p_dotsp_k to a basis p_dotsp_ku_dotsu_l-k of mathcalU where ltextdimmathcalU. item Ext p_dotsp_k to a basis p_dotsp_kw_dotsw_m-k of mathcalW where mtextdimmathcalW. Then p_dotsp_ku_dotsu_l-kw_dotsw_m-k is a basis for mathcalU+mathcalW. itemize abc In particular we have textdimmathcalU+mathcalWtextdim mathcalU+textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
Solution:
Proof. abcliste abc Clearly mathcalU+mathcalWsubseteq SpmathcalUcapmathcalW by the characterization of SpmathcalUcupmathcalW as the set of all linear combinations of elements from mathcalU cup mathcalW. But we also have SpmathcalU cup mathcalWsubseteq mathcalU+mathcalW. Indeed if v in SpmathcalUcupmathcalW then v _i^s a_iu_i+_j^r b_jw_j with a_ib_j in K u_iin U w_jin W. Longrightarrow vin mathcalU+mathcalW. This proves SpmathcalUcupmathcalWsubseteq mathcalU+mathcalW. Putting all together we have mathcalU+mathcalW SpmathcalUcapmathcalW. The rest of the statements in follow immediately. abc We first claim that p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Indeed if a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k+c_w_+...+c_m-kw_m-k. Write v:c_w_+...+c_m-kw_m-k-a_p_+...+_kp_k+b_u_+...+b_l-ku_l-k. Longrightarrow vinmathcalUcapmathcalW. Longrightarrow We can also write valpha_p_+...+alpha_kp_k for some alpha_...alpha_k in K. But vc_w_+...+c_m-kw_m-k. And p_...p_kw_...w_m-k are basis for mathcalW. Longrightarrow alpha_...alpha_k c_...c_m-k and v. But now v-a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k and p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Now clearly p_...p_ku_...u_l-kw_...w_m-k span mathcalU+mathcalW Longrightarrow p_...p_ku_...u_l-kw_...w_m-k is a basis for mathcalU+mathcalW. abc textdimmathcalU+mathcalWk+l-k+m-kl+m-ktextdimmathcalU+ textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
Let mathcalUW subseteq V subspaces. Then: abcliste abc mathcalU+mathcalWtextSpmathcalUcapmathcalW. In particular mathcalU+mathcalWsubseteq V is a subspace and mathcalUmathcalWsubseteq mathcalU+mathcalW are subspaces of mathcalU+mathcalW. abc If mathcalUmathcalW are finite dimensional then so is mathcalU+mathcalW. Moreover one can find a basis for mathcalU+mathcalW as follows: itemize item choose a basis p_dotsp_k for mathcalUcapmathcalW where ktextdimmathcalUcapmathcalW. item Ext p_dotsp_k to a basis p_dotsp_ku_dotsu_l-k of mathcalU where ltextdimmathcalU. item Ext p_dotsp_k to a basis p_dotsp_kw_dotsw_m-k of mathcalW where mtextdimmathcalW. Then p_dotsp_ku_dotsu_l-kw_dotsw_m-k is a basis for mathcalU+mathcalW. itemize abc In particular we have textdimmathcalU+mathcalWtextdim mathcalU+textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
Solution:
Proof. abcliste abc Clearly mathcalU+mathcalWsubseteq SpmathcalUcapmathcalW by the characterization of SpmathcalUcupmathcalW as the set of all linear combinations of elements from mathcalU cup mathcalW. But we also have SpmathcalU cup mathcalWsubseteq mathcalU+mathcalW. Indeed if v in SpmathcalUcupmathcalW then v _i^s a_iu_i+_j^r b_jw_j with a_ib_j in K u_iin U w_jin W. Longrightarrow vin mathcalU+mathcalW. This proves SpmathcalUcupmathcalWsubseteq mathcalU+mathcalW. Putting all together we have mathcalU+mathcalW SpmathcalUcapmathcalW. The rest of the statements in follow immediately. abc We first claim that p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Indeed if a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k+c_w_+...+c_m-kw_m-k. Write v:c_w_+...+c_m-kw_m-k-a_p_+...+_kp_k+b_u_+...+b_l-ku_l-k. Longrightarrow vinmathcalUcapmathcalW. Longrightarrow We can also write valpha_p_+...+alpha_kp_k for some alpha_...alpha_k in K. But vc_w_+...+c_m-kw_m-k. And p_...p_kw_...w_m-k are basis for mathcalW. Longrightarrow alpha_...alpha_k c_...c_m-k and v. But now v-a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k and p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Now clearly p_...p_ku_...u_l-kw_...w_m-k span mathcalU+mathcalW Longrightarrow p_...p_ku_...u_l-kw_...w_m-k is a basis for mathcalU+mathcalW. abc textdimmathcalU+mathcalWk+l-k+m-kl+m-ktextdimmathcalU+ textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
Meta Information
Exercise:
Let mathcalUW subseteq V subspaces. Then: abcliste abc mathcalU+mathcalWtextSpmathcalUcapmathcalW. In particular mathcalU+mathcalWsubseteq V is a subspace and mathcalUmathcalWsubseteq mathcalU+mathcalW are subspaces of mathcalU+mathcalW. abc If mathcalUmathcalW are finite dimensional then so is mathcalU+mathcalW. Moreover one can find a basis for mathcalU+mathcalW as follows: itemize item choose a basis p_dotsp_k for mathcalUcapmathcalW where ktextdimmathcalUcapmathcalW. item Ext p_dotsp_k to a basis p_dotsp_ku_dotsu_l-k of mathcalU where ltextdimmathcalU. item Ext p_dotsp_k to a basis p_dotsp_kw_dotsw_m-k of mathcalW where mtextdimmathcalW. Then p_dotsp_ku_dotsu_l-kw_dotsw_m-k is a basis for mathcalU+mathcalW. itemize abc In particular we have textdimmathcalU+mathcalWtextdim mathcalU+textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
Solution:
Proof. abcliste abc Clearly mathcalU+mathcalWsubseteq SpmathcalUcapmathcalW by the characterization of SpmathcalUcupmathcalW as the set of all linear combinations of elements from mathcalU cup mathcalW. But we also have SpmathcalU cup mathcalWsubseteq mathcalU+mathcalW. Indeed if v in SpmathcalUcupmathcalW then v _i^s a_iu_i+_j^r b_jw_j with a_ib_j in K u_iin U w_jin W. Longrightarrow vin mathcalU+mathcalW. This proves SpmathcalUcupmathcalWsubseteq mathcalU+mathcalW. Putting all together we have mathcalU+mathcalW SpmathcalUcapmathcalW. The rest of the statements in follow immediately. abc We first claim that p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Indeed if a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k+c_w_+...+c_m-kw_m-k. Write v:c_w_+...+c_m-kw_m-k-a_p_+...+_kp_k+b_u_+...+b_l-ku_l-k. Longrightarrow vinmathcalUcapmathcalW. Longrightarrow We can also write valpha_p_+...+alpha_kp_k for some alpha_...alpha_k in K. But vc_w_+...+c_m-kw_m-k. And p_...p_kw_...w_m-k are basis for mathcalW. Longrightarrow alpha_...alpha_k c_...c_m-k and v. But now v-a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k and p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Now clearly p_...p_ku_...u_l-kw_...w_m-k span mathcalU+mathcalW Longrightarrow p_...p_ku_...u_l-kw_...w_m-k is a basis for mathcalU+mathcalW. abc textdimmathcalU+mathcalWk+l-k+m-kl+m-ktextdimmathcalU+ textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
Let mathcalUW subseteq V subspaces. Then: abcliste abc mathcalU+mathcalWtextSpmathcalUcapmathcalW. In particular mathcalU+mathcalWsubseteq V is a subspace and mathcalUmathcalWsubseteq mathcalU+mathcalW are subspaces of mathcalU+mathcalW. abc If mathcalUmathcalW are finite dimensional then so is mathcalU+mathcalW. Moreover one can find a basis for mathcalU+mathcalW as follows: itemize item choose a basis p_dotsp_k for mathcalUcapmathcalW where ktextdimmathcalUcapmathcalW. item Ext p_dotsp_k to a basis p_dotsp_ku_dotsu_l-k of mathcalU where ltextdimmathcalU. item Ext p_dotsp_k to a basis p_dotsp_kw_dotsw_m-k of mathcalW where mtextdimmathcalW. Then p_dotsp_ku_dotsu_l-kw_dotsw_m-k is a basis for mathcalU+mathcalW. itemize abc In particular we have textdimmathcalU+mathcalWtextdim mathcalU+textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
Solution:
Proof. abcliste abc Clearly mathcalU+mathcalWsubseteq SpmathcalUcapmathcalW by the characterization of SpmathcalUcupmathcalW as the set of all linear combinations of elements from mathcalU cup mathcalW. But we also have SpmathcalU cup mathcalWsubseteq mathcalU+mathcalW. Indeed if v in SpmathcalUcupmathcalW then v _i^s a_iu_i+_j^r b_jw_j with a_ib_j in K u_iin U w_jin W. Longrightarrow vin mathcalU+mathcalW. This proves SpmathcalUcupmathcalWsubseteq mathcalU+mathcalW. Putting all together we have mathcalU+mathcalW SpmathcalUcapmathcalW. The rest of the statements in follow immediately. abc We first claim that p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Indeed if a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k+c_w_+...+c_m-kw_m-k. Write v:c_w_+...+c_m-kw_m-k-a_p_+...+_kp_k+b_u_+...+b_l-ku_l-k. Longrightarrow vinmathcalUcapmathcalW. Longrightarrow We can also write valpha_p_+...+alpha_kp_k for some alpha_...alpha_k in K. But vc_w_+...+c_m-kw_m-k. And p_...p_kw_...w_m-k are basis for mathcalW. Longrightarrow alpha_...alpha_k c_...c_m-k and v. But now v-a_p_+...+a_kp_k+b_u_+...+b_l-ku_l-k and p_...p_ku_...u_l-kw_...w_m-k are linearly indepent. Now clearly p_...p_ku_...u_l-kw_...w_m-k span mathcalU+mathcalW Longrightarrow p_...p_ku_...u_l-kw_...w_m-k is a basis for mathcalU+mathcalW. abc textdimmathcalU+mathcalWk+l-k+m-kl+m-ktextdimmathcalU+ textdimmathcalW-textdimmathcalUcapmathcalW. abcliste
Contained in these collections:

