Equality row-rank and col-rank
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 Ain M_mtimes nK. Then textcol-rankAtextrow-rankA.
Solution:
Preparation for the proof bf Lemma . Let T:Vlongrightarrow W be a linear map and S:Wlongrightarrow L:Plongrightarrow V isomorphisms. Then: itemize item textrankScirc TtextrankT. item textrankTcirc LtextrankT. itemize isomorphisms don't change rank Proof of Lemma . itemize item textrankScirc TtextdimSTVtextdimTVtextrankT because S|_TV is an isomorphism between TV and STV. item textrankTcirc LtextdimTLPtextdimTV because LPV. itemize bf Lemma . Let Ain M_ntimes nK. Then A is invertible iff T_A:K^nlongrightarrow K^n is an isomorphism. Proof of Lemma Longrightarrow. Suppose A is invertible. We have T_A^-T_A^-. Proof of Lemma Longleftarrow. Suppose T_A is an isomorphism. We know that AT_epsilon_n^epsilon_n and by a previous result we obtain that A is invertible. bf Lemma . Let A Bin M_mtimes nK s.t. PAQ which means that they are equivalent where Pin textGL_mK Qin textGL_nK.Then: itemize item textcol-rankAtextcol-rankB. item textrow-rankAtextrow-rankB. itemize Proof of Lemma itemize item We have T_BT_Pcirc T_Acirc T_Q and both T_P T_Q are isomorphisms. By Lemma textrankT_BtextrankT_A. But textrankT_Btextcol-rankB textrankT_Atextcol-rankA Longrightarrow textcol-rankBtextcol-rankA. item We have B^TQ^T A^T P^T. Since P and Q are invertible so are also P^T Q^T we have seen P^T^-P^-^T By part of the Lemma we have textcol-rankB^Ttextcol-rankA^T textrow-rankBtextrow-rankA itemize Proof of the theorem. By a previous proposition exists Pin textGL_mK Qin textGL_nK s.t. PAQleftarray@c|c@ matrix I_r matrix & hline & matrix matrix arrayright denote this matrix by B. For B we have textcol-rankBtextrow-rankBr By Lemma textcol-rankAtextcol-rankBrquad textand textrow-rankAtextrow-rankBr &Longrightarrow textcol-rankAtextrow-rankA.
Let Ain M_mtimes nK. Then textcol-rankAtextrow-rankA.
Solution:
Preparation for the proof bf Lemma . Let T:Vlongrightarrow W be a linear map and S:Wlongrightarrow L:Plongrightarrow V isomorphisms. Then: itemize item textrankScirc TtextrankT. item textrankTcirc LtextrankT. itemize isomorphisms don't change rank Proof of Lemma . itemize item textrankScirc TtextdimSTVtextdimTVtextrankT because S|_TV is an isomorphism between TV and STV. item textrankTcirc LtextdimTLPtextdimTV because LPV. itemize bf Lemma . Let Ain M_ntimes nK. Then A is invertible iff T_A:K^nlongrightarrow K^n is an isomorphism. Proof of Lemma Longrightarrow. Suppose A is invertible. We have T_A^-T_A^-. Proof of Lemma Longleftarrow. Suppose T_A is an isomorphism. We know that AT_epsilon_n^epsilon_n and by a previous result we obtain that A is invertible. bf Lemma . Let A Bin M_mtimes nK s.t. PAQ which means that they are equivalent where Pin textGL_mK Qin textGL_nK.Then: itemize item textcol-rankAtextcol-rankB. item textrow-rankAtextrow-rankB. itemize Proof of Lemma itemize item We have T_BT_Pcirc T_Acirc T_Q and both T_P T_Q are isomorphisms. By Lemma textrankT_BtextrankT_A. But textrankT_Btextcol-rankB textrankT_Atextcol-rankA Longrightarrow textcol-rankBtextcol-rankA. item We have B^TQ^T A^T P^T. Since P and Q are invertible so are also P^T Q^T we have seen P^T^-P^-^T By part of the Lemma we have textcol-rankB^Ttextcol-rankA^T textrow-rankBtextrow-rankA itemize Proof of the theorem. By a previous proposition exists Pin textGL_mK Qin textGL_nK s.t. PAQleftarray@c|c@ matrix I_r matrix & hline & matrix matrix arrayright denote this matrix by B. For B we have textcol-rankBtextrow-rankBr By Lemma textcol-rankAtextcol-rankBrquad textand textrow-rankAtextrow-rankBr &Longrightarrow textcol-rankAtextrow-rankA.
Meta Information
Exercise:
Let Ain M_mtimes nK. Then textcol-rankAtextrow-rankA.
Solution:
Preparation for the proof bf Lemma . Let T:Vlongrightarrow W be a linear map and S:Wlongrightarrow L:Plongrightarrow V isomorphisms. Then: itemize item textrankScirc TtextrankT. item textrankTcirc LtextrankT. itemize isomorphisms don't change rank Proof of Lemma . itemize item textrankScirc TtextdimSTVtextdimTVtextrankT because S|_TV is an isomorphism between TV and STV. item textrankTcirc LtextdimTLPtextdimTV because LPV. itemize bf Lemma . Let Ain M_ntimes nK. Then A is invertible iff T_A:K^nlongrightarrow K^n is an isomorphism. Proof of Lemma Longrightarrow. Suppose A is invertible. We have T_A^-T_A^-. Proof of Lemma Longleftarrow. Suppose T_A is an isomorphism. We know that AT_epsilon_n^epsilon_n and by a previous result we obtain that A is invertible. bf Lemma . Let A Bin M_mtimes nK s.t. PAQ which means that they are equivalent where Pin textGL_mK Qin textGL_nK.Then: itemize item textcol-rankAtextcol-rankB. item textrow-rankAtextrow-rankB. itemize Proof of Lemma itemize item We have T_BT_Pcirc T_Acirc T_Q and both T_P T_Q are isomorphisms. By Lemma textrankT_BtextrankT_A. But textrankT_Btextcol-rankB textrankT_Atextcol-rankA Longrightarrow textcol-rankBtextcol-rankA. item We have B^TQ^T A^T P^T. Since P and Q are invertible so are also P^T Q^T we have seen P^T^-P^-^T By part of the Lemma we have textcol-rankB^Ttextcol-rankA^T textrow-rankBtextrow-rankA itemize Proof of the theorem. By a previous proposition exists Pin textGL_mK Qin textGL_nK s.t. PAQleftarray@c|c@ matrix I_r matrix & hline & matrix matrix arrayright denote this matrix by B. For B we have textcol-rankBtextrow-rankBr By Lemma textcol-rankAtextcol-rankBrquad textand textrow-rankAtextrow-rankBr &Longrightarrow textcol-rankAtextrow-rankA.
Let Ain M_mtimes nK. Then textcol-rankAtextrow-rankA.
Solution:
Preparation for the proof bf Lemma . Let T:Vlongrightarrow W be a linear map and S:Wlongrightarrow L:Plongrightarrow V isomorphisms. Then: itemize item textrankScirc TtextrankT. item textrankTcirc LtextrankT. itemize isomorphisms don't change rank Proof of Lemma . itemize item textrankScirc TtextdimSTVtextdimTVtextrankT because S|_TV is an isomorphism between TV and STV. item textrankTcirc LtextdimTLPtextdimTV because LPV. itemize bf Lemma . Let Ain M_ntimes nK. Then A is invertible iff T_A:K^nlongrightarrow K^n is an isomorphism. Proof of Lemma Longrightarrow. Suppose A is invertible. We have T_A^-T_A^-. Proof of Lemma Longleftarrow. Suppose T_A is an isomorphism. We know that AT_epsilon_n^epsilon_n and by a previous result we obtain that A is invertible. bf Lemma . Let A Bin M_mtimes nK s.t. PAQ which means that they are equivalent where Pin textGL_mK Qin textGL_nK.Then: itemize item textcol-rankAtextcol-rankB. item textrow-rankAtextrow-rankB. itemize Proof of Lemma itemize item We have T_BT_Pcirc T_Acirc T_Q and both T_P T_Q are isomorphisms. By Lemma textrankT_BtextrankT_A. But textrankT_Btextcol-rankB textrankT_Atextcol-rankA Longrightarrow textcol-rankBtextcol-rankA. item We have B^TQ^T A^T P^T. Since P and Q are invertible so are also P^T Q^T we have seen P^T^-P^-^T By part of the Lemma we have textcol-rankB^Ttextcol-rankA^T textrow-rankBtextrow-rankA itemize Proof of the theorem. By a previous proposition exists Pin textGL_mK Qin textGL_nK s.t. PAQleftarray@c|c@ matrix I_r matrix & hline & matrix matrix arrayright denote this matrix by B. For B we have textcol-rankBtextrow-rankBr By Lemma textcol-rankAtextcol-rankBrquad textand textrow-rankAtextrow-rankBr &Longrightarrow textcol-rankAtextrow-rankA.
Contained in these collections: