Triagonizablitiy and linear factors
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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 -
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Exercise:
Let V be a finite dimensional vector space over K and Tin textEndV. Then T is triagonizable iff P_Tx splits as a product of linear factors i.e. P_Txlambda-x^l_...lambda_k-x^l_k.
Solution:
Proof of Longrightarrow. If exists a basis mathcalB s.t. T_mathcalB^mathcalB pmatrix lambda_ & & & * & lambda_ & & & & ddots & & & & & lambda_n pmatrix then P_TxtextdetleftT_mathcalB^mathcalB-x Iright textdet pmatrix lambda_-x & & & * & lambda_-x & & & & ddots & & & & & lambda_n-x pmatrix lambda_-x...lambda_n-x. Proof of Longleftarrow. We'll use inductin on ntextdimV and also quotient spaces. bf Recall: If Usubseteq V is a linear subspace we have the quotient space V/U whose elements are equivalence classes v of vin V where vv' if v-v'in U. Recall also that if V is finite dimensional then so is V/U and textdimV/UtextdimV-textdimU. Not to the proof. n: In this case V is -dimensional and T_mathcalB^mathcalBlambda so T is even diagonalizable. Let ngeq . Ase that Longleftarrow of the theorem holds for all vector spaces of textdimleq n and every omorphism of such a space. Let V be an n+-dimensional vector space and Tin textEndV be s.t. P_Tx splits as a product of linear factors. Let lambdain K be a zero of P_TxLongrightarrow exists an eigenvector v_in textKerT-lambda id. Ext v_ to a basis mathcalB_v_w_...w_n+ of V. Then T_mathcalB_^mathcalB_ pmatrix lambda_ & * & * & * & * & * & * vdots& * & * & * & * & * & * pmatrix pmatrix lambda_ & * & hdots & * & & & vdots& & A & & & & pmatrix We have P_Txlambda_-x P_Ax. Define U:textSpv_ and consider V/U. Since TUsubseteq U we obtain a new omorphism overlineT:V/Ulongrightarrow V/U defined by overlineTv:Tv. The vectors w_...w_n+ form a basis for V/U and in this basis overlineT is represented by A. So P_Txlambda_-x P_overlineTx. Since P_Tx splits o a product of linear factors so does also P_overlineTx. By the induction hypothesis exists a basis mathcalCz_...z_n+ for V/U s.t. overlineT_mathcalC^mathcalC pmatrix lambda_ & & & * & & ddots & & & * & lambda_n+ pmatrix Now each element z_i in mathcalC is represented by some w_iin V i.e. z_iv_i. Such a v_i is not unique but we pick one for every i. Define mathcalBv_...v_n+. We claim that mathcalB is a basis for V. Indeed v_...v_n are linearly indepent because if alpha_v_+alpha_v_+...+alpha_nv_n Longrightarrow alpha_v_+...+alpha_n+v_n+-alpha_v_ in U Longrightarrow alpha_v_+...+alpha_n+v_n+ Longrightarrow alpha_z_+...+alpha_n+z_n+ Longrightarrow alpha_...alpha_n+. This shows that v_...v_n+ are linearly indepent. Since textdimVn+ mathcalBv_...v_n+ is a basis for V. We claim that T_mathcalB^mathcalB pmatrix lambda_ & * & * & hdots & * & lambda_ & * & hdots & * vdots & & ddots & &vdots & & & & lambda_n+ pmatrix Indeed Tv_lambda_v_ and for leq ileq n+ we have Tv_ioverlineTv_ioverlineTz_ialppha_iz_+alpha_iz_+...+alpha_i-iz_i-+lambda_iz_iquad alpha_i...alpha_i-i in K Longrightarrow Tv_ialpha_iv_+...+alpha_i-iv_i-+lambda_iv_i Longrightarrow Tv_ialpha_iv_+...+alpha_i-iv_i-+lambda_iv_i+u quad uin U. Write ubeta_i v_. Longrightarrow Tv_ibetaiv_+alpha_iv_+...+alpha_i-iv_i-+lambda_iv_i So T_mathcalB^mathcalB pmatrix lambda_ & beta_ & hdots & beta_i & hdots & beta_n+ & lambda_ & hdots & alpha_i & hdots & * vdots & & ddots & alpha_i-i & & vdots vdots & vdots & & lambda_i & & * vdots & vdots & & & & * vdots & vdots & & vdots & ddots & * & & & & &lambda_n+ pmatrix This shows T is trigonalizable. The induction is now complete.
Let V be a finite dimensional vector space over K and Tin textEndV. Then T is triagonizable iff P_Tx splits as a product of linear factors i.e. P_Txlambda-x^l_...lambda_k-x^l_k.
Solution:
Proof of Longrightarrow. If exists a basis mathcalB s.t. T_mathcalB^mathcalB pmatrix lambda_ & & & * & lambda_ & & & & ddots & & & & & lambda_n pmatrix then P_TxtextdetleftT_mathcalB^mathcalB-x Iright textdet pmatrix lambda_-x & & & * & lambda_-x & & & & ddots & & & & & lambda_n-x pmatrix lambda_-x...lambda_n-x. Proof of Longleftarrow. We'll use inductin on ntextdimV and also quotient spaces. bf Recall: If Usubseteq V is a linear subspace we have the quotient space V/U whose elements are equivalence classes v of vin V where vv' if v-v'in U. Recall also that if V is finite dimensional then so is V/U and textdimV/UtextdimV-textdimU. Not to the proof. n: In this case V is -dimensional and T_mathcalB^mathcalBlambda so T is even diagonalizable. Let ngeq . Ase that Longleftarrow of the theorem holds for all vector spaces of textdimleq n and every omorphism of such a space. Let V be an n+-dimensional vector space and Tin textEndV be s.t. P_Tx splits as a product of linear factors. Let lambdain K be a zero of P_TxLongrightarrow exists an eigenvector v_in textKerT-lambda id. Ext v_ to a basis mathcalB_v_w_...w_n+ of V. Then T_mathcalB_^mathcalB_ pmatrix lambda_ & * & * & * & * & * & * vdots& * & * & * & * & * & * pmatrix pmatrix lambda_ & * & hdots & * & & & vdots& & A & & & & pmatrix We have P_Txlambda_-x P_Ax. Define U:textSpv_ and consider V/U. Since TUsubseteq U we obtain a new omorphism overlineT:V/Ulongrightarrow V/U defined by overlineTv:Tv. The vectors w_...w_n+ form a basis for V/U and in this basis overlineT is represented by A. So P_Txlambda_-x P_overlineTx. Since P_Tx splits o a product of linear factors so does also P_overlineTx. By the induction hypothesis exists a basis mathcalCz_...z_n+ for V/U s.t. overlineT_mathcalC^mathcalC pmatrix lambda_ & & & * & & ddots & & & * & lambda_n+ pmatrix Now each element z_i in mathcalC is represented by some w_iin V i.e. z_iv_i. Such a v_i is not unique but we pick one for every i. Define mathcalBv_...v_n+. We claim that mathcalB is a basis for V. Indeed v_...v_n are linearly indepent because if alpha_v_+alpha_v_+...+alpha_nv_n Longrightarrow alpha_v_+...+alpha_n+v_n+-alpha_v_ in U Longrightarrow alpha_v_+...+alpha_n+v_n+ Longrightarrow alpha_z_+...+alpha_n+z_n+ Longrightarrow alpha_...alpha_n+. This shows that v_...v_n+ are linearly indepent. Since textdimVn+ mathcalBv_...v_n+ is a basis for V. We claim that T_mathcalB^mathcalB pmatrix lambda_ & * & * & hdots & * & lambda_ & * & hdots & * vdots & & ddots & &vdots & & & & lambda_n+ pmatrix Indeed Tv_lambda_v_ and for leq ileq n+ we have Tv_ioverlineTv_ioverlineTz_ialppha_iz_+alpha_iz_+...+alpha_i-iz_i-+lambda_iz_iquad alpha_i...alpha_i-i in K Longrightarrow Tv_ialpha_iv_+...+alpha_i-iv_i-+lambda_iv_i Longrightarrow Tv_ialpha_iv_+...+alpha_i-iv_i-+lambda_iv_i+u quad uin U. Write ubeta_i v_. Longrightarrow Tv_ibetaiv_+alpha_iv_+...+alpha_i-iv_i-+lambda_iv_i So T_mathcalB^mathcalB pmatrix lambda_ & beta_ & hdots & beta_i & hdots & beta_n+ & lambda_ & hdots & alpha_i & hdots & * vdots & & ddots & alpha_i-i & & vdots vdots & vdots & & lambda_i & & * vdots & vdots & & & & * vdots & vdots & & vdots & ddots & * & & & & &lambda_n+ pmatrix This shows T is trigonalizable. The induction is now complete.