Diagonalizability
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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 an n-dimensional vectorspace over K and Tin textEndV. The following statements are equivalent: abcliste abc T is diagonalizable. abc exists a basis for V whose elements are eigenvectors of T. abc The characteristic polynomial P_Tx splits o a product of linear factors and for every eigenvalue of T the geometric and algebraic multiplicities are equal. abc _i^k textdim Eig_Tlambda_itextdimV where lambda_...lambda_kin K are the eigenvalues of T. abc Vcong oplus_i^k textEig_Tlambda_i. abcliste
Solution:
Proof of iff . Already proven. Proof of Longrightarrow . Let mathcalB be a basis in which T_mathcalB^mathcalB has a diagonal form. By changing the order of the elements in mathcalB we can arrange that T_mathcalB^mathcalB has the following shape T_mathcalB^mathcalB pmatrix lambda_ & & & & & & & & & mboxHuge & ddots & & & & & & & & & lambda_ & & & & & & & & & lambda_ & & & & & & & & & ddots & & & & & & & & & lambda_ & & & & & & & & & ddots & & & & & & & & & & lambda_k & & & & & & & & & & ddots & mboxHuge & & & & & & & & &lambda_k pmatrixtext denoted by Lambda. mathcalBv_^...v_l_^v_^...v_l_^...v_^k...v_l_k^k where lambda_...lambda_k are eigenvalues of Tlambda_ineq lambda_jquad forall ineq j and v_^i...v_l_i^i are eigenvectors of lambda_i forall i. P_TxtextdetleftT_mathcalB^mathcalB-xs Iright lambda_-x^l_... lambda_k-x^l_k. * This shows that P_Tx splits as a product of linear factors. It also shows that lambda_...lambda_k are all the eigenvalues of T. It remains to show m_gT;lambda_im_aT;lambda_i forall i. Clearly v_^i...v_l_i^iin textEig_Tlambda_im_gT;lambda_i. But from * we also have l_im_aT;lambda_i. So m_aT;lambda_ileq m_gT;lambda_i forall i. At the same time we always have m_gT;lambda_ileq m_aT;lambda_i Longrightarrow m_gT;lambda_im_aT;lambda_i forall i The proof also shows that l_im_gT;lambda_im_aT;lambda_i. Proof of Longrightarrow . We ase that P_Txlambda_-x_^l_...lambda_k-x^l_k and m_gT;lambda_im_aT;lambda_i forall i. Clearly lambda_...lambda_k are all the eigenvalues of T. By definition l_im_aT;lambda_i. We have ntextdimV_i^k m_aT;lambda_i _i^k m_gT;lambda_i _i^k textdim Eig_Tlambda_i Proof of Longrightarrow . Recall that textEig_Tlambda_+...+textEig_Tlambda_k is a direct . textdimtextEig_Tlambda_+...+textEig_Tlambda_ktextdimtextEig_Tlambda_+....+textdimtextEig_Tlambda_ textdimV Since textEig_Tlambda_+...+textEig_Tlambda_ksubseteq V is a linear subspace we have textEig_Tlambda_+...+textEig_Tlambda_kV. But textEig_Tlambda_+...+textEig_Tlambda_kcond textEig_Tlambda_oplus...oplus textEig_Tlambda_k. Proof of Longrightarrow . If VcondtextEig_Tlambda_oplus...oplus textEig_Tlambda_k then since textEig_Tlambda_+...+textEig_Tlambda_k is a direct and also a linear subspace of V we must have textEig_Tlambda_+...+textEig_Tlambda_kV. Longrightarrow If we choose a basis v_^i...v_l_i^i for textEig_Tlambda_i forall i l_im_gT;lambda_i then v_^...v_l_^...v_^k...v_l_k^k is a basis of V and all the vectors in it are eigenvectors of T.
Let V be an n-dimensional vectorspace over K and Tin textEndV. The following statements are equivalent: abcliste abc T is diagonalizable. abc exists a basis for V whose elements are eigenvectors of T. abc The characteristic polynomial P_Tx splits o a product of linear factors and for every eigenvalue of T the geometric and algebraic multiplicities are equal. abc _i^k textdim Eig_Tlambda_itextdimV where lambda_...lambda_kin K are the eigenvalues of T. abc Vcong oplus_i^k textEig_Tlambda_i. abcliste
Solution:
Proof of iff . Already proven. Proof of Longrightarrow . Let mathcalB be a basis in which T_mathcalB^mathcalB has a diagonal form. By changing the order of the elements in mathcalB we can arrange that T_mathcalB^mathcalB has the following shape T_mathcalB^mathcalB pmatrix lambda_ & & & & & & & & & mboxHuge & ddots & & & & & & & & & lambda_ & & & & & & & & & lambda_ & & & & & & & & & ddots & & & & & & & & & lambda_ & & & & & & & & & ddots & & & & & & & & & & lambda_k & & & & & & & & & & ddots & mboxHuge & & & & & & & & &lambda_k pmatrixtext denoted by Lambda. mathcalBv_^...v_l_^v_^...v_l_^...v_^k...v_l_k^k where lambda_...lambda_k are eigenvalues of Tlambda_ineq lambda_jquad forall ineq j and v_^i...v_l_i^i are eigenvectors of lambda_i forall i. P_TxtextdetleftT_mathcalB^mathcalB-xs Iright lambda_-x^l_... lambda_k-x^l_k. * This shows that P_Tx splits as a product of linear factors. It also shows that lambda_...lambda_k are all the eigenvalues of T. It remains to show m_gT;lambda_im_aT;lambda_i forall i. Clearly v_^i...v_l_i^iin textEig_Tlambda_im_gT;lambda_i. But from * we also have l_im_aT;lambda_i. So m_aT;lambda_ileq m_gT;lambda_i forall i. At the same time we always have m_gT;lambda_ileq m_aT;lambda_i Longrightarrow m_gT;lambda_im_aT;lambda_i forall i The proof also shows that l_im_gT;lambda_im_aT;lambda_i. Proof of Longrightarrow . We ase that P_Txlambda_-x_^l_...lambda_k-x^l_k and m_gT;lambda_im_aT;lambda_i forall i. Clearly lambda_...lambda_k are all the eigenvalues of T. By definition l_im_aT;lambda_i. We have ntextdimV_i^k m_aT;lambda_i _i^k m_gT;lambda_i _i^k textdim Eig_Tlambda_i Proof of Longrightarrow . Recall that textEig_Tlambda_+...+textEig_Tlambda_k is a direct . textdimtextEig_Tlambda_+...+textEig_Tlambda_ktextdimtextEig_Tlambda_+....+textdimtextEig_Tlambda_ textdimV Since textEig_Tlambda_+...+textEig_Tlambda_ksubseteq V is a linear subspace we have textEig_Tlambda_+...+textEig_Tlambda_kV. But textEig_Tlambda_+...+textEig_Tlambda_kcond textEig_Tlambda_oplus...oplus textEig_Tlambda_k. Proof of Longrightarrow . If VcondtextEig_Tlambda_oplus...oplus textEig_Tlambda_k then since textEig_Tlambda_+...+textEig_Tlambda_k is a direct and also a linear subspace of V we must have textEig_Tlambda_+...+textEig_Tlambda_kV. Longrightarrow If we choose a basis v_^i...v_l_i^i for textEig_Tlambda_i forall i l_im_gT;lambda_i then v_^...v_l_^...v_^k...v_l_k^k is a basis of V and all the vectors in it are eigenvectors of T.