Diagonalizable matrices and eigenvectors
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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:
T is diagonalizable iff exists a basis mathcalBv_...v_n consisting of eigenvectors of T i.e. forall i Tv_ilambda_i v_i for some lambda_iin K. Moreover if for some basis mathcalBv_...v_n the matrix T_mathcalB^mathcalB has the shape pmatrix lambda_ & hdots & vdots & lambda_i & vdots & hdots & lambda_n pmatrix. then forall i lambda_i is an eigenvalue of T and v_i is an eigenvector for lambda_i.
Solution:
Proof. Recall that T_mathcalB^mathcalB pmatrix vdots & & vdots Tv__mathcalB & hdots & Tv_n_mathcalB vdots & & vdots pmatrix. and Tv_j_mathcalB pmatrix alpha_j hdots alpha_nj pmatrix where Tv_jalpha_jv_+...+alpha_njv_n. T_mathcalB^mathcalB has the shape above iff alpha_ijquad forall ineq j and alpha_kklambda_kquad k iff Tv_k lambda_k v_k.
T is diagonalizable iff exists a basis mathcalBv_...v_n consisting of eigenvectors of T i.e. forall i Tv_ilambda_i v_i for some lambda_iin K. Moreover if for some basis mathcalBv_...v_n the matrix T_mathcalB^mathcalB has the shape pmatrix lambda_ & hdots & vdots & lambda_i & vdots & hdots & lambda_n pmatrix. then forall i lambda_i is an eigenvalue of T and v_i is an eigenvector for lambda_i.
Solution:
Proof. Recall that T_mathcalB^mathcalB pmatrix vdots & & vdots Tv__mathcalB & hdots & Tv_n_mathcalB vdots & & vdots pmatrix. and Tv_j_mathcalB pmatrix alpha_j hdots alpha_nj pmatrix where Tv_jalpha_jv_+...+alpha_njv_n. T_mathcalB^mathcalB has the shape above iff alpha_ijquad forall ineq j and alpha_kklambda_kquad k iff Tv_k lambda_k v_k.
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Exercise:
T is diagonalizable iff exists a basis mathcalBv_...v_n consisting of eigenvectors of T i.e. forall i Tv_ilambda_i v_i for some lambda_iin K. Moreover if for some basis mathcalBv_...v_n the matrix T_mathcalB^mathcalB has the shape pmatrix lambda_ & hdots & vdots & lambda_i & vdots & hdots & lambda_n pmatrix. then forall i lambda_i is an eigenvalue of T and v_i is an eigenvector for lambda_i.
Solution:
Proof. Recall that T_mathcalB^mathcalB pmatrix vdots & & vdots Tv__mathcalB & hdots & Tv_n_mathcalB vdots & & vdots pmatrix. and Tv_j_mathcalB pmatrix alpha_j hdots alpha_nj pmatrix where Tv_jalpha_jv_+...+alpha_njv_n. T_mathcalB^mathcalB has the shape above iff alpha_ijquad forall ineq j and alpha_kklambda_kquad k iff Tv_k lambda_k v_k.
T is diagonalizable iff exists a basis mathcalBv_...v_n consisting of eigenvectors of T i.e. forall i Tv_ilambda_i v_i for some lambda_iin K. Moreover if for some basis mathcalBv_...v_n the matrix T_mathcalB^mathcalB has the shape pmatrix lambda_ & hdots & vdots & lambda_i & vdots & hdots & lambda_n pmatrix. then forall i lambda_i is an eigenvalue of T and v_i is an eigenvector for lambda_i.
Solution:
Proof. Recall that T_mathcalB^mathcalB pmatrix vdots & & vdots Tv__mathcalB & hdots & Tv_n_mathcalB vdots & & vdots pmatrix. and Tv_j_mathcalB pmatrix alpha_j hdots alpha_nj pmatrix where Tv_jalpha_jv_+...+alpha_njv_n. T_mathcalB^mathcalB has the shape above iff alpha_ijquad forall ineq j and alpha_kklambda_kquad k iff Tv_k lambda_k v_k.
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