Scalar product
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
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Exercise:
If langle rangle_A is a scalar product Longrightarrow A is positive definite. If langle rangle_A is a Hermitian product iff A is positive definite.
Solution:
Proof. We'll prove it for complex matrices and Hermitian products the case of symmetrical matrices & scalar products is very similar and even simpler. Let Ain M_ntimes nmathbbC and suppose that langle vwrangle_A : v^TAoverlinew is a Hermitian product. &Longrightarrow forall vwin mathbbC^n textwe have langle vwrangle_A langle overlinewvrangle_A &Longrightarrow v^TAoverlinewoverlinew^TAoverlinev Longrightarrow v^TAoverlinew overlinew^ToverlineAv Longrightarrow v^tAoverlinewoverlinew^ToverlineAv^T &Longrightarrow v^TAoverlinewv^ToverlineA^Toverlinew. Take ve_i we_j standard basis. Write A pmatrix a_ & hdots & a_j & hdots & a_n vdots & vdots & vdots & vdots & vdots a_n & hdots & a_nj & hdots & a_nn pmatrix. e_i^T A e_j pmatrix s & e_i & s pmatrix A pmatrix vdots e_j vdots pmatrix pmatrix s & e_i & s pmatrix pmatrix a_j vdots a_nj pmatrix Similarly e_i^ToverlineA^Te_j overlinea_ji Longrightarrow overlinea_ji forall ij Longrightarrow overlineA^TA. The fact that A is positive definite is precisely the condition that langle vvrangle_A forall vneq . The opposite statement: A positive definite Longrightarrow langle rangle_A is a Hermitian product.
If langle rangle_A is a scalar product Longrightarrow A is positive definite. If langle rangle_A is a Hermitian product iff A is positive definite.
Solution:
Proof. We'll prove it for complex matrices and Hermitian products the case of symmetrical matrices & scalar products is very similar and even simpler. Let Ain M_ntimes nmathbbC and suppose that langle vwrangle_A : v^TAoverlinew is a Hermitian product. &Longrightarrow forall vwin mathbbC^n textwe have langle vwrangle_A langle overlinewvrangle_A &Longrightarrow v^TAoverlinewoverlinew^TAoverlinev Longrightarrow v^TAoverlinew overlinew^ToverlineAv Longrightarrow v^tAoverlinewoverlinew^ToverlineAv^T &Longrightarrow v^TAoverlinewv^ToverlineA^Toverlinew. Take ve_i we_j standard basis. Write A pmatrix a_ & hdots & a_j & hdots & a_n vdots & vdots & vdots & vdots & vdots a_n & hdots & a_nj & hdots & a_nn pmatrix. e_i^T A e_j pmatrix s & e_i & s pmatrix A pmatrix vdots e_j vdots pmatrix pmatrix s & e_i & s pmatrix pmatrix a_j vdots a_nj pmatrix Similarly e_i^ToverlineA^Te_j overlinea_ji Longrightarrow overlinea_ji forall ij Longrightarrow overlineA^TA. The fact that A is positive definite is precisely the condition that langle vvrangle_A forall vneq . The opposite statement: A positive definite Longrightarrow langle rangle_A is a Hermitian product.
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Exercise:
If langle rangle_A is a scalar product Longrightarrow A is positive definite. If langle rangle_A is a Hermitian product iff A is positive definite.
Solution:
Proof. We'll prove it for complex matrices and Hermitian products the case of symmetrical matrices & scalar products is very similar and even simpler. Let Ain M_ntimes nmathbbC and suppose that langle vwrangle_A : v^TAoverlinew is a Hermitian product. &Longrightarrow forall vwin mathbbC^n textwe have langle vwrangle_A langle overlinewvrangle_A &Longrightarrow v^TAoverlinewoverlinew^TAoverlinev Longrightarrow v^TAoverlinew overlinew^ToverlineAv Longrightarrow v^tAoverlinewoverlinew^ToverlineAv^T &Longrightarrow v^TAoverlinewv^ToverlineA^Toverlinew. Take ve_i we_j standard basis. Write A pmatrix a_ & hdots & a_j & hdots & a_n vdots & vdots & vdots & vdots & vdots a_n & hdots & a_nj & hdots & a_nn pmatrix. e_i^T A e_j pmatrix s & e_i & s pmatrix A pmatrix vdots e_j vdots pmatrix pmatrix s & e_i & s pmatrix pmatrix a_j vdots a_nj pmatrix Similarly e_i^ToverlineA^Te_j overlinea_ji Longrightarrow overlinea_ji forall ij Longrightarrow overlineA^TA. The fact that A is positive definite is precisely the condition that langle vvrangle_A forall vneq . The opposite statement: A positive definite Longrightarrow langle rangle_A is a Hermitian product.
If langle rangle_A is a scalar product Longrightarrow A is positive definite. If langle rangle_A is a Hermitian product iff A is positive definite.
Solution:
Proof. We'll prove it for complex matrices and Hermitian products the case of symmetrical matrices & scalar products is very similar and even simpler. Let Ain M_ntimes nmathbbC and suppose that langle vwrangle_A : v^TAoverlinew is a Hermitian product. &Longrightarrow forall vwin mathbbC^n textwe have langle vwrangle_A langle overlinewvrangle_A &Longrightarrow v^TAoverlinewoverlinew^TAoverlinev Longrightarrow v^TAoverlinew overlinew^ToverlineAv Longrightarrow v^tAoverlinewoverlinew^ToverlineAv^T &Longrightarrow v^TAoverlinewv^ToverlineA^Toverlinew. Take ve_i we_j standard basis. Write A pmatrix a_ & hdots & a_j & hdots & a_n vdots & vdots & vdots & vdots & vdots a_n & hdots & a_nj & hdots & a_nn pmatrix. e_i^T A e_j pmatrix s & e_i & s pmatrix A pmatrix vdots e_j vdots pmatrix pmatrix s & e_i & s pmatrix pmatrix a_j vdots a_nj pmatrix Similarly e_i^ToverlineA^Te_j overlinea_ji Longrightarrow overlinea_ji forall ij Longrightarrow overlineA^TA. The fact that A is positive definite is precisely the condition that langle vvrangle_A forall vneq . The opposite statement: A positive definite Longrightarrow langle rangle_A is a Hermitian product.
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