Gram-Schmidt characteristics
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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 Vlangle rangle be an inner product space over KmathbbR or mathbbC and let ntextdimV. Let v_...v_n be a basis of V. Define a new list of vectors w_...w_n by induction: w_:v_w_jv_j-_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle for j...n. Then abcliste abc w_...w_n is an orthogonal basis for V. abc leftfracw_||w_||...fracw_n||w_n||right is an orthonormal basis for V. abc forall leq jleq n we have textSpv_...v_jtextSpw_...w_jtextSpleftfracw_||w_||...fracw_n||w_n||right. abcliste
Solution:
Proof. We claim that forall leq jleq n w_j is well defined i.e. w_...w_j-neq and moreover w_...w_j is an orthogonal system with textSpw_..w_jtextSpv_...v_j. We prove this by induction on j. For j w_v_ so the claim is obvious v_neq because it is a part of a basis. Let leq jleq n. Ase the claim holds for w_...w_j-. Recall that w_jv_j-_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle w_i. Note that w_j is well defined because w_ineq forall leq ileq j-. We claim that w_jneq . Indeed if w_j then v_ji^j-fraclangle v_jw_i ranglelangle w_iw_i rangle w_i Longrightarrow v_jin textSpw_..w_j-textSpv_...v_j- which is impossible because v_...v_j are linearly indepent. This shows w_jneq . By the induction hypothesis w_...w_j- is an orthogonal system. So we have to prove that w_jperp w_...w_j-. Indeed let leq kleq j- then langle w_jw_k ranglelangle v_jw_k rangle+_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle langle w_iw_krangle langle v_jw_kranglfraclangle v_jw_k ranglelangle w_kw_k rangle langle w_kw_krangle. This shows w_...w_j is an orthogonal system. To complete the induction it remains to show that textSpw_...w_jtextSpv_...v_j. By the definition of w_j we have w_jin textSpw_...w_j-v_jtextSpv_...v_j-v_j. Longrightarrow textSpw_...w_j-w_jsubseteq textSpv_...v_j-v_j. But both these vector spaces have dim j because v_...v_j are linearly indepent and also w_...w_j are linearly indepent being an orthogonal system. Longrightarrow textSpw_...w_j-w_jtextSpv_...v_j-v_j.
Let Vlangle rangle be an inner product space over KmathbbR or mathbbC and let ntextdimV. Let v_...v_n be a basis of V. Define a new list of vectors w_...w_n by induction: w_:v_w_jv_j-_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle for j...n. Then abcliste abc w_...w_n is an orthogonal basis for V. abc leftfracw_||w_||...fracw_n||w_n||right is an orthonormal basis for V. abc forall leq jleq n we have textSpv_...v_jtextSpw_...w_jtextSpleftfracw_||w_||...fracw_n||w_n||right. abcliste
Solution:
Proof. We claim that forall leq jleq n w_j is well defined i.e. w_...w_j-neq and moreover w_...w_j is an orthogonal system with textSpw_..w_jtextSpv_...v_j. We prove this by induction on j. For j w_v_ so the claim is obvious v_neq because it is a part of a basis. Let leq jleq n. Ase the claim holds for w_...w_j-. Recall that w_jv_j-_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle w_i. Note that w_j is well defined because w_ineq forall leq ileq j-. We claim that w_jneq . Indeed if w_j then v_ji^j-fraclangle v_jw_i ranglelangle w_iw_i rangle w_i Longrightarrow v_jin textSpw_..w_j-textSpv_...v_j- which is impossible because v_...v_j are linearly indepent. This shows w_jneq . By the induction hypothesis w_...w_j- is an orthogonal system. So we have to prove that w_jperp w_...w_j-. Indeed let leq kleq j- then langle w_jw_k ranglelangle v_jw_k rangle+_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle langle w_iw_krangle langle v_jw_kranglfraclangle v_jw_k ranglelangle w_kw_k rangle langle w_kw_krangle. This shows w_...w_j is an orthogonal system. To complete the induction it remains to show that textSpw_...w_jtextSpv_...v_j. By the definition of w_j we have w_jin textSpw_...w_j-v_jtextSpv_...v_j-v_j. Longrightarrow textSpw_...w_j-w_jsubseteq textSpv_...v_j-v_j. But both these vector spaces have dim j because v_...v_j are linearly indepent and also w_...w_j are linearly indepent being an orthogonal system. Longrightarrow textSpw_...w_j-w_jtextSpv_...v_j-v_j.
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Exercise:
Let Vlangle rangle be an inner product space over KmathbbR or mathbbC and let ntextdimV. Let v_...v_n be a basis of V. Define a new list of vectors w_...w_n by induction: w_:v_w_jv_j-_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle for j...n. Then abcliste abc w_...w_n is an orthogonal basis for V. abc leftfracw_||w_||...fracw_n||w_n||right is an orthonormal basis for V. abc forall leq jleq n we have textSpv_...v_jtextSpw_...w_jtextSpleftfracw_||w_||...fracw_n||w_n||right. abcliste
Solution:
Proof. We claim that forall leq jleq n w_j is well defined i.e. w_...w_j-neq and moreover w_...w_j is an orthogonal system with textSpw_..w_jtextSpv_...v_j. We prove this by induction on j. For j w_v_ so the claim is obvious v_neq because it is a part of a basis. Let leq jleq n. Ase the claim holds for w_...w_j-. Recall that w_jv_j-_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle w_i. Note that w_j is well defined because w_ineq forall leq ileq j-. We claim that w_jneq . Indeed if w_j then v_ji^j-fraclangle v_jw_i ranglelangle w_iw_i rangle w_i Longrightarrow v_jin textSpw_..w_j-textSpv_...v_j- which is impossible because v_...v_j are linearly indepent. This shows w_jneq . By the induction hypothesis w_...w_j- is an orthogonal system. So we have to prove that w_jperp w_...w_j-. Indeed let leq kleq j- then langle w_jw_k ranglelangle v_jw_k rangle+_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle langle w_iw_krangle langle v_jw_kranglfraclangle v_jw_k ranglelangle w_kw_k rangle langle w_kw_krangle. This shows w_...w_j is an orthogonal system. To complete the induction it remains to show that textSpw_...w_jtextSpv_...v_j. By the definition of w_j we have w_jin textSpw_...w_j-v_jtextSpv_...v_j-v_j. Longrightarrow textSpw_...w_j-w_jsubseteq textSpv_...v_j-v_j. But both these vector spaces have dim j because v_...v_j are linearly indepent and also w_...w_j are linearly indepent being an orthogonal system. Longrightarrow textSpw_...w_j-w_jtextSpv_...v_j-v_j.
Let Vlangle rangle be an inner product space over KmathbbR or mathbbC and let ntextdimV. Let v_...v_n be a basis of V. Define a new list of vectors w_...w_n by induction: w_:v_w_jv_j-_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle for j...n. Then abcliste abc w_...w_n is an orthogonal basis for V. abc leftfracw_||w_||...fracw_n||w_n||right is an orthonormal basis for V. abc forall leq jleq n we have textSpv_...v_jtextSpw_...w_jtextSpleftfracw_||w_||...fracw_n||w_n||right. abcliste
Solution:
Proof. We claim that forall leq jleq n w_j is well defined i.e. w_...w_j-neq and moreover w_...w_j is an orthogonal system with textSpw_..w_jtextSpv_...v_j. We prove this by induction on j. For j w_v_ so the claim is obvious v_neq because it is a part of a basis. Let leq jleq n. Ase the claim holds for w_...w_j-. Recall that w_jv_j-_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle w_i. Note that w_j is well defined because w_ineq forall leq ileq j-. We claim that w_jneq . Indeed if w_j then v_ji^j-fraclangle v_jw_i ranglelangle w_iw_i rangle w_i Longrightarrow v_jin textSpw_..w_j-textSpv_...v_j- which is impossible because v_...v_j are linearly indepent. This shows w_jneq . By the induction hypothesis w_...w_j- is an orthogonal system. So we have to prove that w_jperp w_...w_j-. Indeed let leq kleq j- then langle w_jw_k ranglelangle v_jw_k rangle+_i^j-fraclangle v_jw_i ranglelangle w_iw_i rangle langle w_iw_krangle langle v_jw_kranglfraclangle v_jw_k ranglelangle w_kw_k rangle langle w_kw_krangle. This shows w_...w_j is an orthogonal system. To complete the induction it remains to show that textSpw_...w_jtextSpv_...v_j. By the definition of w_j we have w_jin textSpw_...w_j-v_jtextSpv_...v_j-v_j. Longrightarrow textSpw_...w_j-w_jsubseteq textSpv_...v_j-v_j. But both these vector spaces have dim j because v_...v_j are linearly indepent and also w_...w_j are linearly indepent being an orthogonal system. Longrightarrow textSpw_...w_j-w_jtextSpv_...v_j-v_j.
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