Eigenvalues in 2D
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:
abcliste abc Show that the eigenvalues of a real times matrix bf A leftmatrixa & b c & dmatrixright are given by lambda_ fractaupm sqrttau^-Delta where taua+d is the trace and Deltaad-bc is the determinant of the matrix. abc Find and example of a matrix for which the eigenvalues are purely imaginary. abcliste
Solution:
abcliste abc Eigenvalues are characterised by bf A bf v lambda bf v This can be written more explicitely as leftmatrixa & b c & dmatrixright leftmatrixv_ v_matrixright lambda leftmatrixv_ v_matrixright or leftmatrixa-lambda & b c & d-lambdamatrixright leftmatrixv_ v_matrixright For non-trivial solutions i.e. v_neq or v_neq the determinant of the matrix in the above line has to be zero: det leftmatrixa-lambda & b c & d-lambdamatrixright Using the expression for the determinant of a times matrix we find a-lambdad-lambda-bc lambda^-a+dlambda+ad-bc lambda^-taulambda+Delta The solutions are given by the well-known formula for quadratic s. abc For purely imaginary eigenvalues the trace has to be zero a+d and the determinant has to be positive: ad-bc a-a-bc -a^-bc It follows that b and c have to have different signs and their product has to be greater than a^. A simple example would be the matrix leftmatrix & - & -matrixright with eigenvalues pm i. abcliste
abcliste abc Show that the eigenvalues of a real times matrix bf A leftmatrixa & b c & dmatrixright are given by lambda_ fractaupm sqrttau^-Delta where taua+d is the trace and Deltaad-bc is the determinant of the matrix. abc Find and example of a matrix for which the eigenvalues are purely imaginary. abcliste
Solution:
abcliste abc Eigenvalues are characterised by bf A bf v lambda bf v This can be written more explicitely as leftmatrixa & b c & dmatrixright leftmatrixv_ v_matrixright lambda leftmatrixv_ v_matrixright or leftmatrixa-lambda & b c & d-lambdamatrixright leftmatrixv_ v_matrixright For non-trivial solutions i.e. v_neq or v_neq the determinant of the matrix in the above line has to be zero: det leftmatrixa-lambda & b c & d-lambdamatrixright Using the expression for the determinant of a times matrix we find a-lambdad-lambda-bc lambda^-a+dlambda+ad-bc lambda^-taulambda+Delta The solutions are given by the well-known formula for quadratic s. abc For purely imaginary eigenvalues the trace has to be zero a+d and the determinant has to be positive: ad-bc a-a-bc -a^-bc It follows that b and c have to have different signs and their product has to be greater than a^. A simple example would be the matrix leftmatrix & - & -matrixright with eigenvalues pm i. abcliste
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Exercise:
abcliste abc Show that the eigenvalues of a real times matrix bf A leftmatrixa & b c & dmatrixright are given by lambda_ fractaupm sqrttau^-Delta where taua+d is the trace and Deltaad-bc is the determinant of the matrix. abc Find and example of a matrix for which the eigenvalues are purely imaginary. abcliste
Solution:
abcliste abc Eigenvalues are characterised by bf A bf v lambda bf v This can be written more explicitely as leftmatrixa & b c & dmatrixright leftmatrixv_ v_matrixright lambda leftmatrixv_ v_matrixright or leftmatrixa-lambda & b c & d-lambdamatrixright leftmatrixv_ v_matrixright For non-trivial solutions i.e. v_neq or v_neq the determinant of the matrix in the above line has to be zero: det leftmatrixa-lambda & b c & d-lambdamatrixright Using the expression for the determinant of a times matrix we find a-lambdad-lambda-bc lambda^-a+dlambda+ad-bc lambda^-taulambda+Delta The solutions are given by the well-known formula for quadratic s. abc For purely imaginary eigenvalues the trace has to be zero a+d and the determinant has to be positive: ad-bc a-a-bc -a^-bc It follows that b and c have to have different signs and their product has to be greater than a^. A simple example would be the matrix leftmatrix & - & -matrixright with eigenvalues pm i. abcliste
abcliste abc Show that the eigenvalues of a real times matrix bf A leftmatrixa & b c & dmatrixright are given by lambda_ fractaupm sqrttau^-Delta where taua+d is the trace and Deltaad-bc is the determinant of the matrix. abc Find and example of a matrix for which the eigenvalues are purely imaginary. abcliste
Solution:
abcliste abc Eigenvalues are characterised by bf A bf v lambda bf v This can be written more explicitely as leftmatrixa & b c & dmatrixright leftmatrixv_ v_matrixright lambda leftmatrixv_ v_matrixright or leftmatrixa-lambda & b c & d-lambdamatrixright leftmatrixv_ v_matrixright For non-trivial solutions i.e. v_neq or v_neq the determinant of the matrix in the above line has to be zero: det leftmatrixa-lambda & b c & d-lambdamatrixright Using the expression for the determinant of a times matrix we find a-lambdad-lambda-bc lambda^-a+dlambda+ad-bc lambda^-taulambda+Delta The solutions are given by the well-known formula for quadratic s. abc For purely imaginary eigenvalues the trace has to be zero a+d and the determinant has to be positive: ad-bc a-a-bc -a^-bc It follows that b and c have to have different signs and their product has to be greater than a^. A simple example would be the matrix leftmatrix & - & -matrixright with eigenvalues pm i. abcliste
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