Plotting eigenvalue function along a path with correct coloring
$begingroup$
This question has multiple parts to it. The setup is that I have a matrix that is a function of two parameters a and b. I wish to plot the eigenvalues of this matrix along a general path in the a-b plane and I want these two branches to have the correct coloring. For example, for the simple path for {a,b} from {0,0} to {1,0} the following works:
testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[Evaluate@Eigenvalues[testMat[t, 0]], {t, 0, 1}]
At this point please note that Evaluate
must be included in the second line for the two branches to have different colors. My first question is thus
- Why is Evaluate necessary to get the correct colors for the eigenvalue plots?
Now suppose that I wish to plot the eigenvalues on a path that goes from {0,0} to {0,1} to {1,1}. I implemented this in the following way
testfunc[t_] = Evaluate@Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1],
1 < t < 1.5}}, {0, 0}];
Plot[Eigenvalues@testfunc[t], {t, 0, 1.5}]
However as you can see the two branches have the same color. Somehow Mathematica does not understand that they are two separate plots. Thus my final two questions are:
How do I get the two branches colored separately?
Is there a better way of plotting along a path that Mathematica will find more agreeable?
Thanks in advance!
EDIT: All answers are great but there is a significant problem using First/Last or dot products since the Eigenvalues are listed from largest to smallest while when plotting we are interested in smooth functions (i.e the largest eigenvalue will always be blue and the smaller orange). For example:
testMat2[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a^2}};
Plot[{First@Evaluate@Eigenvalues[testMat2[t, 0]],
Last@Evaluate@Eigenvalues[testMat2[t, 0]]}, {t, 0, 1.6}]
What do I do to fix this?
Edit 2: Inspired by the excellent answers below, the easiest method for me was to create a Table
of eigenvalues and then use Interpolation
to create a vector values function. Now Plot[{interpFunc[t][[1]], interpFunc[t][[2]],...}, {t,0,1}]
works beautifully!
plotting parametric-functions eigenvalues
$endgroup$
add a comment |
$begingroup$
This question has multiple parts to it. The setup is that I have a matrix that is a function of two parameters a and b. I wish to plot the eigenvalues of this matrix along a general path in the a-b plane and I want these two branches to have the correct coloring. For example, for the simple path for {a,b} from {0,0} to {1,0} the following works:
testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[Evaluate@Eigenvalues[testMat[t, 0]], {t, 0, 1}]
At this point please note that Evaluate
must be included in the second line for the two branches to have different colors. My first question is thus
- Why is Evaluate necessary to get the correct colors for the eigenvalue plots?
Now suppose that I wish to plot the eigenvalues on a path that goes from {0,0} to {0,1} to {1,1}. I implemented this in the following way
testfunc[t_] = Evaluate@Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1],
1 < t < 1.5}}, {0, 0}];
Plot[Eigenvalues@testfunc[t], {t, 0, 1.5}]
However as you can see the two branches have the same color. Somehow Mathematica does not understand that they are two separate plots. Thus my final two questions are:
How do I get the two branches colored separately?
Is there a better way of plotting along a path that Mathematica will find more agreeable?
Thanks in advance!
EDIT: All answers are great but there is a significant problem using First/Last or dot products since the Eigenvalues are listed from largest to smallest while when plotting we are interested in smooth functions (i.e the largest eigenvalue will always be blue and the smaller orange). For example:
testMat2[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a^2}};
Plot[{First@Evaluate@Eigenvalues[testMat2[t, 0]],
Last@Evaluate@Eigenvalues[testMat2[t, 0]]}, {t, 0, 1.6}]
What do I do to fix this?
Edit 2: Inspired by the excellent answers below, the easiest method for me was to create a Table
of eigenvalues and then use Interpolation
to create a vector values function. Now Plot[{interpFunc[t][[1]], interpFunc[t][[2]],...}, {t,0,1}]
works beautifully!
plotting parametric-functions eigenvalues
$endgroup$
add a comment |
$begingroup$
This question has multiple parts to it. The setup is that I have a matrix that is a function of two parameters a and b. I wish to plot the eigenvalues of this matrix along a general path in the a-b plane and I want these two branches to have the correct coloring. For example, for the simple path for {a,b} from {0,0} to {1,0} the following works:
testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[Evaluate@Eigenvalues[testMat[t, 0]], {t, 0, 1}]
At this point please note that Evaluate
must be included in the second line for the two branches to have different colors. My first question is thus
- Why is Evaluate necessary to get the correct colors for the eigenvalue plots?
Now suppose that I wish to plot the eigenvalues on a path that goes from {0,0} to {0,1} to {1,1}. I implemented this in the following way
testfunc[t_] = Evaluate@Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1],
1 < t < 1.5}}, {0, 0}];
Plot[Eigenvalues@testfunc[t], {t, 0, 1.5}]
However as you can see the two branches have the same color. Somehow Mathematica does not understand that they are two separate plots. Thus my final two questions are:
How do I get the two branches colored separately?
Is there a better way of plotting along a path that Mathematica will find more agreeable?
Thanks in advance!
EDIT: All answers are great but there is a significant problem using First/Last or dot products since the Eigenvalues are listed from largest to smallest while when plotting we are interested in smooth functions (i.e the largest eigenvalue will always be blue and the smaller orange). For example:
testMat2[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a^2}};
Plot[{First@Evaluate@Eigenvalues[testMat2[t, 0]],
Last@Evaluate@Eigenvalues[testMat2[t, 0]]}, {t, 0, 1.6}]
What do I do to fix this?
Edit 2: Inspired by the excellent answers below, the easiest method for me was to create a Table
of eigenvalues and then use Interpolation
to create a vector values function. Now Plot[{interpFunc[t][[1]], interpFunc[t][[2]],...}, {t,0,1}]
works beautifully!
plotting parametric-functions eigenvalues
$endgroup$
This question has multiple parts to it. The setup is that I have a matrix that is a function of two parameters a and b. I wish to plot the eigenvalues of this matrix along a general path in the a-b plane and I want these two branches to have the correct coloring. For example, for the simple path for {a,b} from {0,0} to {1,0} the following works:
testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[Evaluate@Eigenvalues[testMat[t, 0]], {t, 0, 1}]
At this point please note that Evaluate
must be included in the second line for the two branches to have different colors. My first question is thus
- Why is Evaluate necessary to get the correct colors for the eigenvalue plots?
Now suppose that I wish to plot the eigenvalues on a path that goes from {0,0} to {0,1} to {1,1}. I implemented this in the following way
testfunc[t_] = Evaluate@Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1],
1 < t < 1.5}}, {0, 0}];
Plot[Eigenvalues@testfunc[t], {t, 0, 1.5}]
However as you can see the two branches have the same color. Somehow Mathematica does not understand that they are two separate plots. Thus my final two questions are:
How do I get the two branches colored separately?
Is there a better way of plotting along a path that Mathematica will find more agreeable?
Thanks in advance!
EDIT: All answers are great but there is a significant problem using First/Last or dot products since the Eigenvalues are listed from largest to smallest while when plotting we are interested in smooth functions (i.e the largest eigenvalue will always be blue and the smaller orange). For example:
testMat2[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a^2}};
Plot[{First@Evaluate@Eigenvalues[testMat2[t, 0]],
Last@Evaluate@Eigenvalues[testMat2[t, 0]]}, {t, 0, 1.6}]
What do I do to fix this?
Edit 2: Inspired by the excellent answers below, the easiest method for me was to create a Table
of eigenvalues and then use Interpolation
to create a vector values function. Now Plot[{interpFunc[t][[1]], interpFunc[t][[2]],...}, {t,0,1}]
works beautifully!
plotting parametric-functions eigenvalues
plotting parametric-functions eigenvalues
edited Jan 9 at 9:53
Takoda
asked Dec 27 '18 at 16:49
TakodaTakoda
1477
1477
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
1) Basically, Mathematica has no way of knowing whether to treat the two curves as having the same or distinct colors. Using Evaluate
tells it to use distinct colors. (The underlying reasons relate to the order of evaluation.)
2) Evaluate
has no effect for testfunc
, because it cannot decide which part of Piecewise
to use until t
is provided. Replacing Set
by SetDelayed
does not help. Instead try
Plot[{First@Eigenvalues@testfunc[t], Last@Eigenvalues@testfunc[t]}, {t, 0, 1.5}]
3) Probably not, but I am not sure.
$endgroup$
add a comment |
$begingroup$
The problem is that at the time of the call to Plot
, it is not clear that it is about two function that are to plot. Actually, you tell Mathematica's Plot
command to plot an $mathbb{R}^2$-valued function. You can circumvent this issue, e.g. with ListLinePlot
:
f[t_] := Eigenvalues[Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1], 1 < t <= 1.5}}, {0, 0}]];
tlist = Subdivide[0., 1.5, 250];
ListLinePlot[Transpose[{tlist, #}] & /@ Transpose[testfunc /@ tlist]]
$endgroup$
add a comment |
$begingroup$
I'll add a couple of lines of code without using First, Last
testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[{Eigenvalues[testMat[t, 0]].{1, 0},
Eigenvalues[testMat[t, 0]].{0, 1}}, {t, 0, 1},
PlotStyle -> {Green, Red}]
testfunc[t_] =
Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1],
1 < t < 1.5}}, {0, 0}];
Plot[{Eigenvalues@testfunc[t].{1, 0},
Eigenvalues@testfunc[t].{0, 1}}, {t, 0, 1.5}]
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
1) Basically, Mathematica has no way of knowing whether to treat the two curves as having the same or distinct colors. Using Evaluate
tells it to use distinct colors. (The underlying reasons relate to the order of evaluation.)
2) Evaluate
has no effect for testfunc
, because it cannot decide which part of Piecewise
to use until t
is provided. Replacing Set
by SetDelayed
does not help. Instead try
Plot[{First@Eigenvalues@testfunc[t], Last@Eigenvalues@testfunc[t]}, {t, 0, 1.5}]
3) Probably not, but I am not sure.
$endgroup$
add a comment |
$begingroup$
1) Basically, Mathematica has no way of knowing whether to treat the two curves as having the same or distinct colors. Using Evaluate
tells it to use distinct colors. (The underlying reasons relate to the order of evaluation.)
2) Evaluate
has no effect for testfunc
, because it cannot decide which part of Piecewise
to use until t
is provided. Replacing Set
by SetDelayed
does not help. Instead try
Plot[{First@Eigenvalues@testfunc[t], Last@Eigenvalues@testfunc[t]}, {t, 0, 1.5}]
3) Probably not, but I am not sure.
$endgroup$
add a comment |
$begingroup$
1) Basically, Mathematica has no way of knowing whether to treat the two curves as having the same or distinct colors. Using Evaluate
tells it to use distinct colors. (The underlying reasons relate to the order of evaluation.)
2) Evaluate
has no effect for testfunc
, because it cannot decide which part of Piecewise
to use until t
is provided. Replacing Set
by SetDelayed
does not help. Instead try
Plot[{First@Eigenvalues@testfunc[t], Last@Eigenvalues@testfunc[t]}, {t, 0, 1.5}]
3) Probably not, but I am not sure.
$endgroup$
1) Basically, Mathematica has no way of knowing whether to treat the two curves as having the same or distinct colors. Using Evaluate
tells it to use distinct colors. (The underlying reasons relate to the order of evaluation.)
2) Evaluate
has no effect for testfunc
, because it cannot decide which part of Piecewise
to use until t
is provided. Replacing Set
by SetDelayed
does not help. Instead try
Plot[{First@Eigenvalues@testfunc[t], Last@Eigenvalues@testfunc[t]}, {t, 0, 1.5}]
3) Probably not, but I am not sure.
answered Dec 27 '18 at 18:05
bbgodfreybbgodfrey
44.9k1059110
44.9k1059110
add a comment |
add a comment |
$begingroup$
The problem is that at the time of the call to Plot
, it is not clear that it is about two function that are to plot. Actually, you tell Mathematica's Plot
command to plot an $mathbb{R}^2$-valued function. You can circumvent this issue, e.g. with ListLinePlot
:
f[t_] := Eigenvalues[Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1], 1 < t <= 1.5}}, {0, 0}]];
tlist = Subdivide[0., 1.5, 250];
ListLinePlot[Transpose[{tlist, #}] & /@ Transpose[testfunc /@ tlist]]
$endgroup$
add a comment |
$begingroup$
The problem is that at the time of the call to Plot
, it is not clear that it is about two function that are to plot. Actually, you tell Mathematica's Plot
command to plot an $mathbb{R}^2$-valued function. You can circumvent this issue, e.g. with ListLinePlot
:
f[t_] := Eigenvalues[Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1], 1 < t <= 1.5}}, {0, 0}]];
tlist = Subdivide[0., 1.5, 250];
ListLinePlot[Transpose[{tlist, #}] & /@ Transpose[testfunc /@ tlist]]
$endgroup$
add a comment |
$begingroup$
The problem is that at the time of the call to Plot
, it is not clear that it is about two function that are to plot. Actually, you tell Mathematica's Plot
command to plot an $mathbb{R}^2$-valued function. You can circumvent this issue, e.g. with ListLinePlot
:
f[t_] := Eigenvalues[Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1], 1 < t <= 1.5}}, {0, 0}]];
tlist = Subdivide[0., 1.5, 250];
ListLinePlot[Transpose[{tlist, #}] & /@ Transpose[testfunc /@ tlist]]
$endgroup$
The problem is that at the time of the call to Plot
, it is not clear that it is about two function that are to plot. Actually, you tell Mathematica's Plot
command to plot an $mathbb{R}^2$-valued function. You can circumvent this issue, e.g. with ListLinePlot
:
f[t_] := Eigenvalues[Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1], 1 < t <= 1.5}}, {0, 0}]];
tlist = Subdivide[0., 1.5, 250];
ListLinePlot[Transpose[{tlist, #}] & /@ Transpose[testfunc /@ tlist]]
answered Dec 27 '18 at 18:05
Henrik SchumacherHenrik Schumacher
56.7k577157
56.7k577157
add a comment |
add a comment |
$begingroup$
I'll add a couple of lines of code without using First, Last
testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[{Eigenvalues[testMat[t, 0]].{1, 0},
Eigenvalues[testMat[t, 0]].{0, 1}}, {t, 0, 1},
PlotStyle -> {Green, Red}]
testfunc[t_] =
Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1],
1 < t < 1.5}}, {0, 0}];
Plot[{Eigenvalues@testfunc[t].{1, 0},
Eigenvalues@testfunc[t].{0, 1}}, {t, 0, 1.5}]
$endgroup$
add a comment |
$begingroup$
I'll add a couple of lines of code without using First, Last
testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[{Eigenvalues[testMat[t, 0]].{1, 0},
Eigenvalues[testMat[t, 0]].{0, 1}}, {t, 0, 1},
PlotStyle -> {Green, Red}]
testfunc[t_] =
Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1],
1 < t < 1.5}}, {0, 0}];
Plot[{Eigenvalues@testfunc[t].{1, 0},
Eigenvalues@testfunc[t].{0, 1}}, {t, 0, 1.5}]
$endgroup$
add a comment |
$begingroup$
I'll add a couple of lines of code without using First, Last
testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[{Eigenvalues[testMat[t, 0]].{1, 0},
Eigenvalues[testMat[t, 0]].{0, 1}}, {t, 0, 1},
PlotStyle -> {Green, Red}]
testfunc[t_] =
Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1],
1 < t < 1.5}}, {0, 0}];
Plot[{Eigenvalues@testfunc[t].{1, 0},
Eigenvalues@testfunc[t].{0, 1}}, {t, 0, 1.5}]
$endgroup$
I'll add a couple of lines of code without using First, Last
testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[{Eigenvalues[testMat[t, 0]].{1, 0},
Eigenvalues[testMat[t, 0]].{0, 1}}, {t, 0, 1},
PlotStyle -> {Green, Red}]
testfunc[t_] =
Piecewise[{{testMat[t, 0], 0 <= t <= 1}, {testMat[1, t - 1],
1 < t < 1.5}}, {0, 0}];
Plot[{Eigenvalues@testfunc[t].{1, 0},
Eigenvalues@testfunc[t].{0, 1}}, {t, 0, 1.5}]
answered Dec 27 '18 at 19:21
Alex TrounevAlex Trounev
7,8831521
7,8831521
add a comment |
add a comment |
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