matrix - Verify Cayley-Hamilton Theorem

Consider the matrix:

A = {{-1, -4, -2}, {0, 1, 1}, {-6, -12, 2}}

Now, the characteristic polynomial is found:

p[λ_] = CharacteristicPolynomial[A, λ]

Now, how can I substitute the matrix A into the polynomial to verify that the answer is the zero matrix?

Answer

To verify that matrix is a zero of its characteristic polynomial, The Characteristic polynomial of the matrix is found, then evaluated for the matrix. The result should be the zero matrix.

Clear[x]
a = {{-1, -4, -2}, {0, 1, 1}, {-6, -12, 2}};

n = Length[a];
p = CharacteristicPolynomial[a, x];
(Sum[ Coefficient[p, x, i] MatrixPower[a, i], {i, 0, Exponent[p, x]}]) // MatrixForm

Mathematica graphics

another way:

cl = CoefficientList[p, x];
Sum[MatrixPower[a, j - 1] cl[[j]], {j, 1, Length[cl]}]

Mathematica graphics

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],

Blog

Search This Blog