MATLAB TUTORIAL for the First Course. Part I: Implicit plot

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x=4; y=16;
z = x + j*y

Example: **DESCRIPTION OF PROBLEM GOES HERE** This is a description for some MATLAB code. MATLAB is an extremely useful tool for many different areas in engineering, applied mathematics, computer science, biology, chemistry, and so much more. It is quite amazing at handling matrices, but has lots of competition with other programs such as Mathematica and Maple. Here is a code snippet plotting two lines (y vs. x and z vs. x) on the same graph. Click to view the code!

figure(1)
plot(x, y, 'Color', [1 0 0]) %blue line
hold on
plot(x, z, 'Color', [0 1 0]) %green line

Two n-by-n matrices A and B are called similar if there exists an invertible n-by-n matrix S such that

Theorem: If λ is an eigenvalue of a square matrix A, then its algebraic multiplicity is at least as large as its geometric multiplicity. ▣

Let x₁, x₂, … , x_r be all of the linearly independent eigenvectors associated to λ, so that λ has geometric multiplicity r. Let x_r+1, x_r+2, … , x_n complete this list to a basis for ℜⁿ, and let S be the n×n matrix whose columns are all these vectors x_s, s = 1, 2, … , n. As usual, consider the product of two matrices AS. Because the first r columns of S are eigenvectors, we have

${\bf A\,S} = \begin{bmatrix} \vdots & \vdots&& \vdots & \vdots&& \vdots \\ \lambda{\bf x}_1 & \lambda{\bf x}_2 & \cdots & \lambda{\bf x}_r & ?& \cdots & ? \\ \vdots & \vdots&& \vdots & \vdots&& \vdots \end{bmatrix} .$

Now multiply out S^-1AS. Matrix S is invertible because its columns are a basis for ℜⁿ. We get that the first r columns of S^-1AS are diagonal with &lambda's on the diagonal, but that the rest of the columns are indeterminable. Now S^-1AS has the same characteristic polynomial as A. Indeed,

$\det \left( {\bf S}^{-1} {\bf AS} - \lambda\,{\bf I} \right) = \det \left( {\bf S}^{-1} {\bf AS} - {\bf S}^{-1} \lambda\,{\bf I}{\bf S} \right) = \det \left( {\bf S}^{-1} \left( {\bf A} - \lambda\,{\bf I} \right) {\bf S} \right) = \det \left( {\bf S}^{-1} \right) \det \left( {\bf A} - \lambda\,{\bf I} \right) \det \det \left( {\bf S} \right) = \det \left( {\bf A} - \lambda\,{\bf I} \right)$

because the determinants of S and S^-1 cancel. So the characteristic polynomials of A and S^-1AS are the same. But since the first few columns of S^-1AS has a factor of at least (x - λ)^r, so the algebraic multiplicity is at least as large as the geometric. ◂