Preface
(a*b)/c+13*d \[ {\frac {ab}{c}}+13\,d \]
Example:
**DESCRIPTION OF PROBLEM GOES HERE**
This is a description for some Maple code. Maple 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:
(a*b)/c+13*d
ur code
another line
\[ {\frac {ab}{c}}+13\,d \]
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, …
, xr be all of the
linearly independent eigenvectors associated to λ, so that
λ has geometric multiplicity r. Let
xr+1, xr+2, …
, xn complete this list to a basis for
ℜn, and let S be the n×n
matrix whose columns are all these
vectors xs, 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.
◂
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