Download Physics 3730/6720 – Maple 1b – 1 Linear algebra, Eigenvalues and Eigenvectors

Physics 3730/6720 – Maple 1b –
1
October 8, 2012
Linear algebra, Eigenvalues and Eigenvectors
Maple session
> with(linalg):
_______________________________
> A := matrix([[1,2,3],[3,2,1],[1,0,-1]]);
[1
2
3]
[
]
A := [3
2
1]
[
]
[1
0
-1]
> b := vector([1,2,-1]);
b := [1, 2, -1]
_________________________________
> evalm(3 * b);
[3, 6, -3]
_________________________________
> y := evalm(A &* b);
y := [2, 6, 2]
Comments
The linear algebra procedures are defined in the
linalg package. The simple way to define a matrix is
to list its values, row by row.
Note the square brackets.
Vectors
(column vectors written as row
vectors) are defined by listing
their elements. To do matrix
and vector products and see
the result you need evalm.
(Maple does it, otherwise,
but silently.)
Multiplication by scalars works with *,
but multiplication of matrices with matrices and matrices with vectors requires the
special multiplication operator &*.
> eigenvals(A);
The linear algebra package
1/2
1/2
computes eigenvalues and
0, 1 + 11
, 1 - 11
eigenvectors. The eigenvec_________________________________
tors are listed in brackets (in
> eigenvectors(A);
no particular order). The
1/2
[
1/2
1/2
]
first number in the bracket is
[1 + 11
, 1, {[2 + 11
, 4 + 11
, 1]}], the the eigenvalue (compare
1/2
[
1/2
1/2
]
both results), the second is
[1 - 11
, 1, {[2 - 11
, 4 - 11
, 1]}], the number of times that
[0, 1, {[1, -2, 1]}]
eigenvalue appears (multiplicity), and the third entry
gives the eigenvector itself.
1
Maple session
> inverse(A);
Error, (in inverse) singular matrix
_________________________________
> inverse(1 + A);
[ 0
0
1]
[
]
[-1/7
3/7
-1]
[
]
[3/7
-2/7
0]
> transpose(A);
[1
3
1]
[
]
[2
2
0]
[
]
[3
1
-1]
> linsolve(A,b);
_________________________________
> linsolve(1+A,b);
[-1, 12/7, -1/7]
2
Comments
Since our matrix A has a zero
eigenvalue, it has no inverse,
but the matrix 1 + A has an
inverse, which you get with
inverse.
The transpose.
This is the way to solve Ax =
b, but since A is singular,
Maple refuses to do it. The
matrix 1 + A is not singular, so we can get the answer for a different problem:
(1 + A)x = b.