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Solution to Problem14


*****

Problem14:

Consider the matrices,

> ;

                                   [2    1     2]
                                   [            ]
                              A := [2    2    -2]
                                   [            ]
                                   [3    1     1]


                                        [x1]
                                        [  ]
                                   x := [x2]
                                        [  ]
                                        [x3]

Show that the equation Ax=x can be rewritten as (A-I)x=0 and use this result to solve Ax=x for x. Also solve Ax=4x.

SOLUTION:


It is obvious that we can write Ax=x as (A-I)x = 0. Just substract x from both sides and use the property that Ix=x and the distributivity of matrix multiplication over matrix addition. To find x we can bring the augmented matrix of the homogeneous system (A-I)x=0 into rref (reduced row echelon form). Here is one way to do it with maple,

> A := matrix(3,3,[2,1,2,2,2,-2,3,1,1]); I3 := diag(1,1,1);

                                   [2    1     2]
                                   [            ]
                              A := [2    2    -2]
                                   [            ]
                                   [3    1     1]

                                    [1    0    0]
                                    [           ]
                              I3 := [0    1    0]
                                    [           ]
                                    [0    0    1]
> rref( augment(A-I3, matrix(3,1,[0,0,0])) );
                              [1    0    0    0]
                              [                ]
                              [0    1    0    0]
                              [                ]
                              [0    0    1    0]

this shows that the only solution for x is the trivial solution x=0. i.e. x1=x2=x3=0. Not very interesting but look at the next case. Let us try to find x such that Ax=4x. Now we can find infinitely many solutions... As before we bring into rref the augmented matrix for the homogeneous system (A-4I)x=0.

> rref( augment( A - 4*I3, matrix(3,1,[0,0,0]) ) );

                              [1    0    -1    0]
                              [                 ]
                              [0    1     0    0]
                              [                 ]
                              [0    0     0    0]

this system says that any x such that x1=x3 and that x2=0 will do. In other words t*(1,0,1) for any scalar t will be a solution.

These kinds of problems (i.e. solving Ax=kx for some scalar k) are very important in linear algebra. Non trivial x's (i.e. different from 0) of this kind are known as eigen vectors for the matrix A associated to the eigen value k. We'll study eigen values and eigen vectors later on in this course when we talk about diagonalization of matrices.


Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>
Last modified: Wed Feb 9 15:22:53 EST 2000