Math 325: Differential	Equations    Quiz 3    10/15/97

1. Determine whether the following vector functions are linearly independent.

x1[t_] = {1,  t,  t^2};
x2[t_] = {t,  t^2,t^3};
x3[t_] = {t^2,t^3,t^4};

Solution

We try to solve for the coefficients of a linear combination that will yield the zero vector for all values of t. We plug in three values to get three (of infinitely many possible) equations in the coefficients.

eqns = {c1 x1[0] + c2 x2[0] + c3 x3[0] == {0,0,0},
        c1 x1[1] + c2 x2[1] + c3 x3[1] == {0,0,0},
        c1 x1[2] + c2 x2[2] + c3 x3[2] == {0,0,0}}

Solve[eqns]

Since the only solution is for all coefficients to be zero, the vector functions are linearly independent.

2. Find the eigenvalues and corresponding eigenvectors of the matrix
    {{ 0, 6, 0},
     { 3, 0, 3},
     { 0,-3, 0}}

Solution

A = {{ 0, 6, 0},
     { 3, 0, 3},
     { 0,-3, 0}};

The eigenvalues are the roots of the polynomial given by

Det[A - r IdentityMatrix[3]]

Solve[% == 0]

To find an eigenvector corresponding to the eigenvalue -3, we must solve

(A - (-3)IdentityMatrix[3]).{v1,v2,v3} == {0,0,0}

We solve this by hand using row reduction.

RowReduce[A - (-3)IdentityMatrix[3]] //MatrixForm

In one step:

Solve[(A - (-3)IdentityMatrix[3]).{v1,v2,v3} == {0,0,0}]

So, one eigenvector for the eigenvlaue -3 is V == {-2,1,1}

To find an eigenvector corresponding to the other eigenvalue 0, we solve

(A - 0 IdentityMatrix[3]).{v1,v2,v3} == {0,0,0}

RowReduce[A - 0 IdentityMatrix[3]] //MatrixForm

Solve[(A - 0 IdentityMatrix[3]).{v1,v2,v3} == {0,0,0}]

So, an eigenvector for the eigenvalue 0 is V == {1, 0,-1}.

To find an eigenvector corresponding to the other eigenvalue 3, we solve

(A - 3 IdentityMatrix[3]).{v1,v2,v3} == {0,0,0}

RowReduce[A - 3 IdentityMatrix[3]] //MatrixForm

Solve[(A - 3 IdentityMatrix[3]).{v1,v2,v3} == {0,0,0}]

So, an eigenvector for the eigenvalue 3 is V == {2,1,-1}. Let's check these answers:

A.{-2,1,1} == (-3){-2,1,1}

A.{1, 0,-1} == 0.{1, 0,-1}

A.{2,1,-1} == 3{2,1,-1}