Linear Algebra Handout: Two Solved Examples

February 28, 2003

1. Let W_{1} and W_{2} be the subspaces of R^{3} given by

W_{1} = span {(1, 2, 3), (2, 1, 1)} and W_{2} = span {(1, 0, 1), (3, 0, −1)} .

Find a set of generating vectors for W_{1} ∩ W_{2}.

Response. W_{1} and W_{2} are each spanned by two linearly independent vectors, and for that reason each is a plane through the origin in R^{3}. One therefore expects their intersection to be a line through the origin in R^{3}, i.e., the set of all scalar multiples of a single vector. The strategy adopted is to find equations defining W_{1} and W_{2} and then solve both equations simultaneously to find W_{1} ∩ W_{2}.

For W_{1} one seeks a vector (a, b, c) such that

( \begin{matrix}1 & 2 & 3 \\ 2 & 1 & 1 \end{matrix} ) ( \begin{matrix}a \\ b \\ c \end{matrix} ) = ( \begin{matrix}0 \\ 0 \end{matrix} ) .

Solving the corresponding linear system of equations using row operations, one finds that the most general such vector (a, b, c) is a scalar multiple of (1, −5, 3), and “the” equation of W_{1} is x − 5y + 3z = 0.

A similar method can be used for W_{2}. On the other hand, one sees quickly by inspection that “the” equation of W_{2} is y = 0. Then solving the two equations simultaneously — for example, using row operations on the matrix

( \begin{matrix}1 & −5 & 3 \\ 0 & 1 & 0 \end{matrix} )

— one finds that a basis of W_{1} ∩ W_{2} is given by the single vector (−3, 0, 1), i.e.,

W_{1} ∩ W_{2} = span {(−3, 0, 1)} .

2. Find a particular solution of the linear differential equation

y′′ + 4y = x^{2} .

Response. In the study of differential equations one learns to look for particular solutions in various ways. In this case it will be fruitful to propose

y = a + bx + cx^{2}

as a trial solution. With this y one finds y′′ = 2c, and, therefore,

y′′ + 4y = (4a + 2c) + 4bx + 4cx^{2} .

To have the right-hand side evaluate as x^{2} one needs

4a + 2c = 0
4b = 0
4c = 1

Solving this system for a, b, c one obtains

y = \frac{1}{4}x^{2} − \frac{1}{8} .