Find a Basis and Dimension for these Sets of Conditions

Question

I am trying to solve a question from my textbook. These are the problems I need help with:

Find a basis and the dimension of the subspaces defined by each of the following sets of conditions:

a) $\{f \in Span\{e^x, e^{2x}, e^{3x}\} \mid f(0) = f'(0) = 0\}$

b) $\{f \in Span\{e^x, e^{-x}, \cos(x), \sin(x), 1, x\} \mid f(0) = f'(0) = 0\}$

c) $\{f \in P_3(\mathbb{R}) \mid f(2) = f(-1) = 0\}$, where $P_3(\mathbb{R})$ is the set of all polynomials of degree $\leq 3$.

I know how to obtain the basis for linear systems of equations, by using matrices and their operations, but I am not sure where to start for the above types of questions. Thank you in advance.

Answer 1

For a) denote $V = Span(e^x,e^{2x}, e^{3x})$ and notice that a function $\alpha e^x + \beta e^{2x} + \gamma e^{3x}$ corresponds to the vector $(\alpha, \beta, \gamma)$ in the given basis $\{e^x,e^{2x}, e^{3x}\}$.

Now, a function $f= \alpha e^x + \beta e^{2x} + \gamma e^{3x}\in V$ satisfies $f(0)=0$ iff $0 = \alpha e^0 + \beta e^{0} + \gamma e^{0}$, i.e. $0 = \alpha + \beta + \gamma$ which is equivalent to $\alpha = -\beta - \gamma$, i.e. if it lies in the space $V_1=(-\beta-\gamma, \beta, \gamma)$ for $\beta, \gamma \in F$, where $F$ is the field over which you're asked to work.

Similarly for the derivative you obtain $f'(0)=0$ iff $0 = \alpha e^0 + 2\beta e^{0} + 3\gamma e^{0}$, i.e. $0 = \alpha + 2\beta + 3\gamma$ which is equivalent to $\alpha = -2\beta - 3\gamma$, i.e. if it lies in the space $V_2=(-2\beta-3\gamma, \beta, \gamma)$ for $\beta, \gamma \in F$.

Now just find the intersection $V_1\cap V_2=$<(1,-2,1)> by classical LA methods.

b) and c) are similar.

Answer 2

For (a), consider $V=\operatorname{span}\{e^x,e^{2x},e^{3x}\}$ and the linear map $T\colon V\to\mathbb{R}^2$ defined by $$ T(f)=\begin{bmatrix} f(0)\\f'(0)\end{bmatrix} $$ The subspace you have to find a basis of is the kernel of $T$.

Since $\{e^x,e^{2x},e^{3x}\}$ is a basis, the matrix of $T$ with respect to this basis and the standard basis on $\mathbb{R}^2$ is $$ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \end{bmatrix} $$ which has the RREF $$ \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{bmatrix} $$ so a basis of its null space is the vector $$ \begin{bmatrix} 1 \\ -2 \\ 1\end{bmatrix} $$ Hence a basis for the kernel of $T$ is the function $$ 1\cdot e^x-2\cdot e^{2x}+1\cdot e^{3x} $$

Parts (b) and (c) are done similarly.