숙제2

B. Duality, dual basis.

%\resume{enumerate}

\begin{enumerate}%\setcounter{enumi}{7}
\item
Suppose that for each $x$ in $\mathcal{P}$
the function $y$ is defined by
\begin{enumerate}
\item
$y(x)=\int_{-1}^2 x(t)\,dt$
\item
$y(x)=\int_0^2 (x(t))^2\,dt$
\item
$y(x)=\int_0^1 t^2x(t)\,dt$
\item
$y(x)=\int_0^1 x(t^2)\,dt$
\item
$y(x)=\dfrac{dx}{dt}$
\item
$y(x)=\dfrac{d^2x}{dt^2}\bigg|_{t=1}$
\end{enumerate}

In which of these cases is \(y\) a linear function?

\item If \(y\) is a non-zero linear function on a vector space \(V\), and if \(\alpha\) is an arbitrary scalar, does there necessarily exist a vector \(x\) in \(V\) such that \(y(x)=\alpha\)?

\item Prove that if \(y\) and \(z\) are linear functions (on the same vector space) such that \(y(x)=0\) whenever \(z(x)=0\), then there exists a scalar \(\alpha\) such that \(y=\alpha z\). (Hint: if \(z(x_0)\neq0\), write \(\alpha=y(x_0)/z(x_0)\).)

\item Suppose that \(m<n\) and that \(y_1, \dots, y_m\) are linear functionals on an $n$-dimensional vector space \(V\). Under what conditions on the scalars \(\alpha_1, \dots, \alpha_m\) is it true that there exists a vector \(x\) in \(V\) such that \(y_j(x)=\alpha_j\) for \(j=1,\dots,m\)?

What does this result say about the solutions of linear equations?

\end{enumerate}

[wiki:선형대수숙제: 위로]