0
$\begingroup$

Let $V$ be the space spanned by the two functions $\cos(t)$ and $\sin(t)$. Find the matrix $A$ of the linear transformation T(f(t)) = f''(t) + 7f'(t) + 4f(t) from $V$ into itself with respect to the basis $\{\cos(t), \sin(t)\}$.

$A = \left[\begin{matrix}\color{red}{\square} & \color{red}{\square} \\ \color{red}{\square} & \color{red}{\square} \end{matrix}\right]$

I don't know where to start. Where does A go in regards to the $T(f(t))=\ldots$ equation?

  • 0
    Would be interesting to see, how it acts on the space of all $(\cos(nx),\sin(nx))$!2012-03-28

2 Answers 2

3

You're given a basis which I'll label as $\vec{v}_1 = \cos(t)$ and $\vec{v}_2 = \sin(t)$. This lets you interpret columns of numbers as vectors. The column vector $\begin{bmatrix} a\\b\end{bmatrix}$ literally means the vector $a\cos(t) + b\sin(t)$.

Writing the matrix $A$ relative to this basis is done by computing $A\vec{v}_1$ and expressing it in terms of $\vec{v}_1$ and $\vec{v}_2$ and then doing the same for $A\vec{v}_2$.

In matrix language, $T\vec{v}_1$ corresponds to $A\begin{bmatrix}1\\0\end{bmatrix}$. On the other hand, T\vec{v}_1 = T\cos(t) = \cos(t)''+7\cos(t)' +4\cos(t) = 3\cos(t) -7\sin(t) = \begin{bmatrix}3\\-7\end{bmatrix}

So, the matrix $A$ should have the property that $A\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix} 3\\-7\end{bmatrix}$. This forces $A$ to have the form $A = \begin{bmatrix} 3 & *\\-7 & *\end{bmatrix}$.

Can you take it from here?

0

$\bf Hint:$ First determine where $T$ sends the basis $\{\cos(t),\sin(t)\}$. Then write the images as a linear combination of the basis and consider the matrix that has as columns the coefficients of the linear combination.