q. let P sub 3 denote the subspace consisting of all polynomials of degree no greater than 3. Let E: P sub 3 ---> P sub 3 denote the differentiation operator on P sub 3. Write down the standard matrix representation of E?!

- Aug 13th 2008, 11:37 AMlinearalgebradunceStandard matrix representation of a differentiation operator.
q. let P sub 3 denote the subspace consisting of all polynomials of degree no greater than 3. Let E: P sub 3 ---> P sub 3 denote the differentiation operator on P sub 3. Write down the standard matrix representation of E?!

- Aug 14th 2008, 12:56 AMflyingsquirrel
Hello,

Let $\displaystyle b=(v_1,v_2,v_3,v_4)$ be a basis of $\displaystyle P_3$. The four columns of the matrix representation of $\displaystyle E$ with respect to $\displaystyle b$ are $\displaystyle E(v_1), E(v_2),E(v_3)$ and $\displaystyle E(v_4)$. Usually, one works with $\displaystyle b=(1,X,X^2,X^3)$ so to find, say, the last column of the matrix, one has to compute $\displaystyle E(v_4)=E(X^3)$. As $\displaystyle E(X^3)=\frac{\mathrm{d}}{\mathrm{d}X}\left(X^3\rig ht)=3X^2=3v_3$ one has

$\displaystyle

B=\begin{bmatrix} .& . & . & 0\\

. & . & . & 0 \\

. & . & . & 3 \\

. & . & . & 0\\

\end{bmatrix}$

Does it help ? - Aug 14th 2008, 01:27 AMlinearalgebradunce
yes that seems so simple now, thank you. is there a quick way of deciding whether this resultant matrix is diagonalisable or not?! I know the definition of a diagonalisable matrix is that there exists P'AP=D where P'=inverse P and D is a diagonal matrix, etc as this was in the other question I posted. However this time I do not need to prove or show if the matrix is diagonalisable, simply state yes or no.

- Aug 14th 2008, 04:03 AMflyingsquirrel
$\displaystyle A$ is a strictly triangular matrix so it is nilpotent. As nilpotent matrices are not diagonalizable, $\displaystyle A$ can't be diagonalizable.

If you're not supposed to know this result, let's show it using contradiction : assume that $\displaystyle A$ is diagonalizable. $\displaystyle A$ has one eigenvalue which is 0 and whose multiplicity is 4. (since the matrix is triangular and the four diagonal entries are 0) Let $\displaystyle E_0$ be the eigenspace associated to the eigenvalue 0. $\displaystyle E_0$ is the set $\displaystyle \{p\in P_3\,|\,Ap=0\cdot p=0\}$ so $\displaystyle E_0=\ker A$. For $\displaystyle A$ to be diagonalizable, we need to have $\displaystyle \dim E_0=\dim P_3$. I claim that this is impossible. Can you find out why?