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Math Help - Standard matrix representation of a differentiation operator.

  1. #1
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    Standard 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?!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by linearalgebradunce View Post
    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?!
    Let b=(v_1,v_2,v_3,v_4) be a basis of P_3. The four columns of the matrix representation of E with respect to b are E(v_1), E(v_2),E(v_3) and E(v_4). Usually, one works with b=(1,X,X^2,X^3) so to find, say, the last column of the matrix, one has to compute E(v_4)=E(X^3). As E(X^3)=\frac{\mathrm{d}}{\mathrm{d}X}\left(X^3\rig  ht)=3X^2=3v_3 one has

    <br />
B=\begin{bmatrix} .& . & . & 0\\<br />
. & . & . & 0 \\<br />
. & . & . & 3 \\<br />
. & . & . & 0\\<br />
\end{bmatrix}

    Does it help ?
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    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.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by linearalgebradunce View Post
    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.
    A is a strictly triangular matrix so it is nilpotent. As nilpotent matrices are not diagonalizable, A can't be diagonalizable.

    If you're not supposed to know this result, let's show it using contradiction : assume that A is diagonalizable. 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 E_0 be the eigenspace associated to the eigenvalue 0. E_0 is the set \{p\in P_3\,|\,Ap=0\cdot p=0\} so E_0=\ker A. For A to be diagonalizable, we need to have \dim E_0=\dim P_3. I claim that this is impossible. Can you find out why?
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