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Let $P_3(\mathbb{R})$ be the real vector space consisting of real polynomials of degree $\le2$. Observe the mapping:

$$L: P_3(\mathbb{R}) \to P_3(\mathbb{R})$$ defined by: $$L(\alpha+\beta x + \gamma x^2) = (\alpha + 4\beta + 16\gamma)-(\beta+8\gamma)x+\gamma x^2$$

1) Find the matrix representation $ _V[L]_V $ for L with respect to the basis $V = (1,x,x^2)$

Now let $P_3(\mathbb{R})$ be an inner product space with the inner product: $$ \langle p,q \rangle= p(1)q(1)+p(2)q(2)+p(3)q(3)$$ with the orthogonal basis $W=(w_1,w_2,w_3)$ where: $$ w_1 = 1, w_2 = -2+x, w_3 = 10-12x+3x^2$$ 2) Show that $w_1,w_2,w_3$ are eigenvectors for $L$, and find the corresponding eigenvalues. Explain why $L$ is orthonormal diagonalizable.


1) We know that $ _V[L]_V = ([L(v_1)]_V \ [L(v_2)]_V \ [L(v_3)]_V= \begin{pmatrix} 1 & 0 & 0 \\ 4 & -1 & 0 \\ 1 & -8 & -1\\ \end{pmatrix} = A$

2) The definition of an eigenvector is that an element $v \in V\ \text\ \text{{0}}$ is an eigenvector for $L$ if $L(v) = \lambda \cdot v$. This can be used for the matrix representation as $A\cdot v= \lambda \cdot v$

$A \cdot w_1 = \begin{pmatrix} 1 & 0 & 0 \\ 4 & -1 & 0 \\ 1 & -8 & -1\\ \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix} =\begin{pmatrix} 1 \\ 4 \\ 1 \\ \end{pmatrix} $

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So I am pretty sure that I have computed the matrix representation wrong, but I cannot see what I have done wrong. Could someone point out what I have done wrong with the matrix representation and if this is the right method to find the eigenvectors and eigenvalues for $L$?

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$L(1)=1=1+0x+0x^2$, right? So, the entries of the first column should be $1$, $0$, and $0$.

In fact, your matrix is the transpose of the right one. (well, almost. The entry at the lower right corner is $1$, not $-1$.)

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  • $\begingroup$ Ah, I did it by taking $L(1) = \begin{pmatrix} 1 \\ 0x \\ 0x^2 \\ \end{pmatrix}$ So the matrix is going to be $ \begin{pmatrix} 1 & 4 & 1 \\ 0 & -1 & -8 \\ 0 & 0 & 1\\ \end{pmatrix} $ Now by computing the matrix A with $w_1,w_2,w_3$ I get that: $A\cdot w_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix}$ $A\cdot w_2 = \begin{pmatrix} 2 \\ -1 \\ 0 \\ \end{pmatrix}$ which is good, but for the last one I get: $A\cdot w_3 = \begin{pmatrix} -35 \\ -12 \\ 3 \\ \end{pmatrix}$ $\endgroup$ – Martin Winther May 30 '18 at 13:05
  • $\begingroup$ @MartinWinther I suppose that there is an error in the statement of the problem. The matrix is not diagonalizable. $\endgroup$ – José Carlos Santos May 30 '18 at 13:15
  • $\begingroup$ My bad - stupid mistake on my part (exam stress) - I forgot to write a $16\gamma$ it is fixed now and I get $A\cdot w_3 = \begin{pmatrix} 10 \\ -12 \\ 3 \\ \end{pmatrix}$ Which means that the eigenvalues are -1 and 1 and since $w_1, w_2, w_3$ is an orthogonal basis it is possible to create a orthogonal basis of the eigenvectors, which means that $L$ is orthogonal diagonalizable. $\endgroup$ – Martin Winther May 30 '18 at 13:23

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