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Let $A$:$\mathbb R_5(x)$ $\rightarrow$ $\mathbb R_5(x)$ be linear map so that $A(p)=(3x-1)p'-2p$. Find matrix of operator $A$ using the standard basis and then find all eigenvalues and eigenvector that corresponds to minimal eigenvalue.

This is what I have done so far:

$p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5$

$p'(x)=a_1+2a_2x+3a_3x^2+4a_4x^3+5a_5x^4$

$A(p)=(3x-1)p'-2p= (-2a_0-a_1)+(a_1-2a_2)x+(4a_2-3a_3)x^2+(7a_3-4a_4)x^3+(10a_4-5a_5)x^4+(13a_5)x^5$

Since standard basis is $B=\{1,x,x^2,x^3,x^4,x^5\}$ would matrix of operator $A$ be:

$$[A]_B= \begin{bmatrix} -2a_0-a_1 & 0 & 0 & 0& 0& 0\\ 0 & a_1-2a_2 & 0 & 0& 0& 0\\ 0 & 0 & 4a_2-3a_3 & 0& 0& 0\\ 0 & 0 & 0 & 7a_3-4a_4& 0& 0\\ 0 & 0 & 0 & 0& 10a_4-5a_5& 0\\ 0 & 0 & 0 & 0& 0&13a_5\\ \end{bmatrix} $$ ?

Assuming this is correct, I calculated eigenvalues by finding $$det([A]_B-I\lambda)=0\Rightarrow$$

$(-2a_0-a_1-\lambda)(a_1-2a_2-\lambda)(4a_2-3a_3-\lambda)(7a_3-4a_4-\lambda)(10a_4-5a_5-\lambda)(13a_5-\lambda)=0$

My eigenvalues would be:

$$\lambda_1=-2a_0-a_1$$

$$\lambda_2=a_1-a_2$$

$$\lambda_3=4a_2-3a_3$$

$$\lambda_4=7a_3-4a_4$$

$$\lambda_5=10a_4-5a_5$$

$$\lambda_6=13a_5$$

Is this correct? Should I now assume $a_i=1$, for $i=\overline{0,5}$ and say that $\lambda_1=-3$ is minimal eigenvalue and eigenvector that corresponds to $\lambda_1$ is $[1,0,0,0,0,0]^T$?

Thank you in advance.

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  • $\begingroup$ Oh, my mistake. I did not write it correctly. Thank you. $\endgroup$ – Asleen Jun 3 '16 at 22:41
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You compute the matrix wrongly: the coefficients $a_i$ should not appear.

The columns are obtained by considering the images of the polynomials in the standard basis: \begin{align} A(1)&=-2\\ A(x)&=-1+x\\ A(x^2)&=-2x+4x^2\\ A(x^3)&=-3x^2+7x^3\\ A(x^4)&=-4x^3+10x^4\\ A(x^5)&=-5x^4+13x^5 \end{align} so tha the matrix is $$ \begin{bmatrix} -2 & -1 & 0 & 0 & 0 & 0 \\ 0 & 1 & -2 & 0 & 0 & 0 \\ 0 & 0 & 4 & -3 & 0 & 0 \\ 0 & 0 & 0 & 7 & -4 & 0 \\ 0 & 0 & 0 & 0 & 10 & -5 \\ 0 & 0 & 0 & 0 & 0 & 13 \end{bmatrix} $$ You can read the eigenvalues directly, because the matrix is triangular.

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  • $\begingroup$ I just saw what have I done and I am ashamed. I know I should not answer with 'thanks', but thank you. $\endgroup$ – Asleen Jun 3 '16 at 23:03
  • $\begingroup$ @Asleen It's not shameful to make mistakes, it's shameful not to ask for help! $\endgroup$ – GFauxPas Jun 3 '16 at 23:06
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Here's how you could have set up the problem. Reinterpret polynomials as column vectors. Write

$$\begin{bmatrix} \square & \square & \square & \square & \square & \square \\ \square & \square & \square & \square & \square & \square \\ \square & \square & \square & \square & \square & \square \\ \square & \square & \square & \square & \square & \square \\ \square & \square & \square & \square & \square & \square \\ \square & \square & \square & \square & \square & \square \end{bmatrix} \begin{bmatrix}a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{bmatrix} = \begin{bmatrix} -2a_0-a_1 \\ a_1-2a_2 \\ 4a_2-3a_3 \\ 7a_3-4a_4 \\ 10a_4-5a_5 \\ 13a_5 \end{bmatrix}. $$

From there you figure out what goes in the $\square$s.

(egreg gave you the values if you want to check you can do this.)

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No. The matrix of $A$ is incorrect. You can test your result by simply looking at the case $p(x)=1$: $$ A(1)=-2. $$

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Sorry, but you have a basic and deep mistake. The operator $A$ acts on the polynomial $p$; every $p$! In your calculations, the operator depends on the coefficients of $p$, which are different for different $p$.

The entries of the matrix $[A]_B$ are such that if $B=\{b_0,b_1,b_2,b_3,b_4,b_5\}$, then $$ Ab_k=A_{k0}b_0+A_{k1}b_1+A_{k2}b_2+A_{k3}b_3+A_{k4}b_4+A_{k5}b_5. $$

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  • $\begingroup$ I was afraid I made a terrible mistake. Sadly, it seems I do not understand the problem. $\endgroup$ – Asleen Jun 3 '16 at 22:57

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