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I am not sure about my solution to the following problem:

Which of the following sets are closed under addition?

(i) The set of all vectors in $R^2$ of the form (a, b) where b = $a^2$.

(ii) The set of all 3 × 3 matrices that have the vector $\begin{bmatrix}2&-3&3\end{bmatrix}^T$ as an eigenvector.

(iii) The set of all polynomials in $P_2$ of the form $a_0 + a_1x + a_2x^2$ where $a_0 = a_2$.


Solution:

i) $(c, c^2) + (d, d^2) = (c + d, c^2 + d^2)$

This set is not closed, since $c^2 + d^2 \neq (c + d)^2$

ii) Assume $A_1$ and $A_2$ have the eigenvector.

$A_1 \cdot v + A_2 \cdot v = \lambda_1 \cdot v + \lambda_2 \cdot v$

$(A_1 + A_2) \cdot v = (\lambda_1 + \lambda_2) \cdot v$

The set is closed.

iii) $(a_0 + a_1x + a_2x^2)$ + $(b_0 + b_1x + b_2x^2)$

$(a_0 + b_0) + (a_1 + b_1) \cdot x + (a_2 + b_2) \cdot x^2$

With assumption $a_0 = a_2$:

$(a_2 + b_2) + (a_1 + b_1) \cdot x + (a_2 + b_2) \cdot x^2$

The set is closed.

I assume the answer is ii) and iii) only. Anything wrong with my assumptions?

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  • $\begingroup$ (i) You would need $c^2+d^2$ to be of the form $(c+d)^2$ to actually show it to be closed by addition. But if you put $c=d=1$ you see that it doesn't happen. $\endgroup$ – user565560 Jun 2 '18 at 12:08
  • $\begingroup$ (iii) To wrap up the argument, explicitly mention that because $a_0=a_2$ and $b_0=b_2$, then $(a_0+b_0)=(a_2+b_2)$. $\endgroup$ – user565560 Jun 2 '18 at 12:11
  • $\begingroup$ Oh right! I edited it, so I'm guessing it's right now? Thanks for the help. $\endgroup$ – Shawn S Jun 2 '18 at 12:14
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In (i), $c^2+d^2\not = (c+d)^2$. This is the so called "freshman's binomial"...

In (ii) and (iii) you are alright...

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