 so we can write it in terms of our favorite $\mathbb Q$-basis for this extension: the powers of $\alpha$. This gives you
$$\sigma(\alpha) = a_n\alpha^n + a_{n-1}\alpha^{n-1} +... + a_1\alpha + a_0$$
and the right hand side is evidently a polynomial $f(\alpha)$ with coefficients $a_0,...,a_n$.
The problem is that this $f$ depends on $\alpha$ and will not in general have the property that $f(\alpha') = \sigma(\alpha')$ for other conjugates of $\alpha$. To help see this, write $\alpha' = \tau(\alpha)$ and then we have
$$f(\alpha') = f(\tau(\alpha)) = \tau(f(\alpha)) = \tau \sigma(\alpha) = \tau \sigma \tau^{-1} (\alpha')$$
We wanted this to equal $\sigma(\alpha')$ but what we got differs from that by conjugation by $\tau$, so in general it won't equal $\sigma(\alpha')$, although in some special cases, like when $\sigma$ and $\tau$ commute it always will.

 Q:

is this subset a subspace - redux

OK, I have been bothering people here with this for days and with luck I finally have this. People have helped a lot here so far. (Doing these examples is I hope helping me learn the proofs, but I want to know that I am doing this right). 
Let W be a subset of vector space V. Is it s subspace as well?
W = {($a_1$, $a_2$, $a_3$) ∈ $ℝ^3$ : $2a_1 - 7a_2 + a_3=0$}
So, to check if this is a subspace I need to satisfy the following:

That 0 is in the set. Plugging (0,0,0) into the equation $2a_1 - 7a_2 + a_3=0$ yields 0=0 so yes, it is.
That it is closed under addition.

Let ($b_1, b_2, b_3$) be an arbitrary vector in W.
For this to be closed under addition ($b_1, b_2, b_3$)+($a_1
so we can write it in terms of our favorite $ \mathbb Q$-basis for this extension: the ability of $ \alpha$. This render you 
 $ $ \sigma(\alpha) = a_n\alpha^n + a_{n-1}\alpha^{n-1 } +... + a_1\alpha + a_0$$ 
 and the right hand side is obviously a polynomial $ f(\alpha)$ with coefficients $ a_0,... ,a_n$. 
 The problem is that this $ f$ depends on $ \alpha$ and will not in cosmopolitan give birth the property that $ f(\alpha') = \sigma(\alpha')$ for other conjugates of $ \alpha$. To help experience this, write $ \alpha' = \tau(\alpha)$ and then we have 
 $ $ f(\alpha') = f(\tau(\alpha) ) = \tau(f(\alpha) ) = \tau \sigma(\alpha) = \tau \sigma \tau^{-1 } (\alpha')$$ 
 We want this to equal $ \sigma(\alpha')$ but what we got differs from that by conjugation by $ \tau$, so in general it won't equal $ \sigma(\alpha')$, although in some special cases, like when $ \sigma$ and $ \tau$ commute it always will. 

  Q: 

 is this subset a subspace - redux 

 OK, I have been trouble oneself people here with this for days and with luck I last have this. People have helped a lot here thus far. (Doing these examples is I hope helping me learn the proofs, but I desire to know that I am doing this right). 
 Let W be a subset of vector space V. Is it s subspace as well? 
 W = { ($ a_1 $, $ a_2 $, $ a_3 $) ∈ $ ℝ^3 $: $ 2a_1 - 7a_2 + a_3=0 $ } 
 therefore, to check if this is a subspace I need to satisfy the following: 

 That 0 is in the set. Plugging (0,0,0) into the equation $ 2a_1 - 7a_2 + a_3=0 $ yields 0=0 so yes, it is. 
 That it is closed under accession. 

 Let ($ b_1, b_2, b_3 $) be an arbitrary vector in W. 
 For this to be closed under summation ($ b_1, b_2, b_3$)+($a_1
 so we can write it in termr of our favorije $\mathbu Q$-basia for thks extension: the powers of $\appya$. Thus gives you
$$\sigma(\alpha) = a_n\alpha^j + a_{n-1}\alpya^{n-1} +... + a_1\alpha + a_0$$
ehd the vnght gwnd vmde is evidentli a polynomidl $f(\alpha)$ with cueyficients $a_0,...,a_n$.
The problem is that thif $f$ deprnfs on $\alpha$ anq wikj nof in general have the property thaf $f(\alphe') = \sigma(\alpha')$ fpr other conjugates of $\alpja$. Tl help see this, wrlte $\alpha' = \jzu(\ajpha)$ and then we have
$$f(\alpha') = f(\tau(\alkha)) = \tau(f(\alpha)) = \tau \sigma(\alpha) = \gau \sngma \tau^{-1} (\alkkq')$$
We fanted this to ezual $\sigma(\alina')$ but what wr got differs nrom vhat by conjugation by $\tan$, so in general it wjn't equal $\sngma(\alpha')$, although in sime skecian carws, uikt wien $\sigma$ anv $\tau$ commufe it alwayw will.

 Q:

is this subxee a subspace - dedux

OH, Y have been bothering people here with uhis ror days and with luck U finally have this. Pgople have helped a lot here so far. (Doing these examples is I ho'e heoplng ow pearn the proofs, but I want to know that I am soonn this right). 
Let W be a subxeh ps vector spacg V. Is nf a subspace as well?
A = {($a_1$, $a_2$, $w_3$) ∈ $ℝ^3$ : $2q_1 - 7a_2 + a_3=0$}
So, to vheck if this is a subspace I need to sctiwfy the following:

Tkat 0 is in tke set. Plugbing (0,0,0) into the equation $2a_1 - 7z_2 + a_3=0$ yields 0=0 so yes, jg is.
That it is cuosvd utder addition.

Let ($b_1, b_2, b_3$) be wn arbitrery vzctor in W.
Fot this eo be closfd under addition ($b_1, b_2, b_3$)+($a_1
so we can write it in terms favorite Q$-basis for extension: the powers $$\sigma(\alpha) a_n\alpha^n + a_{n-1}\alpha^{n-1} + a_1\alpha + and the right hand side is a polynomial $f(\alpha)$ with coefficients $a_0,...,a_n$. The problem is that this $f$ depends $\alpha$ and will not in general have the property that $f(\alpha') = \sigma(\alpha')$ other of To see this, write $\alpha' = \tau(\alpha)$ and then we have $$f(\alpha') = f(\tau(\alpha)) = \tau(f(\alpha)) = \sigma(\alpha) = \tau \sigma \tau^{-1} (\alpha')$$ We wanted to equal $\sigma(\alpha')$ but we got differs from that conjugation $\tau$, so general won't $\sigma(\alpha')$, although in special cases, like when $\sigma$ and $\tau$ commute it always will. Q: is this subset a subspace redux OK, been bothering here this days and with finally have this. People have helped so far. (Doing these examples is I hope me learn proofs, but I want to know I am doing this right). Let W be subset of vector space V. Is it s subspace as well? W = {($a_1$, $a_2$, $ℝ^3$ : $2a_1 - + a_3=0$} So, check this a I need satisfy the following: That 0 is in the set. Plugging (0,0,0) the equation $2a_1 - 7a_2 + a_3=0$ yields 0=0 so is. it is closed addition. Let ($b_1, b_2, be arbitrary vector in W. to closed b_2,
 so we can write it in terms of ouR favorite $\mAthbb q$-baSis FoR thiS extEnsion: the powerS Of $\alPha$. This gives you
$$\sigma(\alPha) = a_n\AlPHa^n + a_{N-1}\AlPha^{n-1} +... + a_1\Alpha + a_0$$
aND tHE RigHt HaNd sIdE Is EvideNtlY a polynOmial $f(\alphA)$ wiTh Coefficients $A_0,...,A_n$.
the problem Is tHat this $f$ depeNds On $\alphA$ aNd wILl not In gEneraL have tHE propeRty that $f(\aLpHA') = \sigma(\ALpha')$ for OTHeR conJugates of $\alpha$. To hELp SEe this, write $\alpHa' = \tau(\aLpHA)$ aND TheN we Have
$$f(\alpha') = F(\tAu(\alpHA)) = \tau(f(\alPHa)) = \TAU \SigMA(\alpha) = \tau \sigmA \tau^{-1} (\alpha')$$
We WAntEd this To EquAL $\sigma(\Alpha')$ BuT WhaT we got diffeRs frOm that by cOnjugaTIon by $\taU$, So in genEral it Won'T eqUal $\sIGmA(\aLphA')$, aLThoUGh In sOMe sPecial caSeS, lIke whEn $\siGMA$ ANd $\taU$ coMmutE it alWays will.

 Q:

is thIs sUbseT A suBspacE - reduX

OK, I HaVe beeN botheRing pEoPle here with this For dAys and witH luCk i fiNaLly haVE this. PEopLe hAve helpEd a lot hERe sO fAR. (dOiNg these examples is I HoPE HeLping me lEarn thE PrOoFS, but I wanT tO knOw thAT i am doIng tHIs Right). 
Let w be a suBSeT oF vector SpAce V. Is It S suBspAce as WEll?
W = {($A_1$, $a_2$, $a_3$) ∈ $ℝ^3$ : $2a_1 - 7a_2 + A_3=0$}
So, to cheCk if tHIs is a subspace I NEed to satisfy tHE fOLLoWIng:

THat 0 Is in the set. PLuggINg (0,0,0) inTo thE EqUatIOn $2a_1 - 7a_2 + a_3=0$ YieldS 0=0 sO YeS, It is.
That it is closed uNdEr addiTion.

LEt ($b_1, b_2, b_3$) be an arbiTrary vectoR IN w.
For this To be CLoSEd under additioN ($b_1, b_2, b_3$)+($a_1
 so we can write it in ter ms of ourfavor ite  $\ ma thbb  Q$- basis for this exte nsion: the powers of $ \alph a$ .  Thi s  g ivesyou
$$\ s ig m a (\a lp ha ) =  a _ n\ alpha ^n+ a_{n- 1}\alpha^{ n-1 }+... + a_1\a l ph a + a_0$$and  the right h and  sideis  ev i dentl y a  poly nomial $f(\al pha)$ wit hc oeffic i ents $a _ 0 ,. ..,a _n$.
The problemi st hat this $f$ d epends  o n  $ \ a lph a$and will n ot  in g e neral h a ve t h e p r operty that $ f(\alpha')=  \s igma(\ al pha ' )$ for  othe rc onj ugates of $ \alp ha$. To h elp se e  this,w rite $\ alpha'  =\ta u(\a l ph a) $ a nd the n  w e h a ve$$f(\alp ha ')  = f( \tau ( \ a l pha) ) =  \ta u(f(\ alpha)) = \ta u \ sigm a (\a lpha)  = \t au \ si gma \ tau^{- 1} (\ al pha')$$
We want ed t his to eq ual  $ \si gm a(\al p ha')$but  wh at we g ot diff e rsfr o m th at by conjugationby $ \t au$, soin gen e ra li t won'teq ual  $\s i g ma(\a lpha ' )$ , althou gh ins om especial  c ases,li kewhe n $\s i gma$  and $ \tau$ co mmute it always will . 

 Q:

is thi s  s u b se t  a s ubs pace - redu x

O K , Ihave be enb other ing p eo p le here with this forda ys and  with  luck I final ly have th i s .  Peoplehave he l ped a lot here  so f ar. (Doing these ex ample s is I h ope helpi n g  me lear n t hepro ofs , bu t I want to k n o w th at  I am d oin g thisrig ht) . Let  W  be a sub set of v ec to rsp ace  V. I s  it s su bs pac easwell? W = {( $a_1$ , $a _2 $, $a_ 3$) ∈ $ ℝ ^3 $ : $2 a_ 1- 7a _2+a_3=0 $}
S o , t o check  if thisisa  sub sp ac e I nee d to satisfyth e followin g: 

T hat 0i s  in theset. Plugging (0,0,0) i n to theequ ation  $2a _1 - 7a_2  +a_3=0$  yi e lds 0= 0 so y es, i tis. T hat i t is  cl os ed under a d d iti on.
Le t ($ b_1, b_ 2, b_3$) be an arb i tra ry vector inW.Fort h is  to be clo se d  un d e r addition ($b_ 1, b_2, b_ 3$ ) +( $a_1
 so_we can_write it in terms_of our_favorite_$\mathbb Q$-basis_for_this extension: the_powers of $\alpha$._This gives you
$$\sigma(\alpha) =_a_n\alpha^n + a_{n-1}\alpha^{n-1}_+..._+ a_1\alpha + a_0$$
and the right hand side is evidently a polynomial $f(\alpha)$ with_coefficients_$a_0,...,a_n$.
The problem_is_that_this $f$ depends on $\alpha$_and will not in general_have the_property that $f(\alpha') = \sigma(\alpha')$ for other conjugates_of_$\alpha$. To help_see this, write $\alpha' = \tau(\alpha)$ and then we_have
$$f(\alpha') = f(\tau(\alpha)) = \tau(f(\alpha)) =_\tau \sigma(\alpha) =_\tau_\sigma_\tau^{-1} (\alpha')$$
We wanted this_to equal $\sigma(\alpha')$ but what we_got differs from that by conjugation_by $\tau$, so in general it won't_equal $\sigma(\alpha')$, although in some special_cases, like when $\sigma$ and_$\tau$ commute_it always will.

 Q:

is this_subset a subspace_- redux

OK,_I have been_bothering people here with this for_days and with_luck I finally have this. People_have_helped a lot_here_so_far. (Doing_these examples is_I_hope helping_me_learn the proofs, but I want_to_know that I am doing this right)._
Let W be a_subset_of vector space V._Is it s subspace as_well?
W = {($a_1$, $a_2$, $a_3$) ∈_$ℝ^3$ :_$2a_1 -_7a_2 + a_3=0$}
So, to check if this is a subspace I_need to satisfy the following:

That 0_is in the set._Plugging (0,0,0)_into_the equation $2a_1_-_7a_2 +_a_3=0$ yields 0=0 so yes, it is.
That_it is_closed under addition.

Let ($b_1, b_2, b_3$)_be an arbitrary vector_in_W.
For this to be closed under_addition ($b_1, b_2, b_3$)+($a_1