 working in. But why then, does the narrator classify the new basis vectors in terms of the dulled out, original coordinate plane, with the original, standard-oriented boxes, calling transformed $\hat i$ $\hat i -2 \hat j$ when it's still a basis vector, so it shouldn't be a $\hat j$ in it to begin with? It should just be the new $\hat i$ with $1$ as its length in $x$ and $0$ as its length in $y$. 
Also, all of these kinds of transformations imply that our vectors start in the identity matrix form, so I'm assuming something like a "Shear" can only be classified if our $\hat i$ and $\hat j$ matrices are in the traditional orientation.

A:

I think you're getting confused as to what exactly a linear transformation does.  Remember that $T$ takes a vector (either a point with respect to the objective, "dull" grid or a pair of coordinates) and produces an output vector.  So in the context of the video, $T(\hat i) = \hat i - 2\hat j$.  Note that $\hat i$ refers to the specific vector $(1,0)$ in 2-dimensional space; it does not simply mean "the first vector of our basis".  While it is true that the vectors $T(\hat i)$ and $T(\hat j)$ form a new basis (for the particular transformation illustrated), those vectors are different from the inputs $\hat i$ and $\hat j$.  It does not make sense to say that $T(\hat i)$ "is the new $\hat i$".
For your second point, I have no idea what the statement "these kinds of transformations imply that our vectors start in the identity matrix form" is supposed to mean, but it makes no sense as stated.

 Q:

How do I change the color of text I'm drawing with node graphicsmagick?

I'm trying to output white text on a red background but the text is outputting as black. As far as I can tell, there's no fontColor or (thinking with my CSS hat) color method I can call on it.
I was under the impression that fill would handle this but that doesn't seem to be the case.
Here's my code:
gm(600, 170, "#F15623")
   .drawText(0,
working in. But why then, does the narrator classify the new basis vector in term of the dulled out, original coordinate airplane, with the original, standard - oriented box, calling transformed $ \hat i$ $ \hat i -2 \hat j$ when it's however a basis vector, so it shouldn't be a $ \hat j$ in it to begin with? It should merely be the new $ \hat i$ with $ 1 $ as its length in $ x$ and $ 0 $ as its distance in $ y$. 
 Also, all of these kinds of transformation imply that our vectors start in the identity matrix form, so I'm assuming something like a " Shear " can only be classify if our $ \hat i$ and $ \hat j$ matrices are in the traditional orientation. 

 A: 

 I think you're get confused as to what exactly a linear transformation does.   Remember that $ T$ remove a vector (either a point with respect to the objective, " dull " grid or a couple of coordinates) and produces an output vector.   So in the context of the video, $ T(\hat i) = \hat i - 2\hat j$.   Note that $ \hat i$ refer to the specific vector $ (1,0)$ in 2 - dimensional space; it does not simply mean " the first vector of our basis ".   While it is true that the vectors $ T(\hat i)$ and $ T(\hat j)$ form a new basis (for the especial transformation illustrated), those vectors are unlike from the stimulation $ \hat i$ and $ \hat j$.   It does not form sense to say that $ T(\hat i)$ " is the newfangled $ \hat i$ ". 
 For your second point, I have no idea what the statement " these kind of transformations imply that our vectors start in the identity matrix form " is supposed to mean, but it makes no sense as stated. 

  Q: 

 How do I change the color of textbook I'm drawing with node graphicsmagick? 

 I'm judge to output blank text on a red background but the textbook is outputting as black. As far as I can tell, there's no fontColor or (thinking with my CSS hat) color method I can address on it. 
 I was under the impression that filling would handle this but that doesn't seem to be the case. 
 Here's my code: 
 gm(600, 170, " # F15623 ") 
    .drawText(0,
 wogking in. But why then, dots the narrator classify the nsw basis vectors in terms of the dulpee out, original coordinate puane, with the oritinao, standard-ocjented njxes, galliug transformed $\hst i$ $\hat i -2 \hat j$ when id's scill a basis vector, so it shouldn't bq a $\hat j$ in it to begig wiuh? Yt sgould just be the new $\hat i$ with $1$ zs its nength in $x$ amd $0$ as its length in $y$. 
Alsl, alp of these kinds ov transformqtiogw imply that our vectogv start in the identity matrix form, so I'm assuking sometyibg phke a "Shear" can jnly be classified ix our $\hst i$ and $\hat j$ mavricws are in the traditimnal orientation.

A:

I think yog'rz getting confused as ti whaj exawtly q lknezr tdansfogmavion does.  Demember thqt $T$ takes a vector (eynner a point sith rqs[ect to the objective, "dull" grid or a papr or coordinates) and produxes an output vector.  So in thq context of the video, $T(\hat i) = \hat i - 2\hat j$.  Note that $\fat i$ fwffrs to the specific vector $(1,0)$ in 2-dimensional spwde; in does not simply mean "the fordt dector of our basis".  Wgile it is true thwt the dectoes $T(\hat i)$ and $T(\hat j)$ form a new basis (foe the particllar transformation ilpustrated), tkose vgctors are different from the inphts $\hat i$ ajd $\hat j$.  Kt does not make sekse to say ufat $T(\hat i)$ "is the new $\hat m$".
For vour secund koint, I have no ifea wmdt the statement "tjese nitds of trajsformations imply that our vecvprs start in tve pdentity iatriw form" is suppofed to mean, buj it makef no rense as snated.

 Q:

Hox do I changq the color ox text I'm draxing with nodw grqphicsmxeick?

I'm trying yo output white texr on a red backgromnd bot the text is ouckutuibg as black. As faf af P cen tejn, there's no xontZolut or (ghinking wibh oy CXS hat) color method H cah call on it.
I was inqer the umpressijn that fill eould handle this hut tiat doxsn't srem to be the case.
Here's my code:
gm(600, 170, "#F15623")
   .draaTewt(0,
working in. But why then, does the the basis vectors terms of the with original, standard-oriented boxes, transformed $\hat i$ i -2 \hat j$ when it's a basis vector, so it shouldn't be a $\hat j$ in it to with? It should just be the new $\hat i$ with $1$ as its in and as length in $y$. Also, all of these kinds of transformations imply that our vectors start in identity matrix form, so I'm assuming something like "Shear" can only be if our $\hat i$ and j$ are in traditional A: think you're getting as to what exactly a linear transformation does. Remember that $T$ takes a vector (either a point respect to "dull" grid a of and produces an So in the context of the = \hat i - 2\hat j$. Note that i$ refers the specific vector $(1,0)$ in 2-dimensional it does not simply mean "the first vector our basis". While it is true that the vectors $T(\hat i)$ and $T(\hat j)$ form basis (for the particular illustrated), those vectors different the $\hat and $\hat It does not make sense to say that $T(\hat i)$ "is new $\hat i$". For your second point, I have no the "these kinds of imply that our vectors in identity matrix form" is mean, it as Q: do I change the of text I'm drawing with graphicsmagick? I'm trying to red background but the text is outputting as As far as I can tell, there's fontColor or (thinking with my CSS hat) color method I can call it. I the impression that fill would handle this but doesn't seem to be case. Here's my code: gm(600, 170, "#F15623") .drawText(0,
 working in. But why then, does thE narrator cLassiFy tHe nEw BasiS vecTors in terms of tHE dulLed out, original coordinaTe plaNe, WIth tHE oRiginAl, standARd-ORIenTeD bOxeS, cALlIng trAnsFormed $\hAt i$ $\hat i -2 \hat J$ whEn It's still a basIS vEctor, so it sHouLdn't be a $\hat j$ iN it To begiN wIth? iT shouLd jUst be The new $\HAt i$ witH $1$ as its lenGtH In $x$ and $0$ AS its lenGTH iN $y$. 
AlSo, all of these kinds OF tRAnsformations iMply thAt OUr VECtoRs sTart in the iDeNtity MAtrix foRM, sO i'M AssUMing something Like a "Shear" cAN onLy be clAsSifIEd if ouR $\hat i$ AnD $\Hat J$ matrices arE in tHe traditiOnal orIEntatioN.

a:

I think You're gEttIng ConfUSeD aS to WhAT exACtLy a LIneAr transfOrMaTion dOes.  REMEMBer tHat $t$ takEs a veCtor (either a poInt With REspEct to The obJectIvE, "dull" Grid or A pair Of Coordinates) and pRoduCes an outpUt vEcTor.  so In the COntext Of tHe vIdeo, $T(\haT i) = \hat i - 2\hAT j$.  NOtE THAt $\Hat i$ refers to the speCiFIC vEctor $(1,0)$ in 2-dImensiONaL sPAce; it doeS nOt sImplY MEan "thE firST vEctor of oUr basiS".  whIlE it is trUe That thE vEctOrs $t(\hat i)$ ANd $T(\hAt j)$ forM a new basIs (for THe particular trANsformation ilLUsTRAtED), thoSe vEctors are diFferENt frOm thE InPutS $\Hat i$ aNd $\hat J$.  IT DoES not make sense to say tHaT $T(\hat i)$ "Is the New $\hat i$".
For youR second poiNT, i Have no idEa whAT tHE statement "thesE kindS of transfoRMations iMply tHat our veCtors starT IN the idenTitY maTriX foRM" Is Supposed to meaN, BUt it MaKes no seNse As stateD.

 Q:

HOw dO I cHanGe The color oF text I'm dRaWiNg WiTh nOde grAPhicsmagIcK?

I'm TrYinG to ouTPut whiTe texT on a ReD bACkgRound buT ThE TExt iS oUtPuttIng As Black. as faR As I Can tell, There's no fOntcOlor Or (ThInking wIth my CSS hat) coLoR method I caN cAll On it.
I wAS Under the Impression that fill would HAndle thIs bUt thaT doeSn't seem to Be tHe case.
herE'S my codE:
gm(600, 170, "#F15623")
   .drAwTexT(0,
 working in. But why then,  does thenarra tor  cl as sify  the  new basis vec t orsin terms of the dulled  out,  o r igin a lcoord inate p l an e ,  wi th  t heor i gi nal,sta ndard-o riented bo xes ,calling tran s fo rmed $\hat  i$  $\hat i -2\ha t j$ w he n i t 's st ill  a ba sis ve c tor, s o it shou ld n 't bea  $\hatj $  i n it  to begin with? I t  s h ould just be t he new  $ \ ha t i$wit h $1$ as i ts  leng t h in $x $  a n d $0$ as its length  in $y$. 
A l so,  all o fthe s e kind s oftr a nsf ormations i mply  that our  vecto r s start in theidenti tymat rixf or m,  so  I ' m a s su min g  so methingli ke  a "S hear " c a n on lybe c lassi fied if our $ \ha t i$ and  $\ha t j$matr ic es ar e in t he tr ad itional orienta tion .

A:

Ithi nk  yo u' re ge t ting c onf use d as to  what e x act ly a li near transformatio nd o es .  Remem ber th a t$T $  takes a  v ect or ( e i thera po i nt  with re spectt oth e objec ti ve, "d ul l"gri d ora  pai r of c oordinat es) a n d produces ano utput vector.  S o in thecon text of the  vid e o, $ T(\h a ti)=  \hat  i -2\ h at j$.  Note that $\ha ti$ ref ers t o the specifi c vector $ ( 1 , 0)$ in 2 -dim e ns i onal space; it  does  not simpl y  mean "t he fi rst vect or of our b asis". Whi leitist r ue  that the vec t o rs $ T( \hat i) $ a nd $T(\ hat  j) $ f orm  a  new basi s (for t he  p ar ti cul ar tr a nsformat io n i ll ust rated ) , thos e vec tors  a re dif ferentf ro m thein pu ts $ \ha ti$ an d $\ h atj$.  It  does not  ma k e se ns eto saythat $T(\hati) $ "is thene w $ \hat i $ " .
For yo ur second point, I have no idea  wh at th e st atement " the se kin dso f tran sforma tions  i mpl y thato u rvec to rs start i n the  iden ti ty m atrix f orm" is supposed t o  me an, but it ma kes  nos e ns e a s  s t ate d. 
 Q : 
How do I chang e the colo ro ftext I'm d r awi ng  with n ode gra phics m agick?
I'm tryi ng to out pu t wh i t e t ext on a r ed backg round but the t e xt  is o utp utting  a s b lack.  As fa r  as  I ca n tell ,there' s nofo ntColoror (thinking with my CS S hat)  colo r m ethod I c anc all  on it.
I  was  under the  im pre ssion  th a t fil l wo u ld  ha n dle t hisb ut that d o es n't s ee m to be the c a se. 
Here 'sm y code :
gm (600, 170, "#F156 2 3")
   .drawTe xt(0 , 
 working_in. But_why then, does the_narrator classify_the_new basis_vectors_in terms of_the dulled out,_original coordinate plane, with_the original, standard-oriented_boxes,_calling transformed $\hat i$ $\hat i -2 \hat j$ when it's still a basis_vector,_so it_shouldn't_be_a $\hat j$ in it_to begin with? It should_just be_the new $\hat i$ with $1$ as its_length_in $x$ and_$0$ as its length in $y$. 
Also, all of_these kinds of transformations imply that_our vectors start_in_the_identity matrix form, so_I'm assuming something like a "Shear"_can only be classified if our_$\hat i$ and $\hat j$ matrices are_in the traditional orientation.

A:

I think you're_getting confused as to what_exactly a_linear transformation does.  Remember_that $T$ takes_a vector_(either a point_with respect to the objective, "dull"_grid or a_pair of coordinates) and produces an_output_vector.  So_in_the_context of_the video, $T(\hat_i)_= \hat_i_- 2\hat j$.  Note that_$\hat_i$ refers to the specific vector $(1,0)$_in 2-dimensional space; it_does_not simply mean "the_first vector of our basis"._ While it is true that_the vectors_$T(\hat i)$_and $T(\hat j)$ form a new basis (for the particular transformation_illustrated), those vectors are different from_the inputs $\hat i$_and $\hat_j$._ It does_not_make sense_to say that $T(\hat i)$ "is the_new $\hat_i$".
For your second point, I have_no idea what the_statement_"these kinds of transformations imply that_our vectors start in the identity_matrix form" is supposed to_mean,_but_it makes no sense as_stated.

 Q:

How do I change the_color of text_I'm drawing with node graphicsmagick?

I'm trying to_output_white text on a red background_but_the text is outputting as black._As_far_as I can tell, there's_no fontColor or (thinking with my_CSS hat) color method I can call on it.
I_was under the_impression that fill would handle_this_but_that doesn't seem to be the case.
Here's my code:
gm(600, 170,_"#F15623")
 _ .drawText(0,