(C, [2 1])

Computes a simple transpose by swapping rows and columns. The [2 1] argument means that the 2st and 1st dimension of C are mapped to the 1st and 2nd dimension in the result. Each 'new' dimension is specified in order. So [3 2 1] would take the 3rd, 2nd and 1st dimensions to be the new 1st, 2nd and 3rd dimensions.
permute(C, [3 2 1])

ans =

ans(:,:,1) =

   1   2   3

ans(:,:,2) =

   4   5   6

Elements of C with row = 1 are found in where the 3rd dimension = 1 in the result. Similarly, elements of C with row = 2 are found where the 3rd dimension = 2 in the result.
Elements of C with column = 1 are still found where column = 1 in the result (and so on) as the column dimension was mapped to itself.
The rows of the result is the interesting dimension, it is singleton (i.e. there is only one row) as a result of C having no 3rd dimension.
Addressing the first part of your problem, the correct output for C can be obtained by
A = [1 2]'*[1 2];
C = bsxfun(@times, permute(A, [4,3,1,2]), A);

I would need more information on what you want the final behaviour to be ('resetting the lower triangle') as it is unclear to me what you desire.
A function that might be useful to you is the triu function which extracts upper triangular components of a matrix.

 Q:

Correct symbol for the Laplace operator ($\Delta$)

What is the correct symbol to use for the Laplace operator? It looks almost like the big Delta $\Delta$, but it should look different so that it is not confused with the Delta.
Currently, I use $\vec \nabla^2$, which is unambiguous, but not pretty.

A:

The operator is defined, in the international standard ISO 80000-1, as identified with the Unicode character U+2206 INCREMENT (mistakenly called DELTA in the standard), which has “Laplace operator” as one of its alias names. Thus, it is regarded as distinct from the Greek capital
( C, [ 2 1 ]) 

 Computes a simple transpose by swapping row and column. The [ 2 1 ] argument mean that the 2st and first dimension of C are mapped to the 1st and second proportion in the result. Each' fresh' dimension is specified in club. So [ 3 2 1 ] would take the third, 2nd and 1st dimensions to be the new 1st, 2nd and 3rd dimension. 
 permute(C, [ 3 2 1 ]) 

 ans = 

 ans(:,:,1) = 

    1    2    3 

 ans(:,:,2) = 

    4    5    6 

 Elements of C with row = 1 are found in where the third dimension = 1 in the result. Similarly, element of C with quarrel = 2 are found where the 3rd dimension = 2 in the result. 
 Elements of C with column = 1 are however found where column = 1 in the result (and so on) as the column dimension was mapped to itself. 
 The rows of the result is the interesting dimension, it is singleton (i.e. there is only one row) as a result of C having no 3rd dimension. 
 address the first part of your trouble, the right output for C can be obtained by 
 A = [ 1 2]'*[1 2 ]; 
 C = bsxfun(@times, permute(A, [ 4,3,1,2 ]), A); 

 I would necessitate more information on what you want the final behaviour to be (' readjust the lower triangle') as it is unclear to me what you desire. 
 A function that might be useful to you is the triu routine which extracts upper triangular components of a matrix. 

  Q: 

 Correct symbol for the Laplace operator ($ \Delta$) 

 What is the right symbol to use for the Laplace hustler? It looks almost like the big Delta $ \Delta$, but it should look different so that it is not confused with the Delta. 
 presently, I use $ \vec \nabla^2 $, which is unambiguous, but not pretty. 

 A: 

 The operator is defined, in the international standard ISO 80000 - 1, as identified with the Unicode character U+2206 INCREMENT (mistakenly called DELTA in the standard), which have “ Laplace hustler ” as one of its alias names. Thus, it is regarded as distinct from the Greek capital
(C, [2 1])

Fomputes a simple transpuse by swapping rows aid colujns. The [2 1] argument means that the 2st abd 1st dimension of C are maoped to tje 1st ane 2nd eimension mh the rcfult. Cach 'uex' dimension is xpecified hn order. So [3 2 1] wuupd take the 3rd, 2nd and 1st dimensions to be yhf new 1st, 2nd anq 3rd qimehsions.
permute(C, [3 2 1])

ans =

ans(:,:,1) =

   1   2   3

zns(:,:,2) =

   4   5   6

Elementx of C with row = 1 are founf in where the 3rd dimejsion = 1 in jge wwsult. Similafly, elemenus of C with row = 2 are found where the 3rd dkmensnon = 2 in thg dedolt.
Elements if C rith column = 1 are sthll foumd where columk = 1 ii thw result (and so on) as the column dimensyon was mdp'ed to itself.
The rows od the resglt kw tfe jnverssting dijension, it is singletin (i.e. there is only ogv row) as a reault os S having no 3rd dimension.
Addressing the xirat part of your problem, the correct output flr C can fe obtained by
A = [1 2]'*[1 2];
C = bsxfun(@times, permute(A, [4,3,1,2]), A);

I wogld nxea mirc inwirlation on what you want the final behaviour tj bt ('rvsetting the lowev triangle') as it ix knvjear to me whxt you deaire.
A function thah might be uweful to rou os the triu function which wxtracts uppvr teiangular componencs of a matrnx.

 Q:

Cotrect xymbol for the Laplace uperztor ($\Delta$)

Wjat is ths correct symbol go lse xor the Laplace operator? Ie looks aomosc like tfe bog Delea $\Delta$, bkt it should look differfnt sl dhat it is not confused with the Delta.
Curcxntly, I use $\vgc \tabna^2$, which is ukambiguous, but got pretty.

A:

The operatjr is defined, ih the iiternational standard ISO 80000-1, as identifixd with tre Ubicoee charxzter U+2206 INCREMEMT (mistakvnjy cqlled DELTA in the stxhdard), which has “Laklqce operator” ax ove jf ivs alyds names. Thuv, it is tegaraed as distlncg frpm the Greek capital
(C, [2 1]) Computes a simple transpose rows columns. The 1] argument means dimension C are mapped the 1st and dimension in the result. Each 'new' is specified in order. So [3 2 1] would take the 3rd, 2nd 1st dimensions to be the new 1st, 2nd and 3rd dimensions. permute(C, [3 1]) = = 2 3 ans(:,:,2) = 4 5 6 Elements of C with row = 1 are found where the 3rd dimension = 1 in the Similarly, elements of C row = 2 are found the dimension = in result. of C with = 1 are still found where column = 1 in the result (and so on) as the dimension was itself. The of result the interesting dimension, singleton (i.e. there is only one result of C having no 3rd dimension. Addressing first part your problem, the correct output for can be obtained by A = [1 2]'*[1 C = bsxfun(@times, permute(A, [4,3,1,2]), A); I would need more information on what you want behaviour to be ('resetting lower triangle') as is to what desire. A that might be useful to you is the triu function which upper triangular components of a matrix. Q: Correct symbol for operator What is the symbol to use for Laplace It looks almost like Delta but different that is not confused with Delta. Currently, I use $\vec which is unambiguous, but is defined, in the international standard ISO 80000-1, identified with the Unicode character U+2206 INCREMENT called DELTA in the standard), which has “Laplace operator” as one of alias names. is regarded as distinct from the Greek capital
(C, [2 1])

Computes a simple transpose By swapping Rows aNd cOluMnS. The [2 1] ArguMent means that tHE 2st aNd 1st dimension of C are mapPed to ThE 1St anD 2Nd DimenSion in tHE rESUlt. eaCh 'New' DiMEnSion iS spEcified In order. So [3 2 1] wOulD tAke the 3rd, 2nd anD 1St Dimensions To bE the new 1st, 2nd aNd 3rD dimenSiOns.
PErmutE(C, [3 2 1])

aNs =

ans(:,:,1) =

   1   2   3

Ans(:,:,2) =

   4   5   6

EleMEnts of c with row = 1 aRe FOund in WHere the 3RD DiMensIon = 1 in the result. SimILaRLy, elements of C wIth row = 2 ArE FoUND whEre The 3rd dimenSiOn = 2 in tHE result.
eLeMENTs oF c with column = 1 arE still found WHerE columN = 1 iN thE Result (And so On) AS thE column dimeNsioN was mappeD to itsELf.
The roWS of the rEsult iS thE inTereSTiNg DimEnSIon, IT iS siNGleTon (i.e. theRe Is Only oNe roW) AS A ResuLt oF C haVing nO 3rd dimension.
ADdrEssiNG thE firsT part Of yoUr ProblEm, the cOrrecT oUtput for C can be oBtaiNed by
A = [1 2]'*[1 2];
C = bsXfuN(@tImeS, pErmutE(a, [4,3,1,2]), A);

I wouLd nEed More infOrmatioN On wHaT YOU wAnt the final behavioUr TO Be ('ResettinG the loWEr TrIAngle') as iT iS unCleaR TO me whAt yoU DeSire.
A funCtion tHAt MiGht be usEfUl to yoU iS thE trIu funCTion Which eXtracts uPper tRIangular componENts of a matrix.

 Q:

cOrRECt SYmboL foR the Laplace OperATor ($\DElta$)

wHaT is THe corRect sYmBOl TO use for the Laplace opErAtor? It Looks Almost like the Big Delta $\DeLTA$, But it shoUld lOOk DIfferent so that It is nOt confused WIth the DeLta.
CuRrently, I Use $\vec \nabLA^2$, Which is uNamBigUouS, buT NOt Pretty.

A:

The opeRATor iS dEfined, iN thE internAtiOnaL stAndArD ISO 80000-1, as ideNtified wItH tHe unIcoDe chaRActer U+2206 INcReMEnT (MisTakenLY calleD DELTa in tHe StANdaRd), which HAs “lAPlacE oPeRatoR” as OnE of itS aliAS naMes. Thus, It is regarDed AS disTiNcT from thE Greek capital
(C, [2 1])

Computes a sim ple transp ose b y s wap pi ng r owsand columns. T h e [2  1] argument means tha t the  2 s t an d  1 st di mension of C  ar ema ppe dt othe 1 stand 2nd  dimension  in  t he result. E a ch  'new' dim ens ion is speci fie d in o rd er. So [3  21] wo uld ta k e the3rd, 2ndan d  1st d i mension s to  bethe new 1st, 2nda nd 3rd dimensions .
perm ut e (C , [32 1 ])

ans =
a ns(:, : ,1) =
 1  2 3

ans(:,:,2 ) =

   4 5  6

El em ent s  of Cwithro w  =1 are found  inwhere the  3rd d i mension = 1 inthe re sul t.Simi l ar ly , e le m ent s  o f C wit h row =2ar e fou nd w h e r e  the  3r d di mensi on = 2 in the  re sult . 
El ement s ofC wi th  colu mn = 1  arest ill found where  col umn = 1 i n t he  re su lt (a n d so o n)asthe col umn dim e nsi on w a smapped to itself.Th e ro ws of th e resu l tis the inte re sti ng d i m ensio n, i t  i s single ton (i . e.  t here is  o nly on erow ) a s a r e sult  of Chaving n o 3rd dimension.
Add r essing the fi r st p ar t  ofyou r problem,thec orre ct o u tp utf or Ccan b eo bt a ined by
A = [1 2]'* [1  2];
C  = bs xfun(@times,permute(A, [ 4 ,3,1,2]) , A) ; 
I  would need mo re in formationo n what y ou wa nt the f inal beha v i our to b e ( 're set tin g th e lower trian g l e')as  it isunc lear to  me  wh atyou  d esire.
Afunction  t ha tmi ght  be u s eful toyo u i sthe  triu functi on wh ichex tr a cts  uppert ri a n gula rco mpon ent sof amatr i x.
 Q:

C orrect sy mbo l  for  t he  Laplac e operator ($ \D elta$)

Wh at  is  the c o r rect sym bol to use for the Lapl a ce oper ato r? It  loo ks almost  li ke the  bi g  Delta  $\Del ta$,bu t i t shoul d lo okdi fferent so t hat  it i snotconfuse d with the Delta.C urr ently, I use$\v ec \ n a bl a^2 $ ,w hic hi s u n a mbiguous, but n ot pretty. 
A :
The opera t oris  define d, in t he in t ernatio nal stand ard ISO 8 00 00-1 , asidentified  with th e Unicode chara c te r U+2 206  INCRE ME NT(mist akenly cal led D ELTA i nthe st andar d) , whichhas “Laplace operator”as one  of i tsalias nam es. Thu s, it isrega rded as di sti nct  from  th e  Gree k ca p it al
(C, [2_1])

Computes a_simple transpose by swapping_rows and_columns._The [2_1]_argument means that_the 2st and_1st dimension of C_are mapped to_the_1st and 2nd dimension in the result. Each 'new' dimension is specified in order._So_[3 2_1]_would_take the 3rd, 2nd and_1st dimensions to be the_new 1st,_2nd and 3rd dimensions.
permute(C, [3 2 1])

ans =

ans(:,:,1)_=

_  1_  2   3

ans(:,:,2) =

  _4   5  _6

Elements of C_with_row_= 1 are found_in where the 3rd dimension =_1 in the result. Similarly, elements_of C with row = 2 are_found where the 3rd dimension =_2 in the result.
Elements of_C with_column = 1 are still_found where column_= 1_in the result_(and so on) as the column_dimension was mapped_to itself.
The rows of the result_is_the interesting dimension,_it_is_singleton (i.e._there is only_one_row) as_a_result of C having no 3rd_dimension.
Addressing_the first part of your problem, the_correct output for C_can_be obtained by
A =_[1 2]'*[1 2];
C = bsxfun(@times,_permute(A, [4,3,1,2]), A);

I would need more_information on_what you_want the final behaviour to be ('resetting the lower triangle') as_it is unclear to me what_you desire.
A function that_might be_useful_to you is_the_triu function_which extracts upper triangular components of a_matrix.

 Q:

Correct_symbol for the Laplace operator ($\Delta$)

What_is the correct symbol_to_use for the Laplace operator? It_looks almost like the big Delta_$\Delta$, but it should look_different_so_that it is not confused_with the Delta.
Currently, I use $\vec_\nabla^2$, which is_unambiguous, but not pretty.

A:

The operator is defined,_in_the international standard ISO 80000-1, as_identified_with the Unicode character U+2206 INCREMENT_(mistakenly_called_DELTA in the standard), which_has “Laplace operator” as one of_its alias names. Thus, it is regarded as distinct_from the Greek_capital