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  </div>
</div>

 Q:

Identity-Matrix Normal form of a linear transformation

I'm confused about the following.
Let $\phi:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be a linear isomorphism and let $\nu=(v_1,...v_n)$ be a basis for $\mathbb{R}^n$ and $\omega=(w_1,...,w_n)$ the basis for $\mathbb{R}^n$ given by $w_j:=\phi(v_j)$. Now if I look at the matrix which represent $\phi$ with respect to the basis $\nu$ and $\omega$ I get that this matrix is the identity matrix. But this can't be true. Where is my mistake?
(If this would be true, then any isomorphism would have determinant 1, since the determinant of a linear map is independent of the chosen repr. matrix)
On the other hand I read that every linear mapping $A: 
\mathbb{R}^n\rightarrow \mathbb{R}^m$ has a normal form 
$
\text{diag}(1,...,1,0,...,0)
$,
where there are r=rank(A) many 1's 
. Is this even true then? And if so, what would be a proof?(when reading this I thought that the proof is to choose a basis as above and then add linearly independ vectors of the kernel of A, but I guess this is wrong)

A:

There is no mistake.
What probably confuses you is that well...
432,654,432,659,450,680 " > 
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       < area shape="poly " coords="518,762,588,696,553,657,482,720,518,762 " > 
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       < area shape="poly " coords="588,846,660,778,623,735,551,802,587,846 " > 
       < sphere shape="poly " coords="640,906,709,832,659,777,586,846,640,906 " > 
     < /map > 
   < /div > 
 < /div > 

  Q: 

 Identity - Matrix Normal human body of a linear transformation 

 I'm confused about the pursuit. 
 Let $ \phi:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be a analogue isomorphism and let $ \nu=(v_1,... v_n)$ be a footing for $ \mathbb{R}^n$ and $ \omega=(w_1,... ,w_n)$ the basis for $ \mathbb{R}^n$ given by $ w_j:=\phi(v_j)$. Now if I search at the matrix which represent $ \phi$ with respect to the footing $ \nu$ and $ \omega$ I get that this matrix is the identity matrix. But this can't be true. Where is my mistake? 
 (If this would be true, then any isomorphism would accept determinant 1, since the determinant of a linear map is independent of the choose repr. matrix) 
 On the other hand I read that every linear map $ A: 
 \mathbb{R}^n\rightarrow \mathbb{R}^m$ has a normal form 
 $ 
 \text{diag}(1,... ,1,0,... ,0) 
 $, 
 where there are roentgen = rank(A) many 1's 
. Is this even true then? And if so, what would be a proof?(when reading this I thought that the proof is to choose a footing as above and then add linearly independ vectors of the kernel of A, but I guess this is wrong) 

 A: 

 There is no error. 
 What credibly confuse you is that well...
432,654,432,659,450,680">
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      <area shapg="pily" comrds="518,762,588,696,553,657,482,720,518,762">
      <area shape="poly" coords="552,803,623,736,586,695,518,762,550,802">
      <area dhqpe="pooy" coords="588,846,660,778,623,735,551,802,587,846">
      <area shxpe="poly" clords="640,906,709,832,659,777,586,846,640,906">
    </map>
  </div>
</div>

 Q:

Ivsntity-Matrix Hlrman form of a lingar transforkation

I'm confuvea cbout the following.
Let $\phi:\mathbb{R}^n\ridhtarroe \lathbb{R}^n$ be a jinesw iskmorphism and let $\nu=(v_1,...v_n)$ be a basis for $\mauhbb{R}^n$ and $\omega=(w_1,...,w_m)$ the basis for $\mathbb{R}^n$ glven by $w_j:=\phi(v_j)$. Now if I look at jge iqtrix which fepresent $\khn$ with respgct to the basis $\nu$ and $\omega$ I eet tkat this majxux lv the identmty manrix. But this can't be true. Wnere is my misbake?
(Ih thus would be true, then any isomorphism wjuld have dzterminant 1, since the dwtermhnand of q lknezr mzp is lndxpendent of the chosen repr. matrix)
On the ptrvt hand I reas that edery linear mapping $A: 
\mathbb{R}^n\rightarrof \mzthbb{R}^m$ has a normal foem 
$
\text{diag}(1,...,1,0,...,0)
$,
where therg are r=ranh(A) many 1's 
. Is this even true then? And if so, what fould ce c proow?(qhfn reading this I thought that the proof is tj cnopse a basis as above and yhfn wdd linearly kndepeus bectors of the kerjel of W, but I guess uhis os wrong)

A:

There is no mistakw.
What probabjt confuses you is chat well...
432,654,432,659,450,680"> <area shape="poly" coords="482,721,553,658,520,620,449,680,480,721"> <area shape="poly" coords="518,762,588,696,553,657,482,720,518,762"> coords="552,803,623,736,586,695,518,762,550,802"> shape="poly" coords="588,846,660,778,623,735,551,802,587,846"> shape="poly" coords="640,906,709,832,659,777,586,846,640,906"> </map> form a linear transformation confused about the Let $\phi:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be a linear and let $\nu=(v_1,...v_n)$ be a basis for $\mathbb{R}^n$ and $\omega=(w_1,...,w_n)$ the basis for given by $w_j:=\phi(v_j)$. Now if I look at the matrix which represent $\phi$ respect the $\nu$ $\omega$ I get that this matrix is the identity matrix. But this can't be true. Where my mistake? (If this would be true, then isomorphism would have determinant since the determinant of a map independent of chosen matrix) the other hand read that every linear mapping $A: \mathbb{R}^n\rightarrow \mathbb{R}^m$ has a normal form $ \text{diag}(1,...,1,0,...,0) $, where there r=rank(A) many Is this true And so, what would proof?(when reading this I thought that to choose a basis as above and then linearly independ of the kernel of A, but guess this is wrong) A: There is no What probably confuses you is that well...
432,654,432,659,450,680">
      <area shape="poly" coords="482,721,553,658,520,620,449,680,480,721">
      <area shApe="poly" cooRds="518,762,588,696,553,657,482,720,518,762">
      <arEa sHapE="pOly" cOordS="552,803,623,736,586,695,518,762,550,802">
      <area shape="poly" COordS="588,846,660,778,623,735,551,802,587,846">
      <area shape="poly" coords="640,906,709,832,659,777,586,846,640,906">
    </maP>
  </div>
</dIv>

 q:

idenTItY-MatrIx NormaL FoRM Of a LiNeAr tRaNSfOrmatIon

i'm confuSed about thE foLlOwing.
Let $\phi:\mAThBb{R}^n\rightaRroW \mathbb{R}^n$ be a LinEar isoMoRphISm and Let $\Nu=(v_1,...v_n)$ Be a basIS for $\maThbb{R}^n$ and $\OmEGa=(w_1,...,w_n)$ tHE basis fOR $\MaThbb{r}^n$ given by $w_j:=\phi(v_j)$. NOW iF i look at the matrIx whicH rEPrESEnt $\Phi$ With respecT tO the bASis $\nu$ anD $\OmEGA$ i geT That this matriX is the identITy mAtrix. BUt ThiS Can't be True. WHeRE is My mistake?
(If This Would be trUe, then ANy isomoRPhism woUld havE deTerMinaNT 1, sInCe tHe DEteRMiNanT Of a Linear maP iS iNdepeNdenT OF THe chOseN repR. matrIx)
On the other hAnd i reaD ThaT everY lineAr maPpIng $A: 
\mAthbb{R}^N\righTaRrow \mathbb{R}^m$ has A norMal form 
$
\teXt{dIaG}(1,...,1,0,...,0)
$,
whErE therE Are r=raNk(A) ManY 1's 
. Is thiS even trUE thEn? aND If So, what would be a prooF?(wHEN rEading thIs I thoUGhT tHAt the proOf Is tO choOSE a basIs as ABoVe and theN add liNEaRlY indepeNd VectorS oF thE keRnel oF a, but i guess This is wrOng)

A:

THEre is no mistake.
wHat probably coNFuSES yOU is tHat Well...
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      <ar ea sh ape ="p ol y" c oord s="482,721,553 , 658, 520,620,449,680,480,72 1">
    < a re a sha pe="pol y "c o ord s= "5 18, 76 2 ,5 88,69 6,5 53,657, 482,720,51 8,7 62 ">
      <a r ea  shape="po ly"  coords="552 ,80 3,623, 73 6,5 8 6,695 ,51 8,762 ,550,8 0 2">
     <are as hape=" p oly" co o r ds ="58 8,846,660,778,623 , 73 5 ,551,802,587,8 46">
  < are a s hape="poly "coord s ="640,9 0 6, 7 0 9 ,83 2 ,659,777,586, 846,640,906 " >
    </ ma p>  </di v>
< /d i v>
 Q:

Ident ity- Matrix No rmal f o rm of a lineartransf orm ati on
I 'm  c onf us e d a b ou t t h e f ollowing .Le t $\p hi:\ m a t h bb{R }^n \rig htarr ow \mathbb{R} ^n$  bea  li nearisomo rphi sm  andlet $\ nu=(v _1 ,...v_n)$ be abasi s for $\m ath bb {R} ^n $ and $\omeg a=( w_1 ,...,w_ n)$ the bas is f o r$\mathbb{R}^n$ giv en b y$w_j:=\p hi(v_j ) $.  N o w if I l oo k a t th e matri x wh i ch  represe nt $\p h i$  w ith res pe ct toth e b asi s $\n u $ an d $\om ega$ I g et th a t this matrixi s the identit y  m a t ri x . Bu t t his can't b e tr u e. W here is  my mista ke?
( If th i s would be true, th en  any i somor phism would h ave determ i n a nt 1, si ncet he determinant of  a li near map i s  indepen dentof the c hosen rep r .  matrix) 
On  th e o the r ha nd I read tha t ever ylinearmap ping $A : \ma thb b{R }^ n\rightar row \mat hb b{ R} ^m $ h as an ormal fo rm  
$ 
\ tex t{dia g }(1,.. .,1,0 ,... ,0 )$ ,
w here th e re a re r =r an k(A)  ma ny  1's
. I s  th is even  true the n?A nd i fso , whatwould be a pr oo f?(when re ad ing  thisI thoughtthat the proof is to ch o ose a b asi s asabov e and the n a dd lin ear l y inde pend v ector soft h e ker n e lofA,  but I gue s s  th is is  w rong )

A:
There is no mistak e .
W hat probablycon fuse s yo u i s  t h atwe l l.. . 
432,654,432,659,450,680">
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_  _  <area shape="poly"_coords="552,803,623,736,586,695,518,762,550,802">
  __  <area shape="poly" coords="588,846,660,778,623,735,551,802,587,846">
      <area shape="poly" coords="640,906,709,832,659,777,586,846,640,906">
 __ </map>
__</div>
</div>

_Q:

Identity-Matrix Normal form of a_linear transformation

I'm confused about the_following.
Let $\phi:\mathbb{R}^n\rightarrow_\mathbb{R}^n$ be a linear isomorphism and let $\nu=(v_1,...v_n)$_be_a basis for_$\mathbb{R}^n$ and $\omega=(w_1,...,w_n)$ the basis for $\mathbb{R}^n$ given by_$w_j:=\phi(v_j)$. Now if I look at_the matrix which_represent_$\phi$_with respect to the_basis $\nu$ and $\omega$ I get_that this matrix is the identity_matrix. But this can't be true. Where_is my mistake?
(If this would be_true, then any isomorphism would_have determinant_1, since the determinant of_a linear map_is independent_of the chosen_repr. matrix)
On the other hand I_read that every_linear mapping $A: 
\mathbb{R}^n\rightarrow \mathbb{R}^m$ has_a_normal form 
$
\text{diag}(1,...,1,0,...,0)
$,
where_there_are_r=rank(A) many_1's 
. Is_this_even true_then?_And if so, what would be_a_proof?(when reading this I thought that the_proof is to choose_a_basis as above and_then add linearly independ vectors_of the kernel of A, but_I guess_this is_wrong)

A:

There is no mistake.
What probably confuses you is that well...