Let f : C 2 → C 3 be linear transform defined by:
f ( z 1 z 2 ) = ⎝ ⎛ ( 1 − 2 i ) ∗ z 1 + ( 3 − 4 i ) ∗ z 2 ( 3 − 4 i ) ∗ z 1 + ( 5 + 2 i ) ∗ z 2 ( 1 1 − 3 i ) ∗ z 1 + ( 4 + 2 i ) ∗ z 2 ⎠ ⎞ , where z 1 , z 2 ∈ C
and A = { ( 2 − 5 i 7 + 3 i ) , ( 1 0 − 6 i 3 + 4 i ) } be basis for C 2 and B = ⎩ ⎨ ⎧ ⎝ ⎛ 1 + 2 i 3 + 4 i 5 + 6 i ⎠ ⎞ , ⎝ ⎛ − 7 + 1 0 i 2 − 6 i 1 + 2 i ⎠ ⎞ , ⎝ ⎛ − 6 + 5 i 1 0 + 3 i 5 + i ⎠ ⎞ ⎭ ⎬ ⎫ be a basis for C 3 . .
If M = [ a j k + b j k i ] 3 x 2 represents the linear transform above and j = 1 ∑ 3 k = 1 ∑ 2 ( a j k + b j k i ) = α + β i , find ∥ α + β i ∥ = α 2 + β 2 .
Express the result to four decimal places.
General Case:
Let f : C n → C m be linear transform defined by:
f
⎝
⎜
⎜
⎜
⎜
⎜
⎜
⎛
z
1
z
2
.
.
.
z
n
⎠
⎟
⎟
⎟
⎟
⎟
⎟
⎞
=
⎝
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎛
c
1
1
∗
z
1
+
.
.
.
+
c
1
k
∗
z
k
+
.
.
.
+
c
1
n
∗
z
n
c
2
1
∗
z
1
+
.
.
.
+
c
2
k
∗
z
k
+
.
.
.
+
c
2
n
∗
z
n
.
.
.
c
j
1
∗
z
1
+
.
.
.
+
c
j
k
∗
z
k
+
.
.
.
+
c
j
n
∗
z
n
.
.
.
c
m
1
∗
z
1
+
.
.
.
+
c
m
k
∗
z
k
+
.
.
.
+
c
m
n
∗
z
n
⎠
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎞
,
where
z
k
,
c
j
k
∈
C
for each
(
1
<
=
j
<
=
m
)
and
(
1
<
=
k
<
=
n
)
.
Let V j = ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ v 1 j v 2 j . . . v n j ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ ∈ C n
and A = { V j ∣ ( 1 < = j < = n ) } be a basis for C n .
Let W j = ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ w 1 j w 2 j . . . w m j ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ ∈ C m
and B = { W j ∣ ( 1 < = j < = m ) } be a basis for C m .
You can write a program in any language to find the matrix M = [ a j k + b j k i ] m x n representation of the general linear transform above and ∥ ∑ j = 1 m ∑ k = 1 n ( a j k + b j k i ) ∥ .
You can use the program written to find the matrix M = [ a j k + b j k i ] 3 x 2 that represents the linear transform above and output ∥ ∑ j = 1 3 ∑ k = 1 2 ( a j k + b j k i ) ∥ .
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After doing the problem I supply the program I wrote for the general case which generated the output for this specific problem.
Since f : C 2 → C 3 is a linear transform, for each integer j ∋ ( 1 < = j < = 2 )
f ( z 1 j z 2 j ) = α 1 j ∗ ⎝ ⎛ 1 + 2 i 3 + 4 i 5 + 6 i ⎠ ⎞ + α 2 j ∗ ⎝ ⎛ − 7 + 1 0 i 2 − 6 i 1 + 2 i ⎠ ⎞ + α 3 j ∗ ⎝ ⎛ − 6 + 5 i 1 0 + 3 i 5 + i ⎠ ⎞
= ⎣ ⎡ 1 + 2 i 3 + 4 i 5 + 6 i − 7 + 1 0 i 2 − 6 i 1 + 2 i − 6 + 5 i 1 0 + 3 i 5 + i ⎦ ⎤ ∗ ∣ ∣ ∣ ∣ ∣ ∣ α 1 j α 2 j α 3 j ∣ ∣ ∣ ∣ ∣ ∣
where α 1 j , α 2 j , α 3 j ∈ C and f ( 2 − 5 i 7 + 3 i ) = ⎝ ⎛ 2 5 − 2 8 i 1 5 + 6 i 2 9 − 3 5 i ⎠ ⎞
f ( 1 0 − 6 i 3 + 4 i ) = ⎝ ⎛ 2 3 − 2 6 i 1 3 − 3 2 i 9 6 − 7 4 i ⎠ ⎞
Using the above we can set up the augmented matrix below to solve for the
two 3 X 3 systems of equations.
⎣ ⎢ ⎢ ⎡ 1 + 2 i 3 + 4 i 5 + 6 i − 7 + 1 0 i 2 − 6 i 1 + 2 i − 6 + 5 i 1 0 + 3 i 5 + i 2 5 − 2 8 i 1 5 + 6 i 2 9 − 3 5 i 2 3 − 2 6 i 1 3 − 3 2 i 9 6 − 7 4 i ⎦ ⎥ ⎥ ⎤
( 0 . 2 0 0 0 − 0 . 4 0 0 0 i ) ∗ R O W 1
⎣ ⎢ ⎢ ⎡ 1 . 0 0 0 0 + 0 . 0 0 0 0 i 3 . 0 0 0 0 + 4 . 0 0 0 0 i 5 . 0 0 0 0 + 6 . 0 0 0 0 i 2 . 6 0 0 0 + 4 . 8 0 0 0 i 2 . 0 0 0 0 − 6 . 0 0 0 0 i 1 . 0 0 0 0 + 2 . 0 0 0 0 i 0 . 8 0 0 0 + 3 . 4 0 0 0 i 1 0 . 0 0 0 0 + 3 . 0 0 0 0 i 5 . 0 0 0 0 + 1 . 0 0 0 0 i − 6 . 2 0 0 0 − 1 5 . 6 0 0 0 i 1 5 . 0 0 0 0 + 6 . 0 0 0 0 i 2 9 . 0 0 0 0 − 3 5 . 0 0 0 0 i − 5 . 8 0 0 0 − 1 4 . 4 0 0 0 i 1 3 . 0 0 0 0 − 3 2 . 0 0 0 0 i 9 6 . 0 0 0 0 − 7 4 . 0 0 0 0 i ⎦ ⎥ ⎥ ⎤
( − 3 . 0 0 0 0 − 4 . 0 0 0 0 i ) ∗ R O W 1 + R O W 2
( − 5 . 0 0 0 0 − 6 . 0 0 0 0 i ) ∗ R O W 1 + R O W 3
⎣ ⎢ ⎢ ⎡ 1 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + i 0 . 0 0 0 0 2 . 6 0 0 0 + 4 . 8 0 0 0 i 1 3 . 4 0 0 0 − 3 0 . 8 0 0 0 i 1 6 . 8 0 0 0 − 3 7 . 6 0 0 0 i 0 . 8 0 0 0 + 3 . 4 0 0 0 i 2 1 . 2 0 0 0 − 1 0 . 4 0 0 0 i 2 1 . 4 0 0 0 − 2 0 . 8 0 0 0 i − 6 . 2 0 0 0 − 1 5 . 6 0 0 0 i − 2 8 . 8 0 0 0 + 7 7 . 6 0 0 0 i − 3 3 . 6 0 0 0 + 8 0 . 2 0 0 0 i − 5 . 8 0 0 0 − 1 4 . 4 0 0 0 i − 2 7 . 2 0 0 0 + 3 4 . 4 0 0 0 i 3 8 . 6 0 0 0 + 3 2 . 8 0 0 0 i ⎦ ⎥ ⎥ ⎤
( 0 . 0 1 1 9 + 0 . 0 2 7 3 i ) ∗ R O W 2
⎣ ⎢ ⎢ ⎡ 1 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 2 . 6 0 0 0 + 4 . 8 0 0 0 i 1 . 0 0 0 0 + 0 . 0 0 0 0 i 1 6 . 8 0 0 0 − 3 7 . 6 0 0 0 i 0 . 8 0 0 0 + 3 . 4 0 0 0 i 0 . 5 3 5 7 + 0 . 4 5 5 2 i 2 1 . 4 0 0 0 − 2 0 . 8 0 0 0 i − 6 . 2 0 0 0 − 1 5 . 6 0 0 0 i − 2 . 4 6 0 6 + 0 . 1 3 5 4 i − 3 3 . 6 0 0 0 + 8 0 . 2 0 0 0 i − 5 . 8 0 0 0 − 1 4 . 4 0 0 0 i − 1 . 2 6 2 2 − 0 . 3 3 4 0 i 3 8 . 6 0 0 0 + i 3 2 . 8 0 0 0 i ⎦ ⎥ ⎥ ⎤
( − 2 . 6 0 0 0 − 4 . 8 0 0 0 i ) ∗ R O W 2 + R O W 1
( − 1 6 . 8 0 0 0 + 3 7 . 6 0 0 0 i ) ∗ R O W 2 + R O W 3
⎣ ⎢ ⎢ ⎡ 1 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 1 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 1 . 5 9 2 3 − 0 . 3 5 5 1 i 0 . 5 3 5 7 + 0 . 4 5 5 2 i − 4 . 7 1 7 1 − 8 . 3 0 4 9 i 0 . 8 4 7 5 − 4 . 1 4 1 5 i − 2 . 4 6 0 6 + 0 . 1 3 5 4 i 2 . 6 4 4 9 − 1 4 . 5 9 2 3 i − 4 . 1 2 1 4 − 7 . 4 7 3 1 i − 1 . 2 6 2 2 − 0 . 3 3 4 0 i 7 2 . 3 6 2 5 − 9 . 0 4 7 3 i ⎦ ⎥ ⎥ ⎤
( − 0 . 0 5 1 7 + 0 . 0 9 1 0 i ) ∗ R O W 3
⎣ ⎢ ⎢ ⎡ 1 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 1 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 1 . 5 9 2 3 − 0 . 3 5 5 1 i 0 . 5 3 5 7 + 0 . 4 5 5 2 i 1 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 8 4 7 5 − 4 . 1 4 1 5 i − 2 . 4 6 0 6 + 0 . 1 3 5 4 i 1 . 1 9 1 7 + 0 . 9 9 5 4 i − 4 . 1 2 1 4 − 7 . 4 7 3 1 i − 1 . 2 6 2 2 − 0 . 3 3 4 0 i − 2 . 9 1 8 2 + 7 . 0 5 5 7 i ⎦ ⎥ ⎥ ⎤
( − 1 . 5 9 2 3 + 0 . 3 5 5 1 i ) ∗ R O W 3 + R O W 1
( − 0 . 5 3 5 7 − 0 . 4 5 5 2 i ) ∗ R O W 3 + R O W 2
⎣ ⎢ ⎢ ⎡ 1 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 1 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 0 . 0 0 0 0 + 0 . 0 0 0 0 i 1 . 0 0 0 0 + 0 . 0 0 0 0 i − 1 . 4 0 3 4 − 5 . 3 0 3 2 i − 2 . 6 4 5 9 − 0 . 9 4 0 3 i 1 . 1 9 1 7 + 0 . 9 9 5 4 i − 1 . 9 8 0 3 − 1 9 . 7 4 4 0 i 3 . 5 1 3 2 − 2 . 7 8 5 4 i − 2 . 9 1 8 2 + 7 . 0 5 5 7 i ⎦ ⎥ ⎥ ⎤
⟹
M = ⎣ ⎡ − 1 . 4 0 3 4 − 5 . 3 0 3 2 i − 2 . 6 4 5 9 − 0 . 9 4 0 3 i 1 . 1 9 1 7 + 0 . 9 9 5 4 i − 1 . 9 8 0 3 − 1 9 . 7 4 4 0 i 3 . 5 1 3 2 − 2 . 7 8 5 4 i − 2 . 9 1 8 2 + 7 . 0 5 5 7 i ⎦ ⎤
and,
j = 1 ∑ 3 k = 1 ∑ 2 ( a j k + b j k i ) = − 1 . 7 2 6 5 + 8 . 0 5 1 1 i = α + β i
⟹
∥ α + β i ∥ = α 2 + β 2 = 8 . 2 3 4 1 .
...
I wrote the program in Free Pascal.
program matrixrepresentaionoflineartransform_complex;
{restricted to f:Cn --> Cm}
uses crt;
const maxnum = 100;
type matrix = array[1 .. maxnum,1 .. maxnum] of real;
var coeffa,coeffb,reb,imgb,rea,imga,resol,imgsol:matrix;
function power(base:real; exponent:integer):real;
var n:integer;
product:real;
begin
product:= 1;
for n:= 1 to exponent do
product:= product * base;
power:= product;
end;
procedure getsize;
begin
writeln('Let f:Cn ---> Cm be linear transform ');
writeln('Enter n for Cn');
readln(n);
writeln('Enter m for Cm');
readln(m);
end;
procedure getcoefficients;
var j,k:integer;
begin
for j:= 1 to m do
begin
writeln('For row,', j, ', enter each coefficient of Zj for Cm');
for k:= 1 to n do
begin
read(coeffa[j,k]);
read(coeffb[j,k]);
end;
end;
end;
procedure complexinverse(var reinvz,imginvz:real; rez,imgz:real);
var lengthsqr:real;
begin
lengthsqr:= power(rez,2) + power(imgz,2);
reinvz:= rez/lengthsqr;
imginvz:= -imgz/lengthsqr;
end;
function reproduct(rez1,imgz1,rez2,imgz2:real):real;
begin
reproduct:= rez1 * rez2 - imgz1 * imgz2;
end;
function imgproduct(rez1,imgz1,rez2,imgz2:real):real;
begin
imgproduct:= rez2 * imgz1 + rez1 * imgz2;
end;
procedure getaugmentedmatrix;
var j,k,q,r,s:integer;
begin
assign(myfile,'holdcmatrix.txt');
rewrite(myfile);
writeln('To Enter ', n, ' basis vectors for Cn: ');
for q:= 1 to n do
begin
write('Enter elements of vector ', q, ' : ');
for k:= 1 to n do
begin
read(reb[k,q]);
read(imgb[k,q]);
end;
for j:= 1 to m do
begin
resum:= 0;
imgsum:= 0;
for k:= 1 to n do
begin
resum:= resum + reproduct(coeffa[j,k],coeffb[j,k],reb[k,q],imgb[k,q]);
imgsum:= imgsum + imgproduct(coeffa[j,k],coeffb[j,k],reb[k,q],imgb[k,q]);
end;
revec[j,q]:= resum;
imgvec[j,q]:= imgsum;
end;
end;
writeln('To Enter ', m, ' basis vectors for Cm: ');
for q:= 1 to m do
begin
write('Enter elements of vector ', q, ' : ');
for k:= 1 to m do
begin
read(rea[k,q]);
read(imga[k,q]);
end;
end;
for q:= 1 to m do
begin
for k:= 1 to m do
begin
write(myfile,rea[q,k]:0:4,' ',imga[q,k],' ');
end;
for r:= 1 to n do
begin
write(myfile,revec[q,r]:0:4,' ',imgvec[q,r],' ');
end;
writeln(myfile);
end;
reset(myfile);
for j:= 1 to m do
begin
for k:= 1 to m + n do
begin
read(myfile,rea[j,k]);
read(myfile,imga[j,k]);
end;
end;
close(myfile);
{To show augmented matrix - for show work}
writeln(mywork,' Augmented Matrix');
writeln(mywork);
for q:= 1 to m do
begin
for k:= 1 to m do
begin
write(mywork,rea[q,k]:0:4,' + i', imga[q,k]:0:4,' ');
end;
for r:= 1 to n do
begin
write(mywork,revec[q,r]:0:4,' + i', imgvec[q,r]:0:4,' ');
end;
writeln(mywork);
end;
writeln(mywork);
writeln(mywork);
end;
procedure switch(var c,d:matrix; l,p:integer);
var q:integer;
begin{switch}
for q:= 1 to m + n do
begin{m}
end;{m}
end;{switch}
procedure safeguard(var c,d:matrix; l:integer; var p:integer; var v:integer);
var q:integer;
begin{safeguard}
v:=0;
if (c[l,l] = 0) and (d[l,l] = 0) then
begin{then}
v:= 1;
q:= l;
while ((q + 1) <= n + m) and (v = 1) do
begin{loop}
if (c[q + 1,l] <> 0) or (d[q + 1,l] <> 0) then
begin{then}
end;{then}
q:= q + 1;
end;{loop}
end;{then}
end;{safeguard}
procedure operation1(var a,b:matrix; l:integer);
var q,k:integer;
begin{op1}
for q:= 1 to m + n do
begin{q}
complexinverse(reinvz,imginvz,a[l,l],b[l,l]);
res[l,q]:= reproduct(a[l,q],b[l,q],reinvz,imginvz);
imgs[l,q]:= imgproduct(a[l,q],b[l,q],reinvz,imginvz);
end;{q}
for q:= 1 to m + n do
begin{q}
a[l,q]:= res[l,q];
b[l,q]:= imgs[l,q];
end;{q}
writeln(mywork,'(',reinvz:0:4,' + i', imginvz:0:4,') * ROW ', l);
writeln(mywork);
writeln(mywork);
for q:= 1 to m do
begin
for k:= 1 to m + n do
begin
write(mywork,a[q,k]:0:4,' + i', b[q,k]:0:4,' ');
end;
writeln(mywork);
end;
writeln(mywork);
writeln(mywork);
end;{op1}
procedure operation2(var a,b:matrix; l:integer);
var q,r,k:integer;
begin{op2}
for q:= 1 to m do
begin{q}
if q <> l then
begin{then}
x:= -1 * a[q,l];
y:= -1 * b[q,l];
for r:= 1 to m + n do
begin{r}
res[q,r]:= reproduct(x,y,a[l,r],b[l,r]) + a[q,r];
imgs[q,r]:= imgproduct(x,y,a[l,r],b[l,r]) + b[q,r];
end;{r}
writeln(mywork,'(',x:0:4, ' + i',y:0:4,' ) * ROW ', l,' + ROW ', q);
for r:= 1 to m + n do
begin{r}
a[q,r]:= res[q,r];
b[q,r]:= imgs[q,r];
end;{r}
end;{then}
end;{q}
writeln(mywork);
for q:= 1 to m do
begin
for k:= 1 to m + n do
begin
write(mywork,a[q,k]:0:4,' + i ',b[q,k]:0:4,' ');
end;
writeln(mywork);
end;
writeln(mywork);
writeln(mywork);
end;{op2}
PROCEDURE SHOWDATA;
VAR J,k:INTEGER;
BEGIN
assign(myfile2,'holdsolution6.txt');
rewrite(myfile2);
writeln;
writeln('Matrix Representation of linear transformation f: Cn ---> Cm:');
writeln('-------------------------------------------------------------');
writeln;
for j:= 1 to m do
begin
for k:= m + 1 to m + n do
begin
write(myfile2,rea[j,k]:0:4,' ',imga[j,k],' ');
end;
writeln(myfile2);
end;
reset(myfile2);
for j:= 1 to m do
begin
for k:= 1 to n do
begin read(myfile2,resol[j,k]);
read(myfile2,imgsol[j,k]);
end;
end;
close(myfile2);
for j:= 1 to m do
begin
for k:= 1 to n do
begin
write(resol[j,k]:0:4,' + i', imgsol[j,k]:0:4,' ');
end;
writeln;
end;
{For brillant only.}
for j:= 1 to m do
begin
resum:= 0;
imgsum:= 0;
for k:= 1 to n do
begin
resum:= resum + resol[j,k];
imgsum:= imgsum + imgsol[j,k];
end;
end;
writeln(mywork);
writeln(mywork);
writeln(mywork, 'Sum = ',resum:0:4,' + i', imgsum:0:4);
writeln(mywork, '|Sum| = ',sqrt(power(resum,2) + power(imgsum,2)):0:4);
writeln;
writeln;
writeln('Sum = ',resum:0:4,' + i', imgsum:0:4);
writeln('|Sum| = ',sqrt(power(resum,2) + power(imgsum,2)):0:4);
END;
begin
assign(mywork,'cmywork.txt');
rewrite(mywork);
getsize;
getcoefficients;
getaugmentedmatrix;
for l:= 1 to m do
begin{l}
safeguard(rea,imga,l,p,v);
if v = 0 then
begin{then}
operation1(rea,imga,l);
operation2(rea,imga,l);
end;{then}
end;{l}
showdata;
close(mywork);
readln;
readln;
end.