Matrices
Determine the conditions that (2x2) matrices A , B must fulfill so that : $(A+B)^2 != A^2+2AB+B^2$ and provide a numerical example .
Risposte
"Camillo":
$ I_2 =((1,0),(0,1))$ the identity matrix ( 2x2) .
Ok, your relationship expresses in synthetic way my conditions
$ I_2 =((1,0),(0,1))$ the identity matrix ( 2x2) .
"Camillo":
The conditions required for square matrix A, B to commute is :
$A = alpha.B +beta*I_2 $ . for $(2x2 )$matrices. $AA alpha , beta in RR $
Let's check if the condition indicated by nicasamarciano and this one are equivalent ..
Which is $I_2$?
The conditions required for square matrix A, B to commute is :
$A = alpha.B +beta*I_2 $ . for $(2x2 )$matrices. $AA alpha , beta in RR $
Let's check if the condition indicated by nicasamarciano and this one are equivalent ..
$A = alpha.B +beta*I_2 $ . for $(2x2 )$matrices. $AA alpha , beta in RR $
Let's check if the condition indicated by nicasamarciano and this one are equivalent ..
"Camillo":
[quote="Fioravante Patrone"]
I don't know whether Camillo considers the characterization given above as a solution for his question.
If yes, then:
If not,
Yes of course I do consider a solution for the question , as well as nicasamarciano one
.Further question
: Which are, in synthesis, the conditions required for two square matrices A , B to commute ?[/quote]In synthesis the condition for which $(A+B)^2=A^2+2AB+B^2$ is obviously $AB=BA$ and this condition can be expressed through the following relationship between their coefficients.
That is, fixed $B=((a,b),(c,d))$, the coefficient of the matrix $A=((x,y),(z,t))$ must satisfy these relationships:
${(y/z=b/c),(t-x=(d-a)*(z/c)):}$
"Fioravante Patrone":
I don't know whether Camillo considers the characterization given above as a solution for his question.
If yes, then:
If not,
Yes of course I do consider a solution for the question , as well as nicasamarciano one
.Further question
: Which are, in synthesis, the conditions required for two square matrices A , B to commute ?
"fireball":
I think he means the product of the matrix with itself.
Correct.
"Camillo":
Determine the conditions that (2x2) matrices A , B must fulfill so that : $(A+B)^2 != A^2+2AB+B^2$ and provide a numerical example .
$(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2$
so,
$(A+B)^2 = A^2+2AB+B^2$
iff:
$A^2 + AB + BA + B^2 = A^2+2AB+B^2$
i.e. iff:
$AB = BA$
it is easy to provide a (numerical) example of matrices which do not commute
I don't know whether Camillo considers the characterization given above as a solution for his question.
If yes, then:
If not,
PS: While I was using the preview I saw that nicasamarciano goes on with his calculations. So, maybe his post supersedes mine. Anyway, I had already written mine, so I post it anyway.
$A = ((x, y), (z, t))$
$B = ((a, b), (c, d))$
$(A+B)^2=(A+B)(A+B)=(((a + x)^2 + (b + y)(c +z), (d + t)(b + y)+(a + x)(b + y)),((d + t)(c+z) +(a + x)(c +z),(d + t)^2 +(b + y)(c + z)))$
$A^2=((x^2 + y z, t y + x y), (t z + x z, t^2 + y z))$
$B^2=((a^2 + b c, a b + b d), (a c + c d, b c + d^2))$
$A*B=((a x + c y, b x + d y), (c t + a z, d t + b z))$
Now $(A+B)^2=A^2+B^2+2AB$ $<=>$
1)$(a + x)^2 + (b + y)(c +z)=x^2 + y z+a^2 + b c+2(a x + c y)$
2)$(d + t)(b + y)+(a + x)(b + y)=t y + x y+a b + b d+2(b x+ d y)$
3)$(d + t)(c+z) +(a + x)(c +z)=t z + x z+a c + c d+2(c t + a z)$,
4)$(d + t)^2 + (b + y)(c + z)=t^2 +yz+ b c + d^2 +2(d t + b z)$
Now the first and the fourth equations impose the same condition $bz=cy $, while the second imposes $y(d-a)=b(t-x)$ and the third imposes $z(d-a)=c(t-x)$.
Therefore the final conditions so that $(A+B)^2=A^2+B^2+2AB $ are:
${(y/z=b/c),(t-x=(d-a)*(z/c)):}$
Practically, fixed the matrix $B $, the coefficients of the matrix $A $ have to be selected so that to satisfy the two conditions ${(y/z=b/c),(t-x=(d-a)*(z/c)):} $ to have $(A+B)^2=A^2+B^2+2AB $.
For this reason if only 1 of the two conditions is not satisfied, we have $(A+B)^2!=A^2+B^2+2AB $.
Numerical example
$B=((1,2),(3,4)) $
Now $y/z=2/3 $, therefore we choose for instance $y=4 $and $z=6 $. Besides another condition must be satisfied and that is $t-x=6 $ and for this reason we choose $x=0 $ and $t=6 $ for example.
Making the calculations we find that $(A+B)^2=A^2+B^2+2AB=((55,66),(99,154)) $.
But if for instance $y/z!=2/3 $, that is we choose $y=4 $ and $z=0 $, and the same values for $x$ and $t$, $x=0$ and $t=6$ then $(A+B)^2!=A^2+B^2+2AB $. Infact in this case
$(A+B)^2=((19,66),(33,118))$ while $A^2+B^2+2AB=((7,66),(15,130))$
$B = ((a, b), (c, d))$
$(A+B)^2=(A+B)(A+B)=(((a + x)^2 + (b + y)(c +z), (d + t)(b + y)+(a + x)(b + y)),((d + t)(c+z) +(a + x)(c +z),(d + t)^2 +(b + y)(c + z)))$
$A^2=((x^2 + y z, t y + x y), (t z + x z, t^2 + y z))$
$B^2=((a^2 + b c, a b + b d), (a c + c d, b c + d^2))$
$A*B=((a x + c y, b x + d y), (c t + a z, d t + b z))$
Now $(A+B)^2=A^2+B^2+2AB$ $<=>$
1)$(a + x)^2 + (b + y)(c +z)=x^2 + y z+a^2 + b c+2(a x + c y)$
2)$(d + t)(b + y)+(a + x)(b + y)=t y + x y+a b + b d+2(b x+ d y)$
3)$(d + t)(c+z) +(a + x)(c +z)=t z + x z+a c + c d+2(c t + a z)$,
4)$(d + t)^2 + (b + y)(c + z)=t^2 +yz+ b c + d^2 +2(d t + b z)$
Now the first and the fourth equations impose the same condition $bz=cy $, while the second imposes $y(d-a)=b(t-x)$ and the third imposes $z(d-a)=c(t-x)$.
Therefore the final conditions so that $(A+B)^2=A^2+B^2+2AB $ are:
${(y/z=b/c),(t-x=(d-a)*(z/c)):}$
Practically, fixed the matrix $B $, the coefficients of the matrix $A $ have to be selected so that to satisfy the two conditions ${(y/z=b/c),(t-x=(d-a)*(z/c)):} $ to have $(A+B)^2=A^2+B^2+2AB $.
For this reason if only 1 of the two conditions is not satisfied, we have $(A+B)^2!=A^2+B^2+2AB $.
Numerical example
$B=((1,2),(3,4)) $
Now $y/z=2/3 $, therefore we choose for instance $y=4 $and $z=6 $. Besides another condition must be satisfied and that is $t-x=6 $ and for this reason we choose $x=0 $ and $t=6 $ for example.
Making the calculations we find that $(A+B)^2=A^2+B^2+2AB=((55,66),(99,154)) $.
But if for instance $y/z!=2/3 $, that is we choose $y=4 $ and $z=0 $, and the same values for $x$ and $t$, $x=0$ and $t=6$ then $(A+B)^2!=A^2+B^2+2AB $. Infact in this case
$(A+B)^2=((19,66),(33,118))$ while $A^2+B^2+2AB=((7,66),(15,130))$
I think he means the product of the matrix with itself.
what do you intend with the exponent 2? the square of every element of the matrix or the product of the matrix with itself?
I will try as soon as possible, but I don't know if my capabilities will be enough.
It's since 2001 that I haven't been studying linear algebra.
It's since 2001 that I haven't been studying linear algebra.
Nobody wants to try ?
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