mistake in calculations for ImT and KerT
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?
The linear transformation $T:M_2x2(R) to R_3[x]$ is defined by:
$$Tbeginpmatrixa&b\c&d endpmatrix$$$= (a-d)x^2+(b+c)x+5a-5d$
for every $$beginpmatrixa&b\c&d endpmatrix$$$in M_2x2(R)$
I reasoned in the following way:
For KerT:
a=d
b=-c
kerT:
$$beginpmatrixa&b\-b&a endpmatrix$$=$$abeginpmatrix1&0\0&1 endpmatrix$$+$$bbeginpmatrix0&1\-1&0 endpmatrix$$
So KerT = Sp$$beginpmatrix1&0\0&1 endpmatrix$$,$$beginpmatrix0&1\-1&0 endpmatrix$$
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.
For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.
The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_2x2(R)=4$
linear-algebra linear-transformations
asked Jan 3 at 12:07
daltadalta
627
$endgroup$
add a comment |
$begingroup$
I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?
The linear transformation $T:M_2x2(R) to R_3[x]$ is defined by:
$$Tbeginpmatrixa&b\c&d endpmatrix$$$= (a-d)x^2+(b+c)x+5a-5d$
for every $$beginpmatrixa&b\c&d endpmatrix$$$in M_2x2(R)$
I reasoned in the following way:
For KerT:
a=d
b=-c
kerT:
$$beginpmatrixa&b\-b&a endpmatrix$$=$$abeginpmatrix1&0\0&1 endpmatrix$$+$$bbeginpmatrix0&1\-1&0 endpmatrix$$
So KerT = Sp$$beginpmatrix1&0\0&1 endpmatrix$$,$$beginpmatrix0&1\-1&0 endpmatrix$$
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.
For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.
The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_2x2(R)=4$
linear-algebra linear-transformations
asked Jan 3 at 12:07
daltadalta
627
$endgroup$
$begingroup$
Use$ X$for $ X$.
$endgroup$
– Shaun
Jan 3 at 12:38
add a comment |
$begingroup$
I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?
The linear transformation $T:M_2x2(R) to R_3[x]$ is defined by:
$$Tbeginpmatrixa&b\c&d endpmatrix$$$= (a-d)x^2+(b+c)x+5a-5d$
for every $$beginpmatrixa&b\c&d endpmatrix$$$in M_2x2(R)$
I reasoned in the following way:
For KerT:
a=d
b=-c
kerT:
$$beginpmatrixa&b\-b&a endpmatrix$$=$$abeginpmatrix1&0\0&1 endpmatrix$$+$$bbeginpmatrix0&1\-1&0 endpmatrix$$
So KerT = Sp$$beginpmatrix1&0\0&1 endpmatrix$$,$$beginpmatrix0&1\-1&0 endpmatrix$$
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.
For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.
The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_2x2(R)=4$
linear-algebra linear-transformations
asked Jan 3 at 12:07
daltadalta
627
$endgroup$
I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?
The linear transformation $T:M_2x2(R) to R_3[x]$ is defined by:
$$Tbeginpmatrixa&b\c&d endpmatrix$$$= (a-d)x^2+(b+c)x+5a-5d$
for every $$beginpmatrixa&b\c&d endpmatrix$$$in M_2x2(R)$
I reasoned in the following way:
For KerT:
a=d
b=-c
kerT:
$$beginpmatrixa&b\-b&a endpmatrix$$=$$abeginpmatrix1&0\0&1 endpmatrix$$+$$bbeginpmatrix0&1\-1&0 endpmatrix$$
So KerT = Sp$$beginpmatrix1&0\0&1 endpmatrix$$,$$beginpmatrix0&1\-1&0 endpmatrix$$
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.
For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.
The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_2x2(R)=4$
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 3 at 12:07
daltadalta
627
asked Jan 3 at 12:07
daltadalta
627
edited Jan 3 at 12:36
dalta
asked Jan 3 at 12:07
daltadalta
627
asked Jan 3 at 12:07
daltadalta
627
asked Jan 3 at 12:07
daltadalta
627
627
$begingroup$
Use$ X$for $ X$.
$endgroup$
– Shaun
Jan 3 at 12:38
add a comment |
$begingroup$
Use$ X$for $ X$.
$endgroup$
– Shaun
Jan 3 at 12:38
$begingroup$
Use
$ X$ for $ X$.$endgroup$
– Shaun
Jan 3 at 12:38
$begingroup$
Use
$ X$ for $ X$.$endgroup$
– Shaun
Jan 3 at 12:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.
It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
$$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...
Or for a more general approach, rewrite as (split in all the coefficients):
$$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.
answered Jan 3 at 12:18
StackTDStackTD
22.7k2049
$endgroup$
1
$begingroup$
Thank you! So my calculations for KerT were correct?
$endgroup$
– dalta
Jan 3 at 12:34
1
$begingroup$
Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
$endgroup$
– StackTD
Jan 3 at 12:35
add a comment |
$begingroup$
$$textIm(T)netextSp1,x,x^2$$
This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.
answered Jan 3 at 12:23
Shubham JohriShubham Johri
4,769717
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.
It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
$$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...
Or for a more general approach, rewrite as (split in all the coefficients):
$$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.
answered Jan 3 at 12:18
StackTDStackTD
22.7k2049
$endgroup$
1
$begingroup$
Thank you! So my calculations for KerT were correct?
$endgroup$
– dalta
Jan 3 at 12:34
1
$begingroup$
Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
$endgroup$
– StackTD
Jan 3 at 12:35
add a comment |
$begingroup$
For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.
It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
$$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...
Or for a more general approach, rewrite as (split in all the coefficients):
$$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.
answered Jan 3 at 12:18
StackTDStackTD
22.7k2049
$endgroup$
1
$begingroup$
Thank you! So my calculations for KerT were correct?
$endgroup$
– dalta
Jan 3 at 12:34
1
$begingroup$
Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
$endgroup$
– StackTD
Jan 3 at 12:35
add a comment |
$begingroup$
For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.
It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
$$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...
Or for a more general approach, rewrite as (split in all the coefficients):
$$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.
answered Jan 3 at 12:18
StackTDStackTD
22.7k2049
$endgroup$
For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.
It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
$$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...
Or for a more general approach, rewrite as (split in all the coefficients):
$$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.
answered Jan 3 at 12:18
StackTDStackTD
22.7k2049
edited Jan 3 at 12:23
answered Jan 3 at 12:18
StackTDStackTD
22.7k2049
answered Jan 3 at 12:18
StackTDStackTD
22.7k2049
answered Jan 3 at 12:18
StackTDStackTD
22.7k2049
22.7k2049
1
$begingroup$
Thank you! So my calculations for KerT were correct?
$endgroup$
– dalta
Jan 3 at 12:34
1
$begingroup$
Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
$endgroup$
– StackTD
Jan 3 at 12:35
add a comment |
1
$begingroup$
Thank you! So my calculations for KerT were correct?
$endgroup$
– dalta
Jan 3 at 12:34
1
$begingroup$
Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
$endgroup$
– StackTD
Jan 3 at 12:35
1
1
$begingroup$
Thank you! So my calculations for KerT were correct?
$endgroup$
– dalta
Jan 3 at 12:34
$begingroup$
Thank you! So my calculations for KerT were correct?
$endgroup$
– dalta
Jan 3 at 12:34
1
1
$begingroup$
Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
$endgroup$
– StackTD
Jan 3 at 12:35
$begingroup$
Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
$endgroup$
– StackTD
Jan 3 at 12:35
add a comment |
$begingroup$
$$textIm(T)netextSp1,x,x^2$$
This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.
answered Jan 3 at 12:23
Shubham JohriShubham Johri
4,769717
$endgroup$
add a comment |
$begingroup$
$$textIm(T)netextSp1,x,x^2$$
This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.
answered Jan 3 at 12:23
Shubham JohriShubham Johri
4,769717
$endgroup$
add a comment |
$begingroup$
$$textIm(T)netextSp1,x,x^2$$
This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.
answered Jan 3 at 12:23
Shubham JohriShubham Johri
4,769717
$endgroup$
$$textIm(T)netextSp1,x,x^2$$
This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.
answered Jan 3 at 12:23
Shubham JohriShubham Johri
4,769717
answered Jan 3 at 12:23
Shubham JohriShubham Johri
4,769717
answered Jan 3 at 12:23
Shubham JohriShubham Johri
4,769717
answered Jan 3 at 12:23
Shubham JohriShubham Johri
4,769717
4,769717
add a comment |
add a comment |
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$begingroup$
Use
$ X$for $ X$.$endgroup$
– Shaun
Jan 3 at 12:38