Mixing
finding Ker(T) of a parameter's linear transformation
$begingroup$
I am suppose to find the ker(T) of linear transformation of:
$$
Gbegin{pmatrix}a & d \ c & bend{pmatrix}=
a+frac{b+c}{2}x+frac{b-c}{2}x^2
$$
the form $T:V to W$
My problem is that I don't really know how to read the right side of the equation, is that a vector? A polynomial?
Am I supposed to take the right side of the equation and make it equal with 0?
linear-transformations matrix-equations harmonic-functions
edited Jan 12 at 17:11
egreg
181k1485203
asked Jan 12 at 17:07
ShamesShames
61
$endgroup$
add a comment |
$begingroup$
I am suppose to find the ker(T) of linear transformation of:
$$
Gbegin{pmatrix}a & d \ c & bend{pmatrix}=
a+frac{b+c}{2}x+frac{b-c}{2}x^2
$$
the form $T:V to W$
My problem is that I don't really know how to read the right side of the equation, is that a vector? A polynomial?
Am I supposed to take the right side of the equation and make it equal with 0?
linear-transformations matrix-equations harmonic-functions
edited Jan 12 at 17:11
egreg
181k1485203
asked Jan 12 at 17:07
ShamesShames
61
$endgroup$
1
$begingroup$
Please do not use pictures.
$endgroup$
– Dietrich Burde
Jan 12 at 17:08
$begingroup$
The linear map is called $G$ and not $T$. The domain is the vector space of $2times2$ matrices, the codomain is the vector space of polynomials (of degree at most $2$). You have to prove $G$ is linear, then compute the kernel.
$endgroup$
– egreg
Jan 12 at 17:12
$begingroup$
Yes. You set the right hand side equal to $0$.
$endgroup$
– John Douma
Jan 12 at 17:59
add a comment |
$begingroup$
I am suppose to find the ker(T) of linear transformation of:
$$
Gbegin{pmatrix}a & d \ c & bend{pmatrix}=
a+frac{b+c}{2}x+frac{b-c}{2}x^2
$$
the form $T:V to W$
My problem is that I don't really know how to read the right side of the equation, is that a vector? A polynomial?
Am I supposed to take the right side of the equation and make it equal with 0?
linear-transformations matrix-equations harmonic-functions
edited Jan 12 at 17:11
egreg
181k1485203
asked Jan 12 at 17:07
ShamesShames
61
$endgroup$
I am suppose to find the ker(T) of linear transformation of:
$$
Gbegin{pmatrix}a & d \ c & bend{pmatrix}=
a+frac{b+c}{2}x+frac{b-c}{2}x^2
$$
the form $T:V to W$
My problem is that I don't really know how to read the right side of the equation, is that a vector? A polynomial?
Am I supposed to take the right side of the equation and make it equal with 0?
linear-transformations matrix-equations harmonic-functions
linear-transformations matrix-equations harmonic-functions
edited Jan 12 at 17:11
egreg
181k1485203
asked Jan 12 at 17:07
ShamesShames
61
edited Jan 12 at 17:11
egreg
181k1485203
asked Jan 12 at 17:07
ShamesShames
61
edited Jan 12 at 17:11
egreg
181k1485203
edited Jan 12 at 17:11
egreg
181k1485203
edited Jan 12 at 17:11
egreg
181k1485203
181k1485203
asked Jan 12 at 17:07
ShamesShames
61
asked Jan 12 at 17:07
ShamesShames
61
asked Jan 12 at 17:07
ShamesShames
61
61
1
$begingroup$
Please do not use pictures.
$endgroup$
– Dietrich Burde
Jan 12 at 17:08
$begingroup$
The linear map is called $G$ and not $T$. The domain is the vector space of $2times2$ matrices, the codomain is the vector space of polynomials (of degree at most $2$). You have to prove $G$ is linear, then compute the kernel.
$endgroup$
– egreg
Jan 12 at 17:12
$begingroup$
Yes. You set the right hand side equal to $0$.
$endgroup$
– John Douma
Jan 12 at 17:59
add a comment |
1
$begingroup$
Please do not use pictures.
$endgroup$
– Dietrich Burde
Jan 12 at 17:08
$begingroup$
The linear map is called $G$ and not $T$. The domain is the vector space of $2times2$ matrices, the codomain is the vector space of polynomials (of degree at most $2$). You have to prove $G$ is linear, then compute the kernel.
$endgroup$
– egreg
Jan 12 at 17:12
$begingroup$
Yes. You set the right hand side equal to $0$.
$endgroup$
– John Douma
Jan 12 at 17:59
1
1
$begingroup$
Please do not use pictures.
$endgroup$
– Dietrich Burde
Jan 12 at 17:08
$begingroup$
Please do not use pictures.
$endgroup$
– Dietrich Burde
Jan 12 at 17:08
$begingroup$
The linear map is called $G$ and not $T$. The domain is the vector space of $2times2$ matrices, the codomain is the vector space of polynomials (of degree at most $2$). You have to prove $G$ is linear, then compute the kernel.
$endgroup$
– egreg
Jan 12 at 17:12
$begingroup$
The linear map is called $G$ and not $T$. The domain is the vector space of $2times2$ matrices, the codomain is the vector space of polynomials (of degree at most $2$). You have to prove $G$ is linear, then compute the kernel.
$endgroup$
– egreg
Jan 12 at 17:12
$begingroup$
Yes. You set the right hand side equal to $0$.
$endgroup$
– John Douma
Jan 12 at 17:59
$begingroup$
Yes. You set the right hand side equal to $0$.
$endgroup$
– John Douma
Jan 12 at 17:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The transformation shown is a linear transformation from the space of $2 times 2$ matrices to the space of polynomials of degree two-or-less, i.e., $V$ is the vector space of $2 times 2$ matrices, and $W$ is the space of polynomials of degree two-or-less.
I echo Dietrich's comment: please post equations, not pictures. Search for "MathJax" to find out how to post equations in general.
$endgroup$
add a comment |
$begingroup$
Hint
$$Gbegin{pmatrix}a & d \ c & bend{pmatrix}=
a+frac{b+c}{2}x+frac{b-c}{2}x^2=0iff a=b=c=0$$
answered Jan 12 at 18:03
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
1
$begingroup$
That’s a lot more than a hint. It’s basically the answer to the question.
$endgroup$
– amd
Jan 12 at 21:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The transformation shown is a linear transformation from the space of $2 times 2$ matrices to the space of polynomials of degree two-or-less, i.e., $V$ is the vector space of $2 times 2$ matrices, and $W$ is the space of polynomials of degree two-or-less.
I echo Dietrich's comment: please post equations, not pictures. Search for "MathJax" to find out how to post equations in general.
$endgroup$
add a comment |
$begingroup$
The transformation shown is a linear transformation from the space of $2 times 2$ matrices to the space of polynomials of degree two-or-less, i.e., $V$ is the vector space of $2 times 2$ matrices, and $W$ is the space of polynomials of degree two-or-less.
I echo Dietrich's comment: please post equations, not pictures. Search for "MathJax" to find out how to post equations in general.
$endgroup$
add a comment |
$begingroup$
The transformation shown is a linear transformation from the space of $2 times 2$ matrices to the space of polynomials of degree two-or-less, i.e., $V$ is the vector space of $2 times 2$ matrices, and $W$ is the space of polynomials of degree two-or-less.
I echo Dietrich's comment: please post equations, not pictures. Search for "MathJax" to find out how to post equations in general.
$endgroup$
The transformation shown is a linear transformation from the space of $2 times 2$ matrices to the space of polynomials of degree two-or-less, i.e., $V$ is the vector space of $2 times 2$ matrices, and $W$ is the space of polynomials of degree two-or-less.
I echo Dietrich's comment: please post equations, not pictures. Search for "MathJax" to find out how to post equations in general.
answered Jan 12 at 17:12
community wiki
John Hughes
add a comment |
add a comment |
$begingroup$
Hint
$$Gbegin{pmatrix}a & d \ c & bend{pmatrix}=
a+frac{b+c}{2}x+frac{b-c}{2}x^2=0iff a=b=c=0$$
answered Jan 12 at 18:03
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
1
$begingroup$
That’s a lot more than a hint. It’s basically the answer to the question.
$endgroup$
– amd
Jan 12 at 21:33
add a comment |
$begingroup$
Hint
$$Gbegin{pmatrix}a & d \ c & bend{pmatrix}=
a+frac{b+c}{2}x+frac{b-c}{2}x^2=0iff a=b=c=0$$
answered Jan 12 at 18:03
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
1
$begingroup$
That’s a lot more than a hint. It’s basically the answer to the question.
$endgroup$
– amd
Jan 12 at 21:33
add a comment |
$begingroup$
Hint
$$Gbegin{pmatrix}a & d \ c & bend{pmatrix}=
a+frac{b+c}{2}x+frac{b-c}{2}x^2=0iff a=b=c=0$$
answered Jan 12 at 18:03
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Hint
$$Gbegin{pmatrix}a & d \ c & bend{pmatrix}=
a+frac{b+c}{2}x+frac{b-c}{2}x^2=0iff a=b=c=0$$
answered Jan 12 at 18:03
Mostafa AyazMostafa Ayaz
15.6k3939
edited Jan 13 at 6:19
answered Jan 12 at 18:03
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 18:03
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 18:03
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
1
$begingroup$
That’s a lot more than a hint. It’s basically the answer to the question.
$endgroup$
– amd
Jan 12 at 21:33
add a comment |
1
$begingroup$
That’s a lot more than a hint. It’s basically the answer to the question.
$endgroup$
– amd
Jan 12 at 21:33
1
1
$begingroup$
That’s a lot more than a hint. It’s basically the answer to the question.
$endgroup$
– amd
Jan 12 at 21:33
$begingroup$
That’s a lot more than a hint. It’s basically the answer to the question.
$endgroup$
– amd
Jan 12 at 21:33
add a comment |
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1
$begingroup$
Please do not use pictures.
$endgroup$
– Dietrich Burde
Jan 12 at 17:08
$begingroup$
The linear map is called $G$ and not $T$. The domain is the vector space of $2times2$ matrices, the codomain is the vector space of polynomials (of degree at most $2$). You have to prove $G$ is linear, then compute the kernel.
$endgroup$
– egreg
Jan 12 at 17:12
$begingroup$
Yes. You set the right hand side equal to $0$.
$endgroup$
– John Douma
Jan 12 at 17:59