Let $V$ be a subspace of $mathbb{R}^4$, spanned by $v$ and $u$. Find a linear transformation whose kernel is...
$begingroup$
And the vectors given are $v = (1,0,3,-2)$ and $u = (0,1,4,1)$.
It asks me to find the linear transformation from $mathbb{R}^4$ to $mathbb{R}^2$, where the kernel of that transformation is $V$.
So what I know is that: the transformation I'm trying to find, applied to every vector in the span of $(1,0,3,-2)$ and $(0,1,4,1)$, will give the zero vector.
Please let me know if that interpretation is incorrect.
I've really no idea how to get started on this question. I have the equation $Av = 0$ where $A$ is the matrix of the transformation in question, and v is any vector of the subspace V, but...I don't think that gets me anywhere. Any help is greatly appreciated.
linear-algebra linear-transformations
edited Dec 16 '18 at 2:02
amWhy
1
asked Dec 16 '18 at 1:25
James RonaldJames Ronald
1257
$endgroup$
add a comment |
$begingroup$
And the vectors given are $v = (1,0,3,-2)$ and $u = (0,1,4,1)$.
It asks me to find the linear transformation from $mathbb{R}^4$ to $mathbb{R}^2$, where the kernel of that transformation is $V$.
So what I know is that: the transformation I'm trying to find, applied to every vector in the span of $(1,0,3,-2)$ and $(0,1,4,1)$, will give the zero vector.
Please let me know if that interpretation is incorrect.
I've really no idea how to get started on this question. I have the equation $Av = 0$ where $A$ is the matrix of the transformation in question, and v is any vector of the subspace V, but...I don't think that gets me anywhere. Any help is greatly appreciated.
linear-algebra linear-transformations
edited Dec 16 '18 at 2:02
amWhy
1
asked Dec 16 '18 at 1:25
James RonaldJames Ronald
1257
$endgroup$
$begingroup$
Your interpretation is correct, and more specifically, every vector whose image is the zero vector must be an element of $V$.
$endgroup$
– Anthony Ter
Dec 16 '18 at 2:17
$begingroup$
You can get some relation on the coefficients in a $4times 2$ matrix from the fact that it is multiplied by 2 independent vectors to be 0.
$endgroup$
– NL1992
Dec 16 '18 at 2:18
1
$begingroup$
There is an infinite number of solutions. Hint: the row space of a matrix is the orthogonal complement of its kernel.
$endgroup$
– amd
Dec 16 '18 at 6:02
add a comment |
$begingroup$
And the vectors given are $v = (1,0,3,-2)$ and $u = (0,1,4,1)$.
It asks me to find the linear transformation from $mathbb{R}^4$ to $mathbb{R}^2$, where the kernel of that transformation is $V$.
So what I know is that: the transformation I'm trying to find, applied to every vector in the span of $(1,0,3,-2)$ and $(0,1,4,1)$, will give the zero vector.
Please let me know if that interpretation is incorrect.
I've really no idea how to get started on this question. I have the equation $Av = 0$ where $A$ is the matrix of the transformation in question, and v is any vector of the subspace V, but...I don't think that gets me anywhere. Any help is greatly appreciated.
linear-algebra linear-transformations
edited Dec 16 '18 at 2:02
amWhy
1
asked Dec 16 '18 at 1:25
James RonaldJames Ronald
1257
$endgroup$
And the vectors given are $v = (1,0,3,-2)$ and $u = (0,1,4,1)$.
It asks me to find the linear transformation from $mathbb{R}^4$ to $mathbb{R}^2$, where the kernel of that transformation is $V$.
So what I know is that: the transformation I'm trying to find, applied to every vector in the span of $(1,0,3,-2)$ and $(0,1,4,1)$, will give the zero vector.
Please let me know if that interpretation is incorrect.
I've really no idea how to get started on this question. I have the equation $Av = 0$ where $A$ is the matrix of the transformation in question, and v is any vector of the subspace V, but...I don't think that gets me anywhere. Any help is greatly appreciated.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Dec 16 '18 at 2:02
amWhy
1
asked Dec 16 '18 at 1:25
James RonaldJames Ronald
1257
edited Dec 16 '18 at 2:02
amWhy
1
asked Dec 16 '18 at 1:25
James RonaldJames Ronald
1257
edited Dec 16 '18 at 2:02
amWhy
1
edited Dec 16 '18 at 2:02
amWhy
1
edited Dec 16 '18 at 2:02
amWhy
1
1
asked Dec 16 '18 at 1:25
James RonaldJames Ronald
1257
asked Dec 16 '18 at 1:25
James RonaldJames Ronald
1257
asked Dec 16 '18 at 1:25
James RonaldJames Ronald
1257
1257
$begingroup$
Your interpretation is correct, and more specifically, every vector whose image is the zero vector must be an element of $V$.
$endgroup$
– Anthony Ter
Dec 16 '18 at 2:17
$begingroup$
You can get some relation on the coefficients in a $4times 2$ matrix from the fact that it is multiplied by 2 independent vectors to be 0.
$endgroup$
– NL1992
Dec 16 '18 at 2:18
1
$begingroup$
There is an infinite number of solutions. Hint: the row space of a matrix is the orthogonal complement of its kernel.
$endgroup$
– amd
Dec 16 '18 at 6:02
add a comment |
$begingroup$
Your interpretation is correct, and more specifically, every vector whose image is the zero vector must be an element of $V$.
$endgroup$
– Anthony Ter
Dec 16 '18 at 2:17
$begingroup$
You can get some relation on the coefficients in a $4times 2$ matrix from the fact that it is multiplied by 2 independent vectors to be 0.
$endgroup$
– NL1992
Dec 16 '18 at 2:18
1
$begingroup$
There is an infinite number of solutions. Hint: the row space of a matrix is the orthogonal complement of its kernel.
$endgroup$
– amd
Dec 16 '18 at 6:02
$begingroup$
Your interpretation is correct, and more specifically, every vector whose image is the zero vector must be an element of $V$.
$endgroup$
– Anthony Ter
Dec 16 '18 at 2:17
$begingroup$
Your interpretation is correct, and more specifically, every vector whose image is the zero vector must be an element of $V$.
$endgroup$
– Anthony Ter
Dec 16 '18 at 2:17
$begingroup$
You can get some relation on the coefficients in a $4times 2$ matrix from the fact that it is multiplied by 2 independent vectors to be 0.
$endgroup$
– NL1992
Dec 16 '18 at 2:18
$begingroup$
You can get some relation on the coefficients in a $4times 2$ matrix from the fact that it is multiplied by 2 independent vectors to be 0.
$endgroup$
– NL1992
Dec 16 '18 at 2:18
1
1
$begingroup$
There is an infinite number of solutions. Hint: the row space of a matrix is the orthogonal complement of its kernel.
$endgroup$
– amd
Dec 16 '18 at 6:02
$begingroup$
There is an infinite number of solutions. Hint: the row space of a matrix is the orthogonal complement of its kernel.
$endgroup$
– amd
Dec 16 '18 at 6:02
add a comment |
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$begingroup$
Your interpretation is correct, and more specifically, every vector whose image is the zero vector must be an element of $V$.
$endgroup$
– Anthony Ter
Dec 16 '18 at 2:17
$begingroup$
You can get some relation on the coefficients in a $4times 2$ matrix from the fact that it is multiplied by 2 independent vectors to be 0.
$endgroup$
– NL1992
Dec 16 '18 at 2:18
1
$begingroup$
There is an infinite number of solutions. Hint: the row space of a matrix is the orthogonal complement of its kernel.
$endgroup$
– amd
Dec 16 '18 at 6:02