Hyperplanes and Linear Dependence
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In one of books that I'm studying, I've come across the following proposition:
"Let $x_0, x_1, ...,x_n$ be $n+1$ points in $R^n$ that are contained in no hyperplane. Define
beginalign* y_i=x_i - x_0 endalign* for $i=1,2,...,n$. Then $y_1, ...,y_n$ are linearly independent."
Just in case, a hyperplane $H(p,alpha)$ where $p in R^n$ and $alpha$ is scalar is defined in my book as follows:
$$H(p,alpha) :=biglxin R^n,bigm $$
My book attempted to prove it using the contrapositive argument: if $y_1, ..., y_n$ are assumed linearly dependent, then there are $n$ scalars $p_1, ..., p_n$ such that
$$p_1 y_1 +p_2y_2 + ... +p_ny_n = bar0$$
I'm stuck after this point of the proof. The book says that the vector $p=(p_1,p_2,...,p_n)$ formed by using the scalars $p_1, .. ,p_n$ satisfies that $pcdot y_i = 0$ for $i=1,2,...n$, which in turn implies that $pcdot x_i =p cdot x_0$ thus $x_0, x_1,...,x_n$ are contained by the same hyperplane $H(p,p cdot x_0)$. But I couldn't see that the vector $p$ satisfies $pcdot y_i = 0$ for all $i=1,2,...n$, linear dependence can't imply it. Is the book's proof correct or is there something missing?
Thanks!
linear-algebra
asked Aug 7 at 16:25
Osman Bulut
536
add a comment |Â
up vote
2
down vote
favorite
In one of books that I'm studying, I've come across the following proposition:
"Let $x_0, x_1, ...,x_n$ be $n+1$ points in $R^n$ that are contained in no hyperplane. Define
beginalign* y_i=x_i - x_0 endalign* for $i=1,2,...,n$. Then $y_1, ...,y_n$ are linearly independent."
Just in case, a hyperplane $H(p,alpha)$ where $p in R^n$ and $alpha$ is scalar is defined in my book as follows:
$$H(p,alpha) :=biglxin R^n,bigm $$
My book attempted to prove it using the contrapositive argument: if $y_1, ..., y_n$ are assumed linearly dependent, then there are $n$ scalars $p_1, ..., p_n$ such that
$$p_1 y_1 +p_2y_2 + ... +p_ny_n = bar0$$
I'm stuck after this point of the proof. The book says that the vector $p=(p_1,p_2,...,p_n)$ formed by using the scalars $p_1, .. ,p_n$ satisfies that $pcdot y_i = 0$ for $i=1,2,...n$, which in turn implies that $pcdot x_i =p cdot x_0$ thus $x_0, x_1,...,x_n$ are contained by the same hyperplane $H(p,p cdot x_0)$. But I couldn't see that the vector $p$ satisfies $pcdot y_i = 0$ for all $i=1,2,...n$, linear dependence can't imply it. Is the book's proof correct or is there something missing?
Thanks!
linear-algebra
asked Aug 7 at 16:25
Osman Bulut
536
That statement doesn't follow, as you said. Are you sure you haven't left something out? Consider the $y_i$ as the columns of a matrix $Y$ and consider $p$ as column vector. Then we have $Ap=mathbf0,$ which means that the dot products of $p$ with the transposes of the rows of $Y$ are all $0$, but that's not the claim made in your question. I don't see how to rescue this argument, but maybe I'm missing something simple. If you don't get a resolution here, you should certainly take this question to your instructor.
– saulspatz
Aug 7 at 18:10
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In one of books that I'm studying, I've come across the following proposition:
"Let $x_0, x_1, ...,x_n$ be $n+1$ points in $R^n$ that are contained in no hyperplane. Define
beginalign* y_i=x_i - x_0 endalign* for $i=1,2,...,n$. Then $y_1, ...,y_n$ are linearly independent."
Just in case, a hyperplane $H(p,alpha)$ where $p in R^n$ and $alpha$ is scalar is defined in my book as follows:
$$H(p,alpha) :=biglxin R^n,bigm $$
My book attempted to prove it using the contrapositive argument: if $y_1, ..., y_n$ are assumed linearly dependent, then there are $n$ scalars $p_1, ..., p_n$ such that
$$p_1 y_1 +p_2y_2 + ... +p_ny_n = bar0$$
I'm stuck after this point of the proof. The book says that the vector $p=(p_1,p_2,...,p_n)$ formed by using the scalars $p_1, .. ,p_n$ satisfies that $pcdot y_i = 0$ for $i=1,2,...n$, which in turn implies that $pcdot x_i =p cdot x_0$ thus $x_0, x_1,...,x_n$ are contained by the same hyperplane $H(p,p cdot x_0)$. But I couldn't see that the vector $p$ satisfies $pcdot y_i = 0$ for all $i=1,2,...n$, linear dependence can't imply it. Is the book's proof correct or is there something missing?
Thanks!
linear-algebra
asked Aug 7 at 16:25
Osman Bulut
536
In one of books that I'm studying, I've come across the following proposition:
"Let $x_0, x_1, ...,x_n$ be $n+1$ points in $R^n$ that are contained in no hyperplane. Define
beginalign* y_i=x_i - x_0 endalign* for $i=1,2,...,n$. Then $y_1, ...,y_n$ are linearly independent."
Just in case, a hyperplane $H(p,alpha)$ where $p in R^n$ and $alpha$ is scalar is defined in my book as follows:
$$H(p,alpha) :=biglxin R^n,bigm $$
My book attempted to prove it using the contrapositive argument: if $y_1, ..., y_n$ are assumed linearly dependent, then there are $n$ scalars $p_1, ..., p_n$ such that
$$p_1 y_1 +p_2y_2 + ... +p_ny_n = bar0$$
I'm stuck after this point of the proof. The book says that the vector $p=(p_1,p_2,...,p_n)$ formed by using the scalars $p_1, .. ,p_n$ satisfies that $pcdot y_i = 0$ for $i=1,2,...n$, which in turn implies that $pcdot x_i =p cdot x_0$ thus $x_0, x_1,...,x_n$ are contained by the same hyperplane $H(p,p cdot x_0)$. But I couldn't see that the vector $p$ satisfies $pcdot y_i = 0$ for all $i=1,2,...n$, linear dependence can't imply it. Is the book's proof correct or is there something missing?
Thanks!
linear-algebra
asked Aug 7 at 16:25
Osman Bulut
536
asked Aug 7 at 16:25
Osman Bulut
536
asked Aug 7 at 16:25
Osman Bulut
536
asked Aug 7 at 16:25
Osman Bulut
536
536
That statement doesn't follow, as you said. Are you sure you haven't left something out? Consider the $y_i$ as the columns of a matrix $Y$ and consider $p$ as column vector. Then we have $Ap=mathbf0,$ which means that the dot products of $p$ with the transposes of the rows of $Y$ are all $0$, but that's not the claim made in your question. I don't see how to rescue this argument, but maybe I'm missing something simple. If you don't get a resolution here, you should certainly take this question to your instructor.
– saulspatz
Aug 7 at 18:10
add a comment |Â
That statement doesn't follow, as you said. Are you sure you haven't left something out? Consider the $y_i$ as the columns of a matrix $Y$ and consider $p$ as column vector. Then we have $Ap=mathbf0,$ which means that the dot products of $p$ with the transposes of the rows of $Y$ are all $0$, but that's not the claim made in your question. I don't see how to rescue this argument, but maybe I'm missing something simple. If you don't get a resolution here, you should certainly take this question to your instructor.
– saulspatz
Aug 7 at 18:10
That statement doesn't follow, as you said. Are you sure you haven't left something out? Consider the $y_i$ as the columns of a matrix $Y$ and consider $p$ as column vector. Then we have $Ap=mathbf0,$ which means that the dot products of $p$ with the transposes of the rows of $Y$ are all $0$, but that's not the claim made in your question. I don't see how to rescue this argument, but maybe I'm missing something simple. If you don't get a resolution here, you should certainly take this question to your instructor.
– saulspatz
Aug 7 at 18:10
That statement doesn't follow, as you said. Are you sure you haven't left something out? Consider the $y_i$ as the columns of a matrix $Y$ and consider $p$ as column vector. Then we have $Ap=mathbf0,$ which means that the dot products of $p$ with the transposes of the rows of $Y$ are all $0$, but that's not the claim made in your question. I don't see how to rescue this argument, but maybe I'm missing something simple. If you don't get a resolution here, you should certainly take this question to your instructor.
– saulspatz
Aug 7 at 18:10
add a comment |Â
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That statement doesn't follow, as you said. Are you sure you haven't left something out? Consider the $y_i$ as the columns of a matrix $Y$ and consider $p$ as column vector. Then we have $Ap=mathbf0,$ which means that the dot products of $p$ with the transposes of the rows of $Y$ are all $0$, but that's not the claim made in your question. I don't see how to rescue this argument, but maybe I'm missing something simple. If you don't get a resolution here, you should certainly take this question to your instructor.
– saulspatz
Aug 7 at 18:10