Mixing
Hyperplane in $mathbb{R}^n $ has dimension $n-1$
A hyperplane in $mathbb{R}^n$ is given by
begin{align}
H = {x in mathbb{R}^n : x = p + a_1 y_1 + dots + a_{n-1} y_{n-1}; a_1,dots,a_{n-1} in mathbb{R}}
end{align}
such that $p in mathbb{R}^n$ and $y_1,dots,y_{n-1} in mathbb{R}^n$ are linearly independent. Now the claim is that this hyperplane is $n-1$ dimensional. But why is this so? For example could $p$ be linearly independent from $y_1,dots,y_{n-1}$ and hence the hyperplane would be $n$ dimensional.
real-analysis linear-algebra
edited Nov 20 '18 at 12:10
Bernard
118k639112
add a comment |
A hyperplane in $mathbb{R}^n$ is given by
begin{align}
H = {x in mathbb{R}^n : x = p + a_1 y_1 + dots + a_{n-1} y_{n-1}; a_1,dots,a_{n-1} in mathbb{R}}
end{align}
such that $p in mathbb{R}^n$ and $y_1,dots,y_{n-1} in mathbb{R}^n$ are linearly independent. Now the claim is that this hyperplane is $n-1$ dimensional. But why is this so? For example could $p$ be linearly independent from $y_1,dots,y_{n-1}$ and hence the hyperplane would be $n$ dimensional.
real-analysis linear-algebra
edited Nov 20 '18 at 12:10
Bernard
118k639112
add a comment |
A hyperplane in $mathbb{R}^n$ is given by
begin{align}
H = {x in mathbb{R}^n : x = p + a_1 y_1 + dots + a_{n-1} y_{n-1}; a_1,dots,a_{n-1} in mathbb{R}}
end{align}
such that $p in mathbb{R}^n$ and $y_1,dots,y_{n-1} in mathbb{R}^n$ are linearly independent. Now the claim is that this hyperplane is $n-1$ dimensional. But why is this so? For example could $p$ be linearly independent from $y_1,dots,y_{n-1}$ and hence the hyperplane would be $n$ dimensional.
real-analysis linear-algebra
edited Nov 20 '18 at 12:10
Bernard
118k639112
A hyperplane in $mathbb{R}^n$ is given by
begin{align}
H = {x in mathbb{R}^n : x = p + a_1 y_1 + dots + a_{n-1} y_{n-1}; a_1,dots,a_{n-1} in mathbb{R}}
end{align}
such that $p in mathbb{R}^n$ and $y_1,dots,y_{n-1} in mathbb{R}^n$ are linearly independent. Now the claim is that this hyperplane is $n-1$ dimensional. But why is this so? For example could $p$ be linearly independent from $y_1,dots,y_{n-1}$ and hence the hyperplane would be $n$ dimensional.
real-analysis linear-algebra
real-analysis linear-algebra
edited Nov 20 '18 at 12:10
Bernard
118k639112
edited Nov 20 '18 at 12:10
Bernard
118k639112
edited Nov 20 '18 at 12:10
Bernard
118k639112
edited Nov 20 '18 at 12:10
Bernard
118k639112
edited Nov 20 '18 at 12:10
Bernard
118k639112
118k639112
asked Nov 20 '18 at 11:43
user613181
add a comment |
add a comment |
1 Answer
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Even though $p$ is linearly independent from the set $y_i$, we are not taking linear combinations of $p$ : as you can see , you cannot take scalar combinations of $p$, like $2p$ and so on : these will not belong to the hyperplane.
In other words, $x = p+sum a_iy_i$ is spanned by only $n-1$ elements. It is not a subspace : $0$ will not belong to it if $p neq 0$. However, the dimension of the subspace spanned by the $y_i$ is $n-1$, and this hyperplane is a translate (to $p$) of that subspace, hence as an affine space also has dimension $n-1$.
answered Nov 20 '18 at 11:48
астон вілла олоф мэллбэрг
37.3k33376
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1 Answer
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Even though $p$ is linearly independent from the set $y_i$, we are not taking linear combinations of $p$ : as you can see , you cannot take scalar combinations of $p$, like $2p$ and so on : these will not belong to the hyperplane.
In other words, $x = p+sum a_iy_i$ is spanned by only $n-1$ elements. It is not a subspace : $0$ will not belong to it if $p neq 0$. However, the dimension of the subspace spanned by the $y_i$ is $n-1$, and this hyperplane is a translate (to $p$) of that subspace, hence as an affine space also has dimension $n-1$.
answered Nov 20 '18 at 11:48
астон вілла олоф мэллбэрг
37.3k33376
add a comment |
Even though $p$ is linearly independent from the set $y_i$, we are not taking linear combinations of $p$ : as you can see , you cannot take scalar combinations of $p$, like $2p$ and so on : these will not belong to the hyperplane.
In other words, $x = p+sum a_iy_i$ is spanned by only $n-1$ elements. It is not a subspace : $0$ will not belong to it if $p neq 0$. However, the dimension of the subspace spanned by the $y_i$ is $n-1$, and this hyperplane is a translate (to $p$) of that subspace, hence as an affine space also has dimension $n-1$.
answered Nov 20 '18 at 11:48
астон вілла олоф мэллбэрг
37.3k33376
add a comment |
Even though $p$ is linearly independent from the set $y_i$, we are not taking linear combinations of $p$ : as you can see , you cannot take scalar combinations of $p$, like $2p$ and so on : these will not belong to the hyperplane.
In other words, $x = p+sum a_iy_i$ is spanned by only $n-1$ elements. It is not a subspace : $0$ will not belong to it if $p neq 0$. However, the dimension of the subspace spanned by the $y_i$ is $n-1$, and this hyperplane is a translate (to $p$) of that subspace, hence as an affine space also has dimension $n-1$.
answered Nov 20 '18 at 11:48
астон вілла олоф мэллбэрг
37.3k33376
Even though $p$ is linearly independent from the set $y_i$, we are not taking linear combinations of $p$ : as you can see , you cannot take scalar combinations of $p$, like $2p$ and so on : these will not belong to the hyperplane.
In other words, $x = p+sum a_iy_i$ is spanned by only $n-1$ elements. It is not a subspace : $0$ will not belong to it if $p neq 0$. However, the dimension of the subspace spanned by the $y_i$ is $n-1$, and this hyperplane is a translate (to $p$) of that subspace, hence as an affine space also has dimension $n-1$.
answered Nov 20 '18 at 11:48
астон вілла олоф мэллбэрг
37.3k33376
answered Nov 20 '18 at 11:48
астон вілла олоф мэллбэрг
37.3k33376
answered Nov 20 '18 at 11:48
астон вілла олоф мэллбэрг
37.3k33376
answered Nov 20 '18 at 11:48
астон вілла олоф мэллбэрг
37.3k33376
37.3k33376
add a comment |
add a comment |
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