Hyperplane in $mathbb{R}^n $ has dimension $n-1$











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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.










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    up vote
    1
    down vote

    favorite












    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.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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.










      share|cite|improve this question















      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






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      edited Nov 20 at 12:10









      Bernard

      116k637108




      116k637108










      asked Nov 20 at 11:43









      Diamir

      1204




      1204






















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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 $n-1$.






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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            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 $n-1$.






            share|cite|improve this answer

























              up vote
              3
              down vote



              accepted










              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 $n-1$.






              share|cite|improve this answer























                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                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 $n-1$.






                share|cite|improve this answer












                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 $n-1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 20 at 11:48









                астон вілла олоф мэллбэрг

                36.9k33376




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