Regarding $n$-tuples for defining points in $mathbb{R}^n$












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Is there some way to prove that exactly $n$ numbers are needed to define a point in a space $mathbb{R}^n$?



Using Cartesian systems, this is intuitive for $n=2$ and $n=3$, but how can we prove it for higher dimensions?






linear-algebra







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    read about basis
    – Sandeep Silwal
    Jan 4 at 5:03










  • What exactly do you mean by “a space $mathbb R^n$” here? Depending on how you do that, there might be nothing to prove. That aside, there are some basic theorems that make “dimension” a well-defined concept.
    – amd
    Jan 4 at 5:08








  • 1




    As far as I am concerned, the space $mathbb R^n$ is defined precisely as $underbrace{mathbb Rtimesmathbb Rtimesdotsmtimesmathbb R}_{ntext{ times}}$, which of course tells you that precisely $n$ numbers deetermine a point. Of course, to formally show that some smarter choice of representing variables won't be able to reduce the number or increase it, some linear algebra is required (in particular, the theorem that any basis of a finite dimensional vector space has the same number of vectors).
    – YiFan
    Jan 4 at 6:10


















0














Is there some way to prove that exactly $n$ numbers are needed to define a point in a space $mathbb{R}^n$?



Using Cartesian systems, this is intuitive for $n=2$ and $n=3$, but how can we prove it for higher dimensions?






linear-algebra







share|cite|improve this question









New contributor




LeoVal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    read about basis
    – Sandeep Silwal
    Jan 4 at 5:03










  • What exactly do you mean by “a space $mathbb R^n$” here? Depending on how you do that, there might be nothing to prove. That aside, there are some basic theorems that make “dimension” a well-defined concept.
    – amd
    Jan 4 at 5:08








  • 1




    As far as I am concerned, the space $mathbb R^n$ is defined precisely as $underbrace{mathbb Rtimesmathbb Rtimesdotsmtimesmathbb R}_{ntext{ times}}$, which of course tells you that precisely $n$ numbers deetermine a point. Of course, to formally show that some smarter choice of representing variables won't be able to reduce the number or increase it, some linear algebra is required (in particular, the theorem that any basis of a finite dimensional vector space has the same number of vectors).
    – YiFan
    Jan 4 at 6:10
















0












0








0







Is there some way to prove that exactly $n$ numbers are needed to define a point in a space $mathbb{R}^n$?



Using Cartesian systems, this is intuitive for $n=2$ and $n=3$, but how can we prove it for higher dimensions?






linear-algebra







share|cite|improve this question









New contributor




LeoVal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Is there some way to prove that exactly $n$ numbers are needed to define a point in a space $mathbb{R}^n$?



Using Cartesian systems, this is intuitive for $n=2$ and $n=3$, but how can we prove it for higher dimensions?






linear-algebra




linear-algebra






share|cite|improve this question









New contributor




LeoVal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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share|cite|improve this question




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edited Jan 4 at 5:12









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asked Jan 4 at 4:59









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  • 2




    read about basis
    – Sandeep Silwal
    Jan 4 at 5:03










  • What exactly do you mean by “a space $mathbb R^n$” here? Depending on how you do that, there might be nothing to prove. That aside, there are some basic theorems that make “dimension” a well-defined concept.
    – amd
    Jan 4 at 5:08








  • 1




    As far as I am concerned, the space $mathbb R^n$ is defined precisely as $underbrace{mathbb Rtimesmathbb Rtimesdotsmtimesmathbb R}_{ntext{ times}}$, which of course tells you that precisely $n$ numbers deetermine a point. Of course, to formally show that some smarter choice of representing variables won't be able to reduce the number or increase it, some linear algebra is required (in particular, the theorem that any basis of a finite dimensional vector space has the same number of vectors).
    – YiFan
    Jan 4 at 6:10
















  • 2




    read about basis
    – Sandeep Silwal
    Jan 4 at 5:03










  • What exactly do you mean by “a space $mathbb R^n$” here? Depending on how you do that, there might be nothing to prove. That aside, there are some basic theorems that make “dimension” a well-defined concept.
    – amd
    Jan 4 at 5:08








  • 1




    As far as I am concerned, the space $mathbb R^n$ is defined precisely as $underbrace{mathbb Rtimesmathbb Rtimesdotsmtimesmathbb R}_{ntext{ times}}$, which of course tells you that precisely $n$ numbers deetermine a point. Of course, to formally show that some smarter choice of representing variables won't be able to reduce the number or increase it, some linear algebra is required (in particular, the theorem that any basis of a finite dimensional vector space has the same number of vectors).
    – YiFan
    Jan 4 at 6:10










2




2




read about basis
– Sandeep Silwal
Jan 4 at 5:03




read about basis
– Sandeep Silwal
Jan 4 at 5:03












What exactly do you mean by “a space $mathbb R^n$” here? Depending on how you do that, there might be nothing to prove. That aside, there are some basic theorems that make “dimension” a well-defined concept.
– amd
Jan 4 at 5:08






What exactly do you mean by “a space $mathbb R^n$” here? Depending on how you do that, there might be nothing to prove. That aside, there are some basic theorems that make “dimension” a well-defined concept.
– amd
Jan 4 at 5:08






1




1




As far as I am concerned, the space $mathbb R^n$ is defined precisely as $underbrace{mathbb Rtimesmathbb Rtimesdotsmtimesmathbb R}_{ntext{ times}}$, which of course tells you that precisely $n$ numbers deetermine a point. Of course, to formally show that some smarter choice of representing variables won't be able to reduce the number or increase it, some linear algebra is required (in particular, the theorem that any basis of a finite dimensional vector space has the same number of vectors).
– YiFan
Jan 4 at 6:10






As far as I am concerned, the space $mathbb R^n$ is defined precisely as $underbrace{mathbb Rtimesmathbb Rtimesdotsmtimesmathbb R}_{ntext{ times}}$, which of course tells you that precisely $n$ numbers deetermine a point. Of course, to formally show that some smarter choice of representing variables won't be able to reduce the number or increase it, some linear algebra is required (in particular, the theorem that any basis of a finite dimensional vector space has the same number of vectors).
– YiFan
Jan 4 at 6:10












1 Answer
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This generally follows from the fact that an $n$-dimensional vector space needs a basis set of $n$ vectors to span it. $mathbb{R}^n$ is, of course, $n$ dimensional. The related proofs can be found in any half-decent introductory text on linear algebra.



The end result is that any point in the vector space, such as the $n$-tuples in $mathbb{R}^n$ you're concerned with, can be considered as a vector, and, as that vector is in the span of the basis set, then you can write it as a linear combination of those basis vectors.



As a simple example that touches on what I mean by this, we typically we represent points in $mathbb{R}^n$ by the $n$-tuple $(x_1, x_2, ..., x_n)$. This can be translated as being the sum of the usual basis for $mathbb{R}^n$, which is a unit vector along each axis:




  • $e_1 = (1, 0, 0, ..., 0)$

  • $e_2 = (0, 1, 0, ..., 0)$

  • $e_3 = (0, 0, 1, ..., 0)$

  • $e_k = text{vector of n components, all zero except for a 1 at the k-th place}$

  • $e_n = (0, 0, 0, ..., n)$


Thus, considering the $n$ tuple $(x_1, ..., x_n)$ as a vector,



$$(x_1, ..., x_n) = x_1e_1 + x_2e_2 + ... x_ne_n$$



Of course, this is just one way of presenting this $n$-tuple, dependent on this basis. Change the basis set - the $e_k$'s above - and you will need a different calculation. But since



$$dim(V) = n iff text{V's basis set requires n vectors}$$



for any vector space $V$ (here, $V = mathbb{R}^n$), then each point will always be the linear combination of $n$ vectors.






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    This generally follows from the fact that an $n$-dimensional vector space needs a basis set of $n$ vectors to span it. $mathbb{R}^n$ is, of course, $n$ dimensional. The related proofs can be found in any half-decent introductory text on linear algebra.



    The end result is that any point in the vector space, such as the $n$-tuples in $mathbb{R}^n$ you're concerned with, can be considered as a vector, and, as that vector is in the span of the basis set, then you can write it as a linear combination of those basis vectors.



    As a simple example that touches on what I mean by this, we typically we represent points in $mathbb{R}^n$ by the $n$-tuple $(x_1, x_2, ..., x_n)$. This can be translated as being the sum of the usual basis for $mathbb{R}^n$, which is a unit vector along each axis:




    • $e_1 = (1, 0, 0, ..., 0)$

    • $e_2 = (0, 1, 0, ..., 0)$

    • $e_3 = (0, 0, 1, ..., 0)$

    • $e_k = text{vector of n components, all zero except for a 1 at the k-th place}$

    • $e_n = (0, 0, 0, ..., n)$


    Thus, considering the $n$ tuple $(x_1, ..., x_n)$ as a vector,



    $$(x_1, ..., x_n) = x_1e_1 + x_2e_2 + ... x_ne_n$$



    Of course, this is just one way of presenting this $n$-tuple, dependent on this basis. Change the basis set - the $e_k$'s above - and you will need a different calculation. But since



    $$dim(V) = n iff text{V's basis set requires n vectors}$$



    for any vector space $V$ (here, $V = mathbb{R}^n$), then each point will always be the linear combination of $n$ vectors.






    share|cite|improve this answer


























      2














      This generally follows from the fact that an $n$-dimensional vector space needs a basis set of $n$ vectors to span it. $mathbb{R}^n$ is, of course, $n$ dimensional. The related proofs can be found in any half-decent introductory text on linear algebra.



      The end result is that any point in the vector space, such as the $n$-tuples in $mathbb{R}^n$ you're concerned with, can be considered as a vector, and, as that vector is in the span of the basis set, then you can write it as a linear combination of those basis vectors.



      As a simple example that touches on what I mean by this, we typically we represent points in $mathbb{R}^n$ by the $n$-tuple $(x_1, x_2, ..., x_n)$. This can be translated as being the sum of the usual basis for $mathbb{R}^n$, which is a unit vector along each axis:




      • $e_1 = (1, 0, 0, ..., 0)$

      • $e_2 = (0, 1, 0, ..., 0)$

      • $e_3 = (0, 0, 1, ..., 0)$

      • $e_k = text{vector of n components, all zero except for a 1 at the k-th place}$

      • $e_n = (0, 0, 0, ..., n)$


      Thus, considering the $n$ tuple $(x_1, ..., x_n)$ as a vector,



      $$(x_1, ..., x_n) = x_1e_1 + x_2e_2 + ... x_ne_n$$



      Of course, this is just one way of presenting this $n$-tuple, dependent on this basis. Change the basis set - the $e_k$'s above - and you will need a different calculation. But since



      $$dim(V) = n iff text{V's basis set requires n vectors}$$



      for any vector space $V$ (here, $V = mathbb{R}^n$), then each point will always be the linear combination of $n$ vectors.






      share|cite|improve this answer
























        2












        2








        2






        This generally follows from the fact that an $n$-dimensional vector space needs a basis set of $n$ vectors to span it. $mathbb{R}^n$ is, of course, $n$ dimensional. The related proofs can be found in any half-decent introductory text on linear algebra.



        The end result is that any point in the vector space, such as the $n$-tuples in $mathbb{R}^n$ you're concerned with, can be considered as a vector, and, as that vector is in the span of the basis set, then you can write it as a linear combination of those basis vectors.



        As a simple example that touches on what I mean by this, we typically we represent points in $mathbb{R}^n$ by the $n$-tuple $(x_1, x_2, ..., x_n)$. This can be translated as being the sum of the usual basis for $mathbb{R}^n$, which is a unit vector along each axis:




        • $e_1 = (1, 0, 0, ..., 0)$

        • $e_2 = (0, 1, 0, ..., 0)$

        • $e_3 = (0, 0, 1, ..., 0)$

        • $e_k = text{vector of n components, all zero except for a 1 at the k-th place}$

        • $e_n = (0, 0, 0, ..., n)$


        Thus, considering the $n$ tuple $(x_1, ..., x_n)$ as a vector,



        $$(x_1, ..., x_n) = x_1e_1 + x_2e_2 + ... x_ne_n$$



        Of course, this is just one way of presenting this $n$-tuple, dependent on this basis. Change the basis set - the $e_k$'s above - and you will need a different calculation. But since



        $$dim(V) = n iff text{V's basis set requires n vectors}$$



        for any vector space $V$ (here, $V = mathbb{R}^n$), then each point will always be the linear combination of $n$ vectors.






        share|cite|improve this answer












        This generally follows from the fact that an $n$-dimensional vector space needs a basis set of $n$ vectors to span it. $mathbb{R}^n$ is, of course, $n$ dimensional. The related proofs can be found in any half-decent introductory text on linear algebra.



        The end result is that any point in the vector space, such as the $n$-tuples in $mathbb{R}^n$ you're concerned with, can be considered as a vector, and, as that vector is in the span of the basis set, then you can write it as a linear combination of those basis vectors.



        As a simple example that touches on what I mean by this, we typically we represent points in $mathbb{R}^n$ by the $n$-tuple $(x_1, x_2, ..., x_n)$. This can be translated as being the sum of the usual basis for $mathbb{R}^n$, which is a unit vector along each axis:




        • $e_1 = (1, 0, 0, ..., 0)$

        • $e_2 = (0, 1, 0, ..., 0)$

        • $e_3 = (0, 0, 1, ..., 0)$

        • $e_k = text{vector of n components, all zero except for a 1 at the k-th place}$

        • $e_n = (0, 0, 0, ..., n)$


        Thus, considering the $n$ tuple $(x_1, ..., x_n)$ as a vector,



        $$(x_1, ..., x_n) = x_1e_1 + x_2e_2 + ... x_ne_n$$



        Of course, this is just one way of presenting this $n$-tuple, dependent on this basis. Change the basis set - the $e_k$'s above - and you will need a different calculation. But since



        $$dim(V) = n iff text{V's basis set requires n vectors}$$



        for any vector space $V$ (here, $V = mathbb{R}^n$), then each point will always be the linear combination of $n$ vectors.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 5:11









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