Find locus of points by finding eigenvalues











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Let $boldsymbol{x}=left(begin{matrix}x\ yend{matrix}right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $boldsymbol{M}$, sketch the locus of points $boldsymbol{x}$ that satisfy $$ boldsymbol{x^TMx}=4$$
given that
$$boldsymbol{M}=left(begin{matrix}&5 &sqrt{3}\ &sqrt{3} &3end{matrix}right). $$




I found two eigenvalues to be $lambda_1 = 6$ and $lambda_2=2$, and the corresponding eigenvectors are
$$ boldsymbol{v}_1=left(begin{matrix}sqrt{3}\ 1end{matrix}right)quadtext{ and }quad boldsymbol{v}_2=left(begin{matrix}1\ -sqrt{3}end{matrix}right)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $boldsymbol{x^TMx}=4$.



Any hints?






linear-algebra vector-spaces eigenvalues-eigenvectors







share|cite|improve this question
























  • The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
    – Shubham Johri
    1 hour ago












  • @ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
    – VanDerWarden
    1 hour ago










  • Yes, now the eigenvalues and eigenvectors are fine
    – Shubham Johri
    1 hour ago















up vote
4
down vote

favorite
3













Let $boldsymbol{x}=left(begin{matrix}x\ yend{matrix}right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $boldsymbol{M}$, sketch the locus of points $boldsymbol{x}$ that satisfy $$ boldsymbol{x^TMx}=4$$
given that
$$boldsymbol{M}=left(begin{matrix}&5 &sqrt{3}\ &sqrt{3} &3end{matrix}right). $$




I found two eigenvalues to be $lambda_1 = 6$ and $lambda_2=2$, and the corresponding eigenvectors are
$$ boldsymbol{v}_1=left(begin{matrix}sqrt{3}\ 1end{matrix}right)quadtext{ and }quad boldsymbol{v}_2=left(begin{matrix}1\ -sqrt{3}end{matrix}right)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $boldsymbol{x^TMx}=4$.



Any hints?






linear-algebra vector-spaces eigenvalues-eigenvectors







share|cite|improve this question
























  • The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
    – Shubham Johri
    1 hour ago












  • @ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
    – VanDerWarden
    1 hour ago










  • Yes, now the eigenvalues and eigenvectors are fine
    – Shubham Johri
    1 hour ago













up vote
4
down vote

favorite
3









up vote
4
down vote

favorite
3






3






Let $boldsymbol{x}=left(begin{matrix}x\ yend{matrix}right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $boldsymbol{M}$, sketch the locus of points $boldsymbol{x}$ that satisfy $$ boldsymbol{x^TMx}=4$$
given that
$$boldsymbol{M}=left(begin{matrix}&5 &sqrt{3}\ &sqrt{3} &3end{matrix}right). $$




I found two eigenvalues to be $lambda_1 = 6$ and $lambda_2=2$, and the corresponding eigenvectors are
$$ boldsymbol{v}_1=left(begin{matrix}sqrt{3}\ 1end{matrix}right)quadtext{ and }quad boldsymbol{v}_2=left(begin{matrix}1\ -sqrt{3}end{matrix}right)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $boldsymbol{x^TMx}=4$.



Any hints?






linear-algebra vector-spaces eigenvalues-eigenvectors







share|cite|improve this question
















Let $boldsymbol{x}=left(begin{matrix}x\ yend{matrix}right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $boldsymbol{M}$, sketch the locus of points $boldsymbol{x}$ that satisfy $$ boldsymbol{x^TMx}=4$$
given that
$$boldsymbol{M}=left(begin{matrix}&5 &sqrt{3}\ &sqrt{3} &3end{matrix}right). $$




I found two eigenvalues to be $lambda_1 = 6$ and $lambda_2=2$, and the corresponding eigenvectors are
$$ boldsymbol{v}_1=left(begin{matrix}sqrt{3}\ 1end{matrix}right)quadtext{ and }quad boldsymbol{v}_2=left(begin{matrix}1\ -sqrt{3}end{matrix}right)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $boldsymbol{x^TMx}=4$.



Any hints?






linear-algebra vector-spaces eigenvalues-eigenvectors




linear-algebra vector-spaces eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 38 secs ago









bubba

29.6k32985




29.6k32985










asked 1 hour ago









VanDerWarden

577629




577629












  • The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
    – Shubham Johri
    1 hour ago












  • @ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
    – VanDerWarden
    1 hour ago










  • Yes, now the eigenvalues and eigenvectors are fine
    – Shubham Johri
    1 hour ago


















  • The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
    – Shubham Johri
    1 hour ago












  • @ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
    – VanDerWarden
    1 hour ago










  • Yes, now the eigenvalues and eigenvectors are fine
    – Shubham Johri
    1 hour ago
















The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
– Shubham Johri
1 hour ago






The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
– Shubham Johri
1 hour ago














@ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
– VanDerWarden
1 hour ago




@ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
– VanDerWarden
1 hour ago












Yes, now the eigenvalues and eigenvectors are fine
– Shubham Johri
1 hour ago




Yes, now the eigenvalues and eigenvectors are fine
– Shubham Johri
1 hour ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.



$mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$



$implies6c_1^2+2c_2^2=1$






share|cite|improve this answer






























    up vote
    5
    down vote













    Some hints (as you requested):




    • What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).


    • After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.


    • The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates


    • In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.







    share|cite|improve this answer























    • Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
      – Anvit
      1 hour ago










    • @Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
      – bubba
      1 hour ago












    • @bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
      – AmbretteOrrisey
      57 mins ago











    Your Answer





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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.



    $mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$



    $implies6c_1^2+2c_2^2=1$






    share|cite|improve this answer



























      up vote
      3
      down vote



      accepted










      The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.



      $mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$



      $implies6c_1^2+2c_2^2=1$






      share|cite|improve this answer

























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.



        $mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$



        $implies6c_1^2+2c_2^2=1$






        share|cite|improve this answer














        The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.



        $mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$



        $implies6c_1^2+2c_2^2=1$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        Shubham Johri

        2,648413




        2,648413






















            up vote
            5
            down vote













            Some hints (as you requested):




            • What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).


            • After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.


            • The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates


            • In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.







            share|cite|improve this answer























            • Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
              – Anvit
              1 hour ago










            • @Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
              – bubba
              1 hour ago












            • @bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
              – AmbretteOrrisey
              57 mins ago















            up vote
            5
            down vote













            Some hints (as you requested):




            • What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).


            • After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.


            • The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates


            • In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.







            share|cite|improve this answer























            • Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
              – Anvit
              1 hour ago










            • @Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
              – bubba
              1 hour ago












            • @bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
              – AmbretteOrrisey
              57 mins ago













            up vote
            5
            down vote










            up vote
            5
            down vote









            Some hints (as you requested):




            • What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).


            • After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.


            • The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates


            • In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.







            share|cite|improve this answer














            Some hints (as you requested):




            • What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).


            • After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.


            • The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates


            • In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 min ago

























            answered 1 hour ago









            bubba

            29.6k32985




            29.6k32985












            • Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
              – Anvit
              1 hour ago










            • @Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
              – bubba
              1 hour ago












            • @bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
              – AmbretteOrrisey
              57 mins ago


















            • Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
              – Anvit
              1 hour ago










            • @Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
              – bubba
              1 hour ago












            • @bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
              – AmbretteOrrisey
              57 mins ago
















            Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
            – Anvit
            1 hour ago




            Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
            – Anvit
            1 hour ago












            @Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
            – bubba
            1 hour ago






            @Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
            – bubba
            1 hour ago














            @bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
            – AmbretteOrrisey
            57 mins ago




            @bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
            – AmbretteOrrisey
            57 mins ago


















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