Show that $C={x in mathbb{R}^4 mid x^TAx geq 1}$ is not empty?












2












$begingroup$


Consider the set $C={x in mathbb{R}^4 mid x^TAx geq 1}$ where $A in mathbb{R}^{4 times 4}$ is a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.



Show $C$ is not empty?



My try:



$A$ is symmetric, so it can be written as $A=u Lambda u^T$. Hence,



$$
x^TAx=x^Tu Lambda u^Tx geq 1
$$



So



$$
z^T Lambda z geq 1
$$
where $u^Tx=z in mathbb{R}^4$.
Let $a_1,a_2,a_3,a_4$ be eigenvalues of $A$ so $a_1z_1^2+a_2z_2^2+a_3z_3^2+a_4z_4^2 -1 geq 0$. Since two of the eigenvalues are nonpositive, assume they are zero. Hence, $a_1z_1^2+a_2z_2^2 -1 geq 0$ which is the region outside of an ellipsoid in a 2D plain.
Am I write, or there is a more beautiful way to show it?






linear-algebra inequality convex-analysis symmetric-matrices constraints







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$endgroup$








  • 2




    $begingroup$
    Why don't you just take a vector $x$ with $x^T Ax> 0$ and scale it such that it is in $C$?
    $endgroup$
    – sehigle
    Dec 17 '18 at 0:50










  • $begingroup$
    Suppose we make the eigenvector of a positive eigenvalue long enough. Then it's an element of C isn't it?
    $endgroup$
    – I like Serena
    Dec 17 '18 at 0:52
















2












$begingroup$


Consider the set $C={x in mathbb{R}^4 mid x^TAx geq 1}$ where $A in mathbb{R}^{4 times 4}$ is a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.



Show $C$ is not empty?



My try:



$A$ is symmetric, so it can be written as $A=u Lambda u^T$. Hence,



$$
x^TAx=x^Tu Lambda u^Tx geq 1
$$



So



$$
z^T Lambda z geq 1
$$
where $u^Tx=z in mathbb{R}^4$.
Let $a_1,a_2,a_3,a_4$ be eigenvalues of $A$ so $a_1z_1^2+a_2z_2^2+a_3z_3^2+a_4z_4^2 -1 geq 0$. Since two of the eigenvalues are nonpositive, assume they are zero. Hence, $a_1z_1^2+a_2z_2^2 -1 geq 0$ which is the region outside of an ellipsoid in a 2D plain.
Am I write, or there is a more beautiful way to show it?






linear-algebra inequality convex-analysis symmetric-matrices constraints







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Why don't you just take a vector $x$ with $x^T Ax> 0$ and scale it such that it is in $C$?
    $endgroup$
    – sehigle
    Dec 17 '18 at 0:50










  • $begingroup$
    Suppose we make the eigenvector of a positive eigenvalue long enough. Then it's an element of C isn't it?
    $endgroup$
    – I like Serena
    Dec 17 '18 at 0:52














2












2








2


1



$begingroup$


Consider the set $C={x in mathbb{R}^4 mid x^TAx geq 1}$ where $A in mathbb{R}^{4 times 4}$ is a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.



Show $C$ is not empty?



My try:



$A$ is symmetric, so it can be written as $A=u Lambda u^T$. Hence,



$$
x^TAx=x^Tu Lambda u^Tx geq 1
$$



So



$$
z^T Lambda z geq 1
$$
where $u^Tx=z in mathbb{R}^4$.
Let $a_1,a_2,a_3,a_4$ be eigenvalues of $A$ so $a_1z_1^2+a_2z_2^2+a_3z_3^2+a_4z_4^2 -1 geq 0$. Since two of the eigenvalues are nonpositive, assume they are zero. Hence, $a_1z_1^2+a_2z_2^2 -1 geq 0$ which is the region outside of an ellipsoid in a 2D plain.
Am I write, or there is a more beautiful way to show it?






linear-algebra inequality convex-analysis symmetric-matrices constraints







share|cite|improve this question









$endgroup$




Consider the set $C={x in mathbb{R}^4 mid x^TAx geq 1}$ where $A in mathbb{R}^{4 times 4}$ is a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.



Show $C$ is not empty?



My try:



$A$ is symmetric, so it can be written as $A=u Lambda u^T$. Hence,



$$
x^TAx=x^Tu Lambda u^Tx geq 1
$$



So



$$
z^T Lambda z geq 1
$$
where $u^Tx=z in mathbb{R}^4$.
Let $a_1,a_2,a_3,a_4$ be eigenvalues of $A$ so $a_1z_1^2+a_2z_2^2+a_3z_3^2+a_4z_4^2 -1 geq 0$. Since two of the eigenvalues are nonpositive, assume they are zero. Hence, $a_1z_1^2+a_2z_2^2 -1 geq 0$ which is the region outside of an ellipsoid in a 2D plain.
Am I write, or there is a more beautiful way to show it?






linear-algebra inequality convex-analysis symmetric-matrices constraints




linear-algebra inequality convex-analysis symmetric-matrices constraints






share|cite|improve this question













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asked Dec 17 '18 at 0:41









SepideSepide

3038




3038








  • 2




    $begingroup$
    Why don't you just take a vector $x$ with $x^T Ax> 0$ and scale it such that it is in $C$?
    $endgroup$
    – sehigle
    Dec 17 '18 at 0:50










  • $begingroup$
    Suppose we make the eigenvector of a positive eigenvalue long enough. Then it's an element of C isn't it?
    $endgroup$
    – I like Serena
    Dec 17 '18 at 0:52














  • 2




    $begingroup$
    Why don't you just take a vector $x$ with $x^T Ax> 0$ and scale it such that it is in $C$?
    $endgroup$
    – sehigle
    Dec 17 '18 at 0:50










  • $begingroup$
    Suppose we make the eigenvector of a positive eigenvalue long enough. Then it's an element of C isn't it?
    $endgroup$
    – I like Serena
    Dec 17 '18 at 0:52








2




2




$begingroup$
Why don't you just take a vector $x$ with $x^T Ax> 0$ and scale it such that it is in $C$?
$endgroup$
– sehigle
Dec 17 '18 at 0:50




$begingroup$
Why don't you just take a vector $x$ with $x^T Ax> 0$ and scale it such that it is in $C$?
$endgroup$
– sehigle
Dec 17 '18 at 0:50












$begingroup$
Suppose we make the eigenvector of a positive eigenvalue long enough. Then it's an element of C isn't it?
$endgroup$
– I like Serena
Dec 17 '18 at 0:52




$begingroup$
Suppose we make the eigenvector of a positive eigenvalue long enough. Then it's an element of C isn't it?
$endgroup$
– I like Serena
Dec 17 '18 at 0:52










2 Answers
2






active

oldest

votes


















2












$begingroup$

You can't simply assume that the non-positive eigenvalues of $A$ are zero.



Note that to show that $C$ is non-empty, you only need to show that there is at least one vector in $C$.



Let $lambda$ be the one of the positive eigenvalues of $A$, and let $v$ be a corresponding eigenvector with unit norm.



If $x = alpha v$ for some constant $alpha$, then how big is $x^TAx$? Can you find a large enough $alpha$ such that $x = alpha v in C?$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let $x in R^4$ be the eigenvector corresponding to one of the positive eigenvalues $lambda_1$. Then $x^TAx = lambda_1 x^Tx > 0$. Now all you have to do is multiply $x$ by a normalizing scalar to get some $y in R^4$ and you have $y^TAy = 1 Rightarrow y in C$.






    share|cite|improve this answer











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      2 Answers
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      active

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      2 Answers
      2






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      2












      $begingroup$

      You can't simply assume that the non-positive eigenvalues of $A$ are zero.



      Note that to show that $C$ is non-empty, you only need to show that there is at least one vector in $C$.



      Let $lambda$ be the one of the positive eigenvalues of $A$, and let $v$ be a corresponding eigenvector with unit norm.



      If $x = alpha v$ for some constant $alpha$, then how big is $x^TAx$? Can you find a large enough $alpha$ such that $x = alpha v in C?$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        You can't simply assume that the non-positive eigenvalues of $A$ are zero.



        Note that to show that $C$ is non-empty, you only need to show that there is at least one vector in $C$.



        Let $lambda$ be the one of the positive eigenvalues of $A$, and let $v$ be a corresponding eigenvector with unit norm.



        If $x = alpha v$ for some constant $alpha$, then how big is $x^TAx$? Can you find a large enough $alpha$ such that $x = alpha v in C?$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          You can't simply assume that the non-positive eigenvalues of $A$ are zero.



          Note that to show that $C$ is non-empty, you only need to show that there is at least one vector in $C$.



          Let $lambda$ be the one of the positive eigenvalues of $A$, and let $v$ be a corresponding eigenvector with unit norm.



          If $x = alpha v$ for some constant $alpha$, then how big is $x^TAx$? Can you find a large enough $alpha$ such that $x = alpha v in C?$






          share|cite|improve this answer









          $endgroup$



          You can't simply assume that the non-positive eigenvalues of $A$ are zero.



          Note that to show that $C$ is non-empty, you only need to show that there is at least one vector in $C$.



          Let $lambda$ be the one of the positive eigenvalues of $A$, and let $v$ be a corresponding eigenvector with unit norm.



          If $x = alpha v$ for some constant $alpha$, then how big is $x^TAx$? Can you find a large enough $alpha$ such that $x = alpha v in C?$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 0:50









          JimmyK4542JimmyK4542

          41.1k245106




          41.1k245106























              0












              $begingroup$

              Let $x in R^4$ be the eigenvector corresponding to one of the positive eigenvalues $lambda_1$. Then $x^TAx = lambda_1 x^Tx > 0$. Now all you have to do is multiply $x$ by a normalizing scalar to get some $y in R^4$ and you have $y^TAy = 1 Rightarrow y in C$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Let $x in R^4$ be the eigenvector corresponding to one of the positive eigenvalues $lambda_1$. Then $x^TAx = lambda_1 x^Tx > 0$. Now all you have to do is multiply $x$ by a normalizing scalar to get some $y in R^4$ and you have $y^TAy = 1 Rightarrow y in C$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let $x in R^4$ be the eigenvector corresponding to one of the positive eigenvalues $lambda_1$. Then $x^TAx = lambda_1 x^Tx > 0$. Now all you have to do is multiply $x$ by a normalizing scalar to get some $y in R^4$ and you have $y^TAy = 1 Rightarrow y in C$.






                  share|cite|improve this answer











                  $endgroup$



                  Let $x in R^4$ be the eigenvector corresponding to one of the positive eigenvalues $lambda_1$. Then $x^TAx = lambda_1 x^Tx > 0$. Now all you have to do is multiply $x$ by a normalizing scalar to get some $y in R^4$ and you have $y^TAy = 1 Rightarrow y in C$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 17 '18 at 0:57

























                  answered Dec 17 '18 at 0:51









                  svailsvail

                  12




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