Conditions for a certain parametric matrix to be positive definite











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The task:




Let's consider matrix $A=begin{bmatrix}1&a&a\a&1&a\a&a&1 end{bmatrix}$. Show that $A>0$ if and only if $-1<2a<2$.




Solution attempt: By definition $A$ is positive definite iff $forall_{zinmathbb{R}^3}z^{operatorname{T}}Az>0$.



For $z=begin{bmatrix}z_1\z_2\z_3end{bmatrix}$, $$left(z^{operatorname{T}}Aright)z=left(z_1^2+az_1z_2+az_1z_3right)+left(az_1z_2+z_2^2+az_2z_3right)+left(az_1z_3+az_2z_3+z_3^2right)\=z_1^2+z_2^2+z_3^2+2aleft(z_1z_2+z_1z_3+z_2z_3right)>0\iff\z_1^2+z_2^2+z_3^2>-2aleft(z_1z_2+z_1z_3+z_2z_3right)$$



This is trivially true for $z_1z_2+z_1z_3+z_2z_3geq0$, so without loss of generality let's assume that $z_1z_2+z_1z_3+z_2z_3<0$, and thus $-(z_1z_2+z_1z_3+z_2z_3)>0$. Therefore we need to prove that:



$$left({Largeforall_{z_1,z_2,z_3}}-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2aright)iff -1<2a<2$$



And this seems false to me. This is because $-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}$ is always $geq0$. For this reason we have a trivial counterexample, eg when $2a=-1000$ then $2anotin(-1;2)$ but nonetheless $forall_{z_1,z_2,z_3}-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2a$.



I can't find my error; could you kindly help me please?










share|cite|improve this question




















  • 1




    Do you know what the chracteristic polynomial of a matrix is? Its roots are the eigenvalues. In this case, one can compute the eigenvalues of the given matrix explicitly using this approach. A matrix is positive definite if and only if all its eigenvalues are positive and real.
    – астон вілла олоф мэллбэрг
    Nov 13 at 17:10












  • @астонвіллаолофмэллбэрг I know what the characteristic polynomial is, but I don't know what is the correlation between eigenvalues and positive definiteness...
    – gaazkam
    Nov 13 at 17:13






  • 1




    Your error is when you say "this is trivially true for $z_1z_2+z_1z_3+z_2z_3 geq 0$ ": that's not true for $a<0$
    – krirkrirk
    Nov 13 at 17:14










  • @астонвіллаолофмэллбэрг OK: I checked Wikipedia: All eigenvalues are positive iff matrix is positive definite. Thank you, I'll try to solve it that way
    – gaazkam
    Nov 13 at 17:20










  • Yes, I think the CP is the easiest way of doing this.
    – астон вілла олоф мэллбэрг
    Nov 13 at 17:23















up vote
1
down vote

favorite












The task:




Let's consider matrix $A=begin{bmatrix}1&a&a\a&1&a\a&a&1 end{bmatrix}$. Show that $A>0$ if and only if $-1<2a<2$.




Solution attempt: By definition $A$ is positive definite iff $forall_{zinmathbb{R}^3}z^{operatorname{T}}Az>0$.



For $z=begin{bmatrix}z_1\z_2\z_3end{bmatrix}$, $$left(z^{operatorname{T}}Aright)z=left(z_1^2+az_1z_2+az_1z_3right)+left(az_1z_2+z_2^2+az_2z_3right)+left(az_1z_3+az_2z_3+z_3^2right)\=z_1^2+z_2^2+z_3^2+2aleft(z_1z_2+z_1z_3+z_2z_3right)>0\iff\z_1^2+z_2^2+z_3^2>-2aleft(z_1z_2+z_1z_3+z_2z_3right)$$



This is trivially true for $z_1z_2+z_1z_3+z_2z_3geq0$, so without loss of generality let's assume that $z_1z_2+z_1z_3+z_2z_3<0$, and thus $-(z_1z_2+z_1z_3+z_2z_3)>0$. Therefore we need to prove that:



$$left({Largeforall_{z_1,z_2,z_3}}-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2aright)iff -1<2a<2$$



And this seems false to me. This is because $-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}$ is always $geq0$. For this reason we have a trivial counterexample, eg when $2a=-1000$ then $2anotin(-1;2)$ but nonetheless $forall_{z_1,z_2,z_3}-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2a$.



I can't find my error; could you kindly help me please?










share|cite|improve this question




















  • 1




    Do you know what the chracteristic polynomial of a matrix is? Its roots are the eigenvalues. In this case, one can compute the eigenvalues of the given matrix explicitly using this approach. A matrix is positive definite if and only if all its eigenvalues are positive and real.
    – астон вілла олоф мэллбэрг
    Nov 13 at 17:10












  • @астонвіллаолофмэллбэрг I know what the characteristic polynomial is, but I don't know what is the correlation between eigenvalues and positive definiteness...
    – gaazkam
    Nov 13 at 17:13






  • 1




    Your error is when you say "this is trivially true for $z_1z_2+z_1z_3+z_2z_3 geq 0$ ": that's not true for $a<0$
    – krirkrirk
    Nov 13 at 17:14










  • @астонвіллаолофмэллбэрг OK: I checked Wikipedia: All eigenvalues are positive iff matrix is positive definite. Thank you, I'll try to solve it that way
    – gaazkam
    Nov 13 at 17:20










  • Yes, I think the CP is the easiest way of doing this.
    – астон вілла олоф мэллбэрг
    Nov 13 at 17:23













up vote
1
down vote

favorite









up vote
1
down vote

favorite











The task:




Let's consider matrix $A=begin{bmatrix}1&a&a\a&1&a\a&a&1 end{bmatrix}$. Show that $A>0$ if and only if $-1<2a<2$.




Solution attempt: By definition $A$ is positive definite iff $forall_{zinmathbb{R}^3}z^{operatorname{T}}Az>0$.



For $z=begin{bmatrix}z_1\z_2\z_3end{bmatrix}$, $$left(z^{operatorname{T}}Aright)z=left(z_1^2+az_1z_2+az_1z_3right)+left(az_1z_2+z_2^2+az_2z_3right)+left(az_1z_3+az_2z_3+z_3^2right)\=z_1^2+z_2^2+z_3^2+2aleft(z_1z_2+z_1z_3+z_2z_3right)>0\iff\z_1^2+z_2^2+z_3^2>-2aleft(z_1z_2+z_1z_3+z_2z_3right)$$



This is trivially true for $z_1z_2+z_1z_3+z_2z_3geq0$, so without loss of generality let's assume that $z_1z_2+z_1z_3+z_2z_3<0$, and thus $-(z_1z_2+z_1z_3+z_2z_3)>0$. Therefore we need to prove that:



$$left({Largeforall_{z_1,z_2,z_3}}-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2aright)iff -1<2a<2$$



And this seems false to me. This is because $-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}$ is always $geq0$. For this reason we have a trivial counterexample, eg when $2a=-1000$ then $2anotin(-1;2)$ but nonetheless $forall_{z_1,z_2,z_3}-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2a$.



I can't find my error; could you kindly help me please?










share|cite|improve this question















The task:




Let's consider matrix $A=begin{bmatrix}1&a&a\a&1&a\a&a&1 end{bmatrix}$. Show that $A>0$ if and only if $-1<2a<2$.




Solution attempt: By definition $A$ is positive definite iff $forall_{zinmathbb{R}^3}z^{operatorname{T}}Az>0$.



For $z=begin{bmatrix}z_1\z_2\z_3end{bmatrix}$, $$left(z^{operatorname{T}}Aright)z=left(z_1^2+az_1z_2+az_1z_3right)+left(az_1z_2+z_2^2+az_2z_3right)+left(az_1z_3+az_2z_3+z_3^2right)\=z_1^2+z_2^2+z_3^2+2aleft(z_1z_2+z_1z_3+z_2z_3right)>0\iff\z_1^2+z_2^2+z_3^2>-2aleft(z_1z_2+z_1z_3+z_2z_3right)$$



This is trivially true for $z_1z_2+z_1z_3+z_2z_3geq0$, so without loss of generality let's assume that $z_1z_2+z_1z_3+z_2z_3<0$, and thus $-(z_1z_2+z_1z_3+z_2z_3)>0$. Therefore we need to prove that:



$$left({Largeforall_{z_1,z_2,z_3}}-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2aright)iff -1<2a<2$$



And this seems false to me. This is because $-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}$ is always $geq0$. For this reason we have a trivial counterexample, eg when $2a=-1000$ then $2anotin(-1;2)$ but nonetheless $forall_{z_1,z_2,z_3}-frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2a$.



I can't find my error; could you kindly help me please?







vectors matrix-equations positive-definite






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edited Nov 13 at 16:54









Henning Makholm

235k16299534




235k16299534










asked Nov 13 at 16:52









gaazkam

418313




418313








  • 1




    Do you know what the chracteristic polynomial of a matrix is? Its roots are the eigenvalues. In this case, one can compute the eigenvalues of the given matrix explicitly using this approach. A matrix is positive definite if and only if all its eigenvalues are positive and real.
    – астон вілла олоф мэллбэрг
    Nov 13 at 17:10












  • @астонвіллаолофмэллбэрг I know what the characteristic polynomial is, but I don't know what is the correlation between eigenvalues and positive definiteness...
    – gaazkam
    Nov 13 at 17:13






  • 1




    Your error is when you say "this is trivially true for $z_1z_2+z_1z_3+z_2z_3 geq 0$ ": that's not true for $a<0$
    – krirkrirk
    Nov 13 at 17:14










  • @астонвіллаолофмэллбэрг OK: I checked Wikipedia: All eigenvalues are positive iff matrix is positive definite. Thank you, I'll try to solve it that way
    – gaazkam
    Nov 13 at 17:20










  • Yes, I think the CP is the easiest way of doing this.
    – астон вілла олоф мэллбэрг
    Nov 13 at 17:23














  • 1




    Do you know what the chracteristic polynomial of a matrix is? Its roots are the eigenvalues. In this case, one can compute the eigenvalues of the given matrix explicitly using this approach. A matrix is positive definite if and only if all its eigenvalues are positive and real.
    – астон вілла олоф мэллбэрг
    Nov 13 at 17:10












  • @астонвіллаолофмэллбэрг I know what the characteristic polynomial is, but I don't know what is the correlation between eigenvalues and positive definiteness...
    – gaazkam
    Nov 13 at 17:13






  • 1




    Your error is when you say "this is trivially true for $z_1z_2+z_1z_3+z_2z_3 geq 0$ ": that's not true for $a<0$
    – krirkrirk
    Nov 13 at 17:14










  • @астонвіллаолофмэллбэрг OK: I checked Wikipedia: All eigenvalues are positive iff matrix is positive definite. Thank you, I'll try to solve it that way
    – gaazkam
    Nov 13 at 17:20










  • Yes, I think the CP is the easiest way of doing this.
    – астон вілла олоф мэллбэрг
    Nov 13 at 17:23








1




1




Do you know what the chracteristic polynomial of a matrix is? Its roots are the eigenvalues. In this case, one can compute the eigenvalues of the given matrix explicitly using this approach. A matrix is positive definite if and only if all its eigenvalues are positive and real.
– астон вілла олоф мэллбэрг
Nov 13 at 17:10






Do you know what the chracteristic polynomial of a matrix is? Its roots are the eigenvalues. In this case, one can compute the eigenvalues of the given matrix explicitly using this approach. A matrix is positive definite if and only if all its eigenvalues are positive and real.
– астон вілла олоф мэллбэрг
Nov 13 at 17:10














@астонвіллаолофмэллбэрг I know what the characteristic polynomial is, but I don't know what is the correlation between eigenvalues and positive definiteness...
– gaazkam
Nov 13 at 17:13




@астонвіллаолофмэллбэрг I know what the characteristic polynomial is, but I don't know what is the correlation between eigenvalues and positive definiteness...
– gaazkam
Nov 13 at 17:13




1




1




Your error is when you say "this is trivially true for $z_1z_2+z_1z_3+z_2z_3 geq 0$ ": that's not true for $a<0$
– krirkrirk
Nov 13 at 17:14




Your error is when you say "this is trivially true for $z_1z_2+z_1z_3+z_2z_3 geq 0$ ": that's not true for $a<0$
– krirkrirk
Nov 13 at 17:14












@астонвіллаолофмэллбэрг OK: I checked Wikipedia: All eigenvalues are positive iff matrix is positive definite. Thank you, I'll try to solve it that way
– gaazkam
Nov 13 at 17:20




@астонвіллаолофмэллбэрг OK: I checked Wikipedia: All eigenvalues are positive iff matrix is positive definite. Thank you, I'll try to solve it that way
– gaazkam
Nov 13 at 17:20












Yes, I think the CP is the easiest way of doing this.
– астон вілла олоф мэллбэрг
Nov 13 at 17:23




Yes, I think the CP is the easiest way of doing this.
– астон вілла олоф мэллбэрг
Nov 13 at 17:23










2 Answers
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$z_1^2+z_2^2+z_3^2+2aleft(z_1z_2+z_1z_3+z_2z_3right)>0iff z_1^2+z_2^2+z_3^2>-2aleft(z_1z_2+z_1z_3+z_2z_3right)$



This is trivially true for $z_1z_2+z_1z_3+z_2z_3geq0$




Not, it is not, since $a$ might be negative.



I suppose showing $A$ is definite positive could be done your way, but it's way too exhausting for me. I'd go with using the fact that $$Atext{ definite positive } iff text{ its eigenvalues are } > 0$$



Now choose your favorite way to find the eigenvalues of $A$. Note that the sum of the rows are $2a+1$ so this is one of them ; which by the way leads to $-1<2a$. Now your turn...






share|cite|improve this answer




























    up vote
    1
    down vote













    Let's determine the eigenvalues of $A$. If we can find the conditions for which those eigenvalues are all positive, then we can solve the question.



    Let $u in mathbb R^3$ be the vector with $1$ for all its components. Then note that $uu^T$ is the $3times 3$ matrix with $1$ for all its entries.
    Therefore $$A=(1-a)I+auu^T$$
    Now, notice that $$Au=(1-a)u+au (u^Tu)=(1-a)u+3au=(1+2a)u$$
    So $u$ is an eigenvector of $A$ with eigenvalue $(1+2a)$.



    Now, let's determine the other eigenvalues of $A$. Suppose $xinmathbb R^3$ is orthogonal to $u$, that is $u^Tx=0$, then
    $$Ax=(1-a)x+auu^Tx=(1-a)x$$
    so $x$ is an eigenvector of $A$ with eigenvalue $(1-a)$. Since the space of vectors $x$ orthogonal to $u$ is of dimension 2, we have now determined that $(1-a)$ is an eigenvalue of multiplicity 2 for $A$.



    In summary, the spectrum of $A$ is $(1+2a)$ (multiplicity 1) and $(1-a)$ (multiplicity 2). Consequently, $A$ is positive definite iff $$1+2a>0 text{ and } 1-a >0$$ that is iff $-1 < 2a < 2$$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      $z_1^2+z_2^2+z_3^2+2aleft(z_1z_2+z_1z_3+z_2z_3right)>0iff z_1^2+z_2^2+z_3^2>-2aleft(z_1z_2+z_1z_3+z_2z_3right)$



      This is trivially true for $z_1z_2+z_1z_3+z_2z_3geq0$




      Not, it is not, since $a$ might be negative.



      I suppose showing $A$ is definite positive could be done your way, but it's way too exhausting for me. I'd go with using the fact that $$Atext{ definite positive } iff text{ its eigenvalues are } > 0$$



      Now choose your favorite way to find the eigenvalues of $A$. Note that the sum of the rows are $2a+1$ so this is one of them ; which by the way leads to $-1<2a$. Now your turn...






      share|cite|improve this answer

























        up vote
        2
        down vote














        $z_1^2+z_2^2+z_3^2+2aleft(z_1z_2+z_1z_3+z_2z_3right)>0iff z_1^2+z_2^2+z_3^2>-2aleft(z_1z_2+z_1z_3+z_2z_3right)$



        This is trivially true for $z_1z_2+z_1z_3+z_2z_3geq0$




        Not, it is not, since $a$ might be negative.



        I suppose showing $A$ is definite positive could be done your way, but it's way too exhausting for me. I'd go with using the fact that $$Atext{ definite positive } iff text{ its eigenvalues are } > 0$$



        Now choose your favorite way to find the eigenvalues of $A$. Note that the sum of the rows are $2a+1$ so this is one of them ; which by the way leads to $-1<2a$. Now your turn...






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote










          $z_1^2+z_2^2+z_3^2+2aleft(z_1z_2+z_1z_3+z_2z_3right)>0iff z_1^2+z_2^2+z_3^2>-2aleft(z_1z_2+z_1z_3+z_2z_3right)$



          This is trivially true for $z_1z_2+z_1z_3+z_2z_3geq0$




          Not, it is not, since $a$ might be negative.



          I suppose showing $A$ is definite positive could be done your way, but it's way too exhausting for me. I'd go with using the fact that $$Atext{ definite positive } iff text{ its eigenvalues are } > 0$$



          Now choose your favorite way to find the eigenvalues of $A$. Note that the sum of the rows are $2a+1$ so this is one of them ; which by the way leads to $-1<2a$. Now your turn...






          share|cite|improve this answer













          $z_1^2+z_2^2+z_3^2+2aleft(z_1z_2+z_1z_3+z_2z_3right)>0iff z_1^2+z_2^2+z_3^2>-2aleft(z_1z_2+z_1z_3+z_2z_3right)$



          This is trivially true for $z_1z_2+z_1z_3+z_2z_3geq0$




          Not, it is not, since $a$ might be negative.



          I suppose showing $A$ is definite positive could be done your way, but it's way too exhausting for me. I'd go with using the fact that $$Atext{ definite positive } iff text{ its eigenvalues are } > 0$$



          Now choose your favorite way to find the eigenvalues of $A$. Note that the sum of the rows are $2a+1$ so this is one of them ; which by the way leads to $-1<2a$. Now your turn...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 13 at 17:20









          krirkrirk

          1,462518




          1,462518






















              up vote
              1
              down vote













              Let's determine the eigenvalues of $A$. If we can find the conditions for which those eigenvalues are all positive, then we can solve the question.



              Let $u in mathbb R^3$ be the vector with $1$ for all its components. Then note that $uu^T$ is the $3times 3$ matrix with $1$ for all its entries.
              Therefore $$A=(1-a)I+auu^T$$
              Now, notice that $$Au=(1-a)u+au (u^Tu)=(1-a)u+3au=(1+2a)u$$
              So $u$ is an eigenvector of $A$ with eigenvalue $(1+2a)$.



              Now, let's determine the other eigenvalues of $A$. Suppose $xinmathbb R^3$ is orthogonal to $u$, that is $u^Tx=0$, then
              $$Ax=(1-a)x+auu^Tx=(1-a)x$$
              so $x$ is an eigenvector of $A$ with eigenvalue $(1-a)$. Since the space of vectors $x$ orthogonal to $u$ is of dimension 2, we have now determined that $(1-a)$ is an eigenvalue of multiplicity 2 for $A$.



              In summary, the spectrum of $A$ is $(1+2a)$ (multiplicity 1) and $(1-a)$ (multiplicity 2). Consequently, $A$ is positive definite iff $$1+2a>0 text{ and } 1-a >0$$ that is iff $-1 < 2a < 2$$






              share|cite|improve this answer

























                up vote
                1
                down vote













                Let's determine the eigenvalues of $A$. If we can find the conditions for which those eigenvalues are all positive, then we can solve the question.



                Let $u in mathbb R^3$ be the vector with $1$ for all its components. Then note that $uu^T$ is the $3times 3$ matrix with $1$ for all its entries.
                Therefore $$A=(1-a)I+auu^T$$
                Now, notice that $$Au=(1-a)u+au (u^Tu)=(1-a)u+3au=(1+2a)u$$
                So $u$ is an eigenvector of $A$ with eigenvalue $(1+2a)$.



                Now, let's determine the other eigenvalues of $A$. Suppose $xinmathbb R^3$ is orthogonal to $u$, that is $u^Tx=0$, then
                $$Ax=(1-a)x+auu^Tx=(1-a)x$$
                so $x$ is an eigenvector of $A$ with eigenvalue $(1-a)$. Since the space of vectors $x$ orthogonal to $u$ is of dimension 2, we have now determined that $(1-a)$ is an eigenvalue of multiplicity 2 for $A$.



                In summary, the spectrum of $A$ is $(1+2a)$ (multiplicity 1) and $(1-a)$ (multiplicity 2). Consequently, $A$ is positive definite iff $$1+2a>0 text{ and } 1-a >0$$ that is iff $-1 < 2a < 2$$






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                  Let's determine the eigenvalues of $A$. If we can find the conditions for which those eigenvalues are all positive, then we can solve the question.



                  Let $u in mathbb R^3$ be the vector with $1$ for all its components. Then note that $uu^T$ is the $3times 3$ matrix with $1$ for all its entries.
                  Therefore $$A=(1-a)I+auu^T$$
                  Now, notice that $$Au=(1-a)u+au (u^Tu)=(1-a)u+3au=(1+2a)u$$
                  So $u$ is an eigenvector of $A$ with eigenvalue $(1+2a)$.



                  Now, let's determine the other eigenvalues of $A$. Suppose $xinmathbb R^3$ is orthogonal to $u$, that is $u^Tx=0$, then
                  $$Ax=(1-a)x+auu^Tx=(1-a)x$$
                  so $x$ is an eigenvector of $A$ with eigenvalue $(1-a)$. Since the space of vectors $x$ orthogonal to $u$ is of dimension 2, we have now determined that $(1-a)$ is an eigenvalue of multiplicity 2 for $A$.



                  In summary, the spectrum of $A$ is $(1+2a)$ (multiplicity 1) and $(1-a)$ (multiplicity 2). Consequently, $A$ is positive definite iff $$1+2a>0 text{ and } 1-a >0$$ that is iff $-1 < 2a < 2$$






                  share|cite|improve this answer












                  Let's determine the eigenvalues of $A$. If we can find the conditions for which those eigenvalues are all positive, then we can solve the question.



                  Let $u in mathbb R^3$ be the vector with $1$ for all its components. Then note that $uu^T$ is the $3times 3$ matrix with $1$ for all its entries.
                  Therefore $$A=(1-a)I+auu^T$$
                  Now, notice that $$Au=(1-a)u+au (u^Tu)=(1-a)u+3au=(1+2a)u$$
                  So $u$ is an eigenvector of $A$ with eigenvalue $(1+2a)$.



                  Now, let's determine the other eigenvalues of $A$. Suppose $xinmathbb R^3$ is orthogonal to $u$, that is $u^Tx=0$, then
                  $$Ax=(1-a)x+auu^Tx=(1-a)x$$
                  so $x$ is an eigenvector of $A$ with eigenvalue $(1-a)$. Since the space of vectors $x$ orthogonal to $u$ is of dimension 2, we have now determined that $(1-a)$ is an eigenvalue of multiplicity 2 for $A$.



                  In summary, the spectrum of $A$ is $(1+2a)$ (multiplicity 1) and $(1-a)$ (multiplicity 2). Consequently, $A$ is positive definite iff $$1+2a>0 text{ and } 1-a >0$$ that is iff $-1 < 2a < 2$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 13 at 18:24









                  Stefan Lafon

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