Questions on symmetric matrices and skew-symmetric matrices












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$begingroup$


Let $A$ be a $3times 3;$ symmetric matrix. Let $U$ be the set of all $3times 3;$ skew-symmetric matrices. Let $T : Uto U$ be defined as $T(B)=AB+BA.$

Prove that $T$ is bijective iff the sum of eigenvalues of $A$ is not an eigenvalue of $A.$



This question makes me stuck for days, since I cannot relate the eigenvalues with the map. Can anyone help me?






linear-algebra matrices eigenvalues-eigenvectors linear-transformations symmetric-matrices







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    0












    $begingroup$


    Let $A$ be a $3times 3;$ symmetric matrix. Let $U$ be the set of all $3times 3;$ skew-symmetric matrices. Let $T : Uto U$ be defined as $T(B)=AB+BA.$

    Prove that $T$ is bijective iff the sum of eigenvalues of $A$ is not an eigenvalue of $A.$



    This question makes me stuck for days, since I cannot relate the eigenvalues with the map. Can anyone help me?






    linear-algebra matrices eigenvalues-eigenvectors linear-transformations symmetric-matrices







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Let $A$ be a $3times 3;$ symmetric matrix. Let $U$ be the set of all $3times 3;$ skew-symmetric matrices. Let $T : Uto U$ be defined as $T(B)=AB+BA.$

      Prove that $T$ is bijective iff the sum of eigenvalues of $A$ is not an eigenvalue of $A.$



      This question makes me stuck for days, since I cannot relate the eigenvalues with the map. Can anyone help me?






      linear-algebra matrices eigenvalues-eigenvectors linear-transformations symmetric-matrices







      share|cite|improve this question











      $endgroup$




      Let $A$ be a $3times 3;$ symmetric matrix. Let $U$ be the set of all $3times 3;$ skew-symmetric matrices. Let $T : Uto U$ be defined as $T(B)=AB+BA.$

      Prove that $T$ is bijective iff the sum of eigenvalues of $A$ is not an eigenvalue of $A.$



      This question makes me stuck for days, since I cannot relate the eigenvalues with the map. Can anyone help me?






      linear-algebra matrices eigenvalues-eigenvectors linear-transformations symmetric-matrices




      linear-algebra matrices eigenvalues-eigenvectors linear-transformations symmetric-matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 18:28









      user376343

      3,9584829




      3,9584829










      asked Dec 8 '18 at 15:31









      deep12345deep12345

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      152






















          1 Answer
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          $begingroup$

          I am assuming that you are working over the real field. So, $A$ is diagonalizable and you can assume without loss of generality that $A$ is diagonal:$$A=begin{bmatrix}alpha&0&0\0&beta&0\0&0&gammaend{bmatrix}$$and $alpha$, $beta$, and $gamma$ are the eigenvalues of $A$. So, if$$X=begin{bmatrix}0&x&-z\-x&0&y\z&-y&0end{bmatrix},$$then$$T(X)=begin{bmatrix}0 & alpha x+beta x & -alpha z-gamma z \ -alpha x-beta x & 0 & beta y+gamma y \ alpha z+gamma z & -beta y-gamma y & 0end{bmatrix}$$and therefore the eigenvalues are $alpha+beta$, $alpha+gamma$, and $beta+gamma$. Thereforebegin{align}Ttext{ is bijective}&iff Ttext{ is injective}\&iff0text{ is not an eigenvalue of }T\&iffalpha+beta,alpha+gamma,beta+gammaneq0\&iffalpha+beta+gammaneqgammatext{, }alpha+beta+gammaneqbetatext{, and }alpha+beta+gammaneqalpha\&iffalpha+beta+gammatext{ is not an eigenvalue of }A.end{align}






          share|cite|improve this answer











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          • $begingroup$
            Actually your example is wrong since -1+1+2=2 is an eigenvalue. What do you mean by eigenvalues of T are sum of pairs of eigenvalue of A?
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:18












          • $begingroup$
            I misundrstood the condition! I thought that it was “the sum of two eigenvalues is not an eigenvalue”. But then, yes, the statment is correct. I will edit my answer and provide a proof.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:26










          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:37










          • $begingroup$
            Never thought of brute force to compute T directly, though the answer turns out to be surprisingly simple. Thanks!
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:44










          • $begingroup$
            What happens for dimensions other than $3times 3$? For $2times 2$ I get $$T(X) = begin{bmatrix} 0 & alpha x+beta x \ -alpha x-beta x & 0 end{bmatrix}$$
            $endgroup$
            – Jeppe Stig Nielsen
            Dec 8 '18 at 17:17











          Your Answer





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          $begingroup$

          I am assuming that you are working over the real field. So, $A$ is diagonalizable and you can assume without loss of generality that $A$ is diagonal:$$A=begin{bmatrix}alpha&0&0\0&beta&0\0&0&gammaend{bmatrix}$$and $alpha$, $beta$, and $gamma$ are the eigenvalues of $A$. So, if$$X=begin{bmatrix}0&x&-z\-x&0&y\z&-y&0end{bmatrix},$$then$$T(X)=begin{bmatrix}0 & alpha x+beta x & -alpha z-gamma z \ -alpha x-beta x & 0 & beta y+gamma y \ alpha z+gamma z & -beta y-gamma y & 0end{bmatrix}$$and therefore the eigenvalues are $alpha+beta$, $alpha+gamma$, and $beta+gamma$. Thereforebegin{align}Ttext{ is bijective}&iff Ttext{ is injective}\&iff0text{ is not an eigenvalue of }T\&iffalpha+beta,alpha+gamma,beta+gammaneq0\&iffalpha+beta+gammaneqgammatext{, }alpha+beta+gammaneqbetatext{, and }alpha+beta+gammaneqalpha\&iffalpha+beta+gammatext{ is not an eigenvalue of }A.end{align}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Actually your example is wrong since -1+1+2=2 is an eigenvalue. What do you mean by eigenvalues of T are sum of pairs of eigenvalue of A?
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:18












          • $begingroup$
            I misundrstood the condition! I thought that it was “the sum of two eigenvalues is not an eigenvalue”. But then, yes, the statment is correct. I will edit my answer and provide a proof.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:26










          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:37










          • $begingroup$
            Never thought of brute force to compute T directly, though the answer turns out to be surprisingly simple. Thanks!
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:44










          • $begingroup$
            What happens for dimensions other than $3times 3$? For $2times 2$ I get $$T(X) = begin{bmatrix} 0 & alpha x+beta x \ -alpha x-beta x & 0 end{bmatrix}$$
            $endgroup$
            – Jeppe Stig Nielsen
            Dec 8 '18 at 17:17
















          1












          $begingroup$

          I am assuming that you are working over the real field. So, $A$ is diagonalizable and you can assume without loss of generality that $A$ is diagonal:$$A=begin{bmatrix}alpha&0&0\0&beta&0\0&0&gammaend{bmatrix}$$and $alpha$, $beta$, and $gamma$ are the eigenvalues of $A$. So, if$$X=begin{bmatrix}0&x&-z\-x&0&y\z&-y&0end{bmatrix},$$then$$T(X)=begin{bmatrix}0 & alpha x+beta x & -alpha z-gamma z \ -alpha x-beta x & 0 & beta y+gamma y \ alpha z+gamma z & -beta y-gamma y & 0end{bmatrix}$$and therefore the eigenvalues are $alpha+beta$, $alpha+gamma$, and $beta+gamma$. Thereforebegin{align}Ttext{ is bijective}&iff Ttext{ is injective}\&iff0text{ is not an eigenvalue of }T\&iffalpha+beta,alpha+gamma,beta+gammaneq0\&iffalpha+beta+gammaneqgammatext{, }alpha+beta+gammaneqbetatext{, and }alpha+beta+gammaneqalpha\&iffalpha+beta+gammatext{ is not an eigenvalue of }A.end{align}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Actually your example is wrong since -1+1+2=2 is an eigenvalue. What do you mean by eigenvalues of T are sum of pairs of eigenvalue of A?
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:18












          • $begingroup$
            I misundrstood the condition! I thought that it was “the sum of two eigenvalues is not an eigenvalue”. But then, yes, the statment is correct. I will edit my answer and provide a proof.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:26










          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:37










          • $begingroup$
            Never thought of brute force to compute T directly, though the answer turns out to be surprisingly simple. Thanks!
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:44










          • $begingroup$
            What happens for dimensions other than $3times 3$? For $2times 2$ I get $$T(X) = begin{bmatrix} 0 & alpha x+beta x \ -alpha x-beta x & 0 end{bmatrix}$$
            $endgroup$
            – Jeppe Stig Nielsen
            Dec 8 '18 at 17:17














          1












          1








          1





          $begingroup$

          I am assuming that you are working over the real field. So, $A$ is diagonalizable and you can assume without loss of generality that $A$ is diagonal:$$A=begin{bmatrix}alpha&0&0\0&beta&0\0&0&gammaend{bmatrix}$$and $alpha$, $beta$, and $gamma$ are the eigenvalues of $A$. So, if$$X=begin{bmatrix}0&x&-z\-x&0&y\z&-y&0end{bmatrix},$$then$$T(X)=begin{bmatrix}0 & alpha x+beta x & -alpha z-gamma z \ -alpha x-beta x & 0 & beta y+gamma y \ alpha z+gamma z & -beta y-gamma y & 0end{bmatrix}$$and therefore the eigenvalues are $alpha+beta$, $alpha+gamma$, and $beta+gamma$. Thereforebegin{align}Ttext{ is bijective}&iff Ttext{ is injective}\&iff0text{ is not an eigenvalue of }T\&iffalpha+beta,alpha+gamma,beta+gammaneq0\&iffalpha+beta+gammaneqgammatext{, }alpha+beta+gammaneqbetatext{, and }alpha+beta+gammaneqalpha\&iffalpha+beta+gammatext{ is not an eigenvalue of }A.end{align}






          share|cite|improve this answer











          $endgroup$



          I am assuming that you are working over the real field. So, $A$ is diagonalizable and you can assume without loss of generality that $A$ is diagonal:$$A=begin{bmatrix}alpha&0&0\0&beta&0\0&0&gammaend{bmatrix}$$and $alpha$, $beta$, and $gamma$ are the eigenvalues of $A$. So, if$$X=begin{bmatrix}0&x&-z\-x&0&y\z&-y&0end{bmatrix},$$then$$T(X)=begin{bmatrix}0 & alpha x+beta x & -alpha z-gamma z \ -alpha x-beta x & 0 & beta y+gamma y \ alpha z+gamma z & -beta y-gamma y & 0end{bmatrix}$$and therefore the eigenvalues are $alpha+beta$, $alpha+gamma$, and $beta+gamma$. Thereforebegin{align}Ttext{ is bijective}&iff Ttext{ is injective}\&iff0text{ is not an eigenvalue of }T\&iffalpha+beta,alpha+gamma,beta+gammaneq0\&iffalpha+beta+gammaneqgammatext{, }alpha+beta+gammaneqbetatext{, and }alpha+beta+gammaneqalpha\&iffalpha+beta+gammatext{ is not an eigenvalue of }A.end{align}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 8 '18 at 16:36

























          answered Dec 8 '18 at 15:59









          José Carlos SantosJosé Carlos Santos

          168k22132236




          168k22132236












          • $begingroup$
            Actually your example is wrong since -1+1+2=2 is an eigenvalue. What do you mean by eigenvalues of T are sum of pairs of eigenvalue of A?
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:18












          • $begingroup$
            I misundrstood the condition! I thought that it was “the sum of two eigenvalues is not an eigenvalue”. But then, yes, the statment is correct. I will edit my answer and provide a proof.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:26










          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:37










          • $begingroup$
            Never thought of brute force to compute T directly, though the answer turns out to be surprisingly simple. Thanks!
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:44










          • $begingroup$
            What happens for dimensions other than $3times 3$? For $2times 2$ I get $$T(X) = begin{bmatrix} 0 & alpha x+beta x \ -alpha x-beta x & 0 end{bmatrix}$$
            $endgroup$
            – Jeppe Stig Nielsen
            Dec 8 '18 at 17:17


















          • $begingroup$
            Actually your example is wrong since -1+1+2=2 is an eigenvalue. What do you mean by eigenvalues of T are sum of pairs of eigenvalue of A?
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:18












          • $begingroup$
            I misundrstood the condition! I thought that it was “the sum of two eigenvalues is not an eigenvalue”. But then, yes, the statment is correct. I will edit my answer and provide a proof.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:26










          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 16:37










          • $begingroup$
            Never thought of brute force to compute T directly, though the answer turns out to be surprisingly simple. Thanks!
            $endgroup$
            – deep12345
            Dec 8 '18 at 16:44










          • $begingroup$
            What happens for dimensions other than $3times 3$? For $2times 2$ I get $$T(X) = begin{bmatrix} 0 & alpha x+beta x \ -alpha x-beta x & 0 end{bmatrix}$$
            $endgroup$
            – Jeppe Stig Nielsen
            Dec 8 '18 at 17:17
















          $begingroup$
          Actually your example is wrong since -1+1+2=2 is an eigenvalue. What do you mean by eigenvalues of T are sum of pairs of eigenvalue of A?
          $endgroup$
          – deep12345
          Dec 8 '18 at 16:18






          $begingroup$
          Actually your example is wrong since -1+1+2=2 is an eigenvalue. What do you mean by eigenvalues of T are sum of pairs of eigenvalue of A?
          $endgroup$
          – deep12345
          Dec 8 '18 at 16:18














          $begingroup$
          I misundrstood the condition! I thought that it was “the sum of two eigenvalues is not an eigenvalue”. But then, yes, the statment is correct. I will edit my answer and provide a proof.
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 16:26




          $begingroup$
          I misundrstood the condition! I thought that it was “the sum of two eigenvalues is not an eigenvalue”. But then, yes, the statment is correct. I will edit my answer and provide a proof.
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 16:26












          $begingroup$
          I've edited my answer. What do you think now?
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 16:37




          $begingroup$
          I've edited my answer. What do you think now?
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 16:37












          $begingroup$
          Never thought of brute force to compute T directly, though the answer turns out to be surprisingly simple. Thanks!
          $endgroup$
          – deep12345
          Dec 8 '18 at 16:44




          $begingroup$
          Never thought of brute force to compute T directly, though the answer turns out to be surprisingly simple. Thanks!
          $endgroup$
          – deep12345
          Dec 8 '18 at 16:44












          $begingroup$
          What happens for dimensions other than $3times 3$? For $2times 2$ I get $$T(X) = begin{bmatrix} 0 & alpha x+beta x \ -alpha x-beta x & 0 end{bmatrix}$$
          $endgroup$
          – Jeppe Stig Nielsen
          Dec 8 '18 at 17:17




          $begingroup$
          What happens for dimensions other than $3times 3$? For $2times 2$ I get $$T(X) = begin{bmatrix} 0 & alpha x+beta x \ -alpha x-beta x & 0 end{bmatrix}$$
          $endgroup$
          – Jeppe Stig Nielsen
          Dec 8 '18 at 17:17


















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