Regarding a real $n times n$ matrix $A$ satisfying $A^6 = -A^2$












1












$begingroup$


The Problem:



Suppose $A$ represents a real $n times n$ matrix satisfying $A^6 = -A^2$.



(a) Prove that if $A$ is symmetric, then $A = 0$.



(b) Prove that if $n$ is odd, then $A$ is not invertible.



(c) Give an example of an invertible $A$ with $n$ even (e.g., $n = 2$).



My Approach:



For (a): I'm just going through all of the properties of real symmetric matrices and seeing if anything pops out. Let $lambda_1, ..., lambda_n$ be the eigenvalues of $A$. Then we have that $A = QDQ^{-1}$, where $D$ is diagonal (with $lambda_1, ..., lambda_n$ as its diagonal entries) and $Q$ is a real orthogonal matrix (with the eigenvectors of $A$ as its columns). So, since $A^6 = -A^2$, it follows that $D^6 = -D^2$; and so $lambda_i^6 = -(lambda_i^2)$, for each $i in {1,2,...,n}$. But since $lambda_1, ..., lambda_n in mathbb{R}$, it must then be that $lambda_1 = cdots =lambda_n = 0$. Well, that means that $D = 0$, but that doesn't necessarily mean that $A = 0$... or does it?



For (b): (I think I've got this one.) Since $A^6 = -A^2$, it follows that $text{det}(A)^6 = (-1)^ntext{det}(A)^2$. But the LHS of this equation is positive and the RHS is negative (since $n$ is odd). Thus $text{det}(A)$ must be $0$; and so $A$ is not invertible.



For (c): I can easily come up with examples satisfying $A^6 = A^2$; e.g.,
begin{pmatrix}
0 & 1 \
-1 & 0 \
end{pmatrix}
But that minus sign is killing me. Of course, I'm just using trial-and-error here; I'd certainly prefer a more "informed" approach to finding such a matrix...






linear-algebra matrices matrix-equations







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    The Problem:



    Suppose $A$ represents a real $n times n$ matrix satisfying $A^6 = -A^2$.



    (a) Prove that if $A$ is symmetric, then $A = 0$.



    (b) Prove that if $n$ is odd, then $A$ is not invertible.



    (c) Give an example of an invertible $A$ with $n$ even (e.g., $n = 2$).



    My Approach:



    For (a): I'm just going through all of the properties of real symmetric matrices and seeing if anything pops out. Let $lambda_1, ..., lambda_n$ be the eigenvalues of $A$. Then we have that $A = QDQ^{-1}$, where $D$ is diagonal (with $lambda_1, ..., lambda_n$ as its diagonal entries) and $Q$ is a real orthogonal matrix (with the eigenvectors of $A$ as its columns). So, since $A^6 = -A^2$, it follows that $D^6 = -D^2$; and so $lambda_i^6 = -(lambda_i^2)$, for each $i in {1,2,...,n}$. But since $lambda_1, ..., lambda_n in mathbb{R}$, it must then be that $lambda_1 = cdots =lambda_n = 0$. Well, that means that $D = 0$, but that doesn't necessarily mean that $A = 0$... or does it?



    For (b): (I think I've got this one.) Since $A^6 = -A^2$, it follows that $text{det}(A)^6 = (-1)^ntext{det}(A)^2$. But the LHS of this equation is positive and the RHS is negative (since $n$ is odd). Thus $text{det}(A)$ must be $0$; and so $A$ is not invertible.



    For (c): I can easily come up with examples satisfying $A^6 = A^2$; e.g.,
    begin{pmatrix}
    0 & 1 \
    -1 & 0 \
    end{pmatrix}
    But that minus sign is killing me. Of course, I'm just using trial-and-error here; I'd certainly prefer a more "informed" approach to finding such a matrix...






    linear-algebra matrices matrix-equations







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      The Problem:



      Suppose $A$ represents a real $n times n$ matrix satisfying $A^6 = -A^2$.



      (a) Prove that if $A$ is symmetric, then $A = 0$.



      (b) Prove that if $n$ is odd, then $A$ is not invertible.



      (c) Give an example of an invertible $A$ with $n$ even (e.g., $n = 2$).



      My Approach:



      For (a): I'm just going through all of the properties of real symmetric matrices and seeing if anything pops out. Let $lambda_1, ..., lambda_n$ be the eigenvalues of $A$. Then we have that $A = QDQ^{-1}$, where $D$ is diagonal (with $lambda_1, ..., lambda_n$ as its diagonal entries) and $Q$ is a real orthogonal matrix (with the eigenvectors of $A$ as its columns). So, since $A^6 = -A^2$, it follows that $D^6 = -D^2$; and so $lambda_i^6 = -(lambda_i^2)$, for each $i in {1,2,...,n}$. But since $lambda_1, ..., lambda_n in mathbb{R}$, it must then be that $lambda_1 = cdots =lambda_n = 0$. Well, that means that $D = 0$, but that doesn't necessarily mean that $A = 0$... or does it?



      For (b): (I think I've got this one.) Since $A^6 = -A^2$, it follows that $text{det}(A)^6 = (-1)^ntext{det}(A)^2$. But the LHS of this equation is positive and the RHS is negative (since $n$ is odd). Thus $text{det}(A)$ must be $0$; and so $A$ is not invertible.



      For (c): I can easily come up with examples satisfying $A^6 = A^2$; e.g.,
      begin{pmatrix}
      0 & 1 \
      -1 & 0 \
      end{pmatrix}
      But that minus sign is killing me. Of course, I'm just using trial-and-error here; I'd certainly prefer a more "informed" approach to finding such a matrix...






      linear-algebra matrices matrix-equations







      share|cite|improve this question











      $endgroup$




      The Problem:



      Suppose $A$ represents a real $n times n$ matrix satisfying $A^6 = -A^2$.



      (a) Prove that if $A$ is symmetric, then $A = 0$.



      (b) Prove that if $n$ is odd, then $A$ is not invertible.



      (c) Give an example of an invertible $A$ with $n$ even (e.g., $n = 2$).



      My Approach:



      For (a): I'm just going through all of the properties of real symmetric matrices and seeing if anything pops out. Let $lambda_1, ..., lambda_n$ be the eigenvalues of $A$. Then we have that $A = QDQ^{-1}$, where $D$ is diagonal (with $lambda_1, ..., lambda_n$ as its diagonal entries) and $Q$ is a real orthogonal matrix (with the eigenvectors of $A$ as its columns). So, since $A^6 = -A^2$, it follows that $D^6 = -D^2$; and so $lambda_i^6 = -(lambda_i^2)$, for each $i in {1,2,...,n}$. But since $lambda_1, ..., lambda_n in mathbb{R}$, it must then be that $lambda_1 = cdots =lambda_n = 0$. Well, that means that $D = 0$, but that doesn't necessarily mean that $A = 0$... or does it?



      For (b): (I think I've got this one.) Since $A^6 = -A^2$, it follows that $text{det}(A)^6 = (-1)^ntext{det}(A)^2$. But the LHS of this equation is positive and the RHS is negative (since $n$ is odd). Thus $text{det}(A)$ must be $0$; and so $A$ is not invertible.



      For (c): I can easily come up with examples satisfying $A^6 = A^2$; e.g.,
      begin{pmatrix}
      0 & 1 \
      -1 & 0 \
      end{pmatrix}
      But that minus sign is killing me. Of course, I'm just using trial-and-error here; I'd certainly prefer a more "informed" approach to finding such a matrix...






      linear-algebra matrices matrix-equations




      linear-algebra matrices matrix-equations






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      edited Jan 13 at 17:02









      José Carlos Santos

      168k23132236




      168k23132236










      asked Jan 13 at 16:50









      thisisourconcerndudethisisourconcerndude

      1,1271123




      1,1271123






















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          $begingroup$

          (a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.



          (b) Right.



          (c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.






          share|cite|improve this answer











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            $begingroup$

            (a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.



            (b) Right.



            (c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              (a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.



              (b) Right.



              (c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                (a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.



                (b) Right.



                (c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.






                share|cite|improve this answer











                $endgroup$



                (a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.



                (b) Right.



                (c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 13 at 17:11

























                answered Jan 13 at 16:57









                José Carlos SantosJosé Carlos Santos

                168k23132236




                168k23132236






























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