Solution to this problem is incorrect about a diagonalizable matrix A?












1














I have this matrix A



$$ begin{bmatrix}
1 & c & 0 & 0 & 0 \
0 & 1 & c & 0 & 0 \
0 & 0 & 1 & a & 0\
0 & 0 & 0 & b & c\
0 & 0 & 0 & 0 & b
end{bmatrix}
$$




(d) (4 points) Find all values of a, b, c, if any, such that A is
diagonalizable but not orthogonally diagonalizable. Justify your
answer.



Solution: By spectral theorem, we know that A is orthogonally
diagonalizable iff A is symmetric. So we know that at least one of a
or c must be nonzero so that A is not symmetric. If b !=1, we know
by part (c) that c=0 for the matrix to be diagonalizable. Therefore, a != 0.
If b = 1, then the matrix is only diagonalizable iff a = c = 0 which
is a contradiction as then A will be symmetric. Therefore, the
requirements are the b != 1, c = 0, and a != 0.




However, b must be 0. For example, you cannot have b = 2. If b = 2, then the dimension of the kernel of this matrix is only $1$, so the eigenvalue $1$ has $gemu(1) < almu(1)$, so this matrix would not be diagonalizable. Is there anything wrong with my reasoning?






linear-algebra matrices diagonalization







share|cite|improve this question





























    1














    I have this matrix A



    $$ begin{bmatrix}
    1 & c & 0 & 0 & 0 \
    0 & 1 & c & 0 & 0 \
    0 & 0 & 1 & a & 0\
    0 & 0 & 0 & b & c\
    0 & 0 & 0 & 0 & b
    end{bmatrix}
    $$




    (d) (4 points) Find all values of a, b, c, if any, such that A is
    diagonalizable but not orthogonally diagonalizable. Justify your
    answer.



    Solution: By spectral theorem, we know that A is orthogonally
    diagonalizable iff A is symmetric. So we know that at least one of a
    or c must be nonzero so that A is not symmetric. If b !=1, we know
    by part (c) that c=0 for the matrix to be diagonalizable. Therefore, a != 0.
    If b = 1, then the matrix is only diagonalizable iff a = c = 0 which
    is a contradiction as then A will be symmetric. Therefore, the
    requirements are the b != 1, c = 0, and a != 0.




    However, b must be 0. For example, you cannot have b = 2. If b = 2, then the dimension of the kernel of this matrix is only $1$, so the eigenvalue $1$ has $gemu(1) < almu(1)$, so this matrix would not be diagonalizable. Is there anything wrong with my reasoning?






    linear-algebra matrices diagonalization







    share|cite|improve this question



























      1












      1








      1







      I have this matrix A



      $$ begin{bmatrix}
      1 & c & 0 & 0 & 0 \
      0 & 1 & c & 0 & 0 \
      0 & 0 & 1 & a & 0\
      0 & 0 & 0 & b & c\
      0 & 0 & 0 & 0 & b
      end{bmatrix}
      $$




      (d) (4 points) Find all values of a, b, c, if any, such that A is
      diagonalizable but not orthogonally diagonalizable. Justify your
      answer.



      Solution: By spectral theorem, we know that A is orthogonally
      diagonalizable iff A is symmetric. So we know that at least one of a
      or c must be nonzero so that A is not symmetric. If b !=1, we know
      by part (c) that c=0 for the matrix to be diagonalizable. Therefore, a != 0.
      If b = 1, then the matrix is only diagonalizable iff a = c = 0 which
      is a contradiction as then A will be symmetric. Therefore, the
      requirements are the b != 1, c = 0, and a != 0.




      However, b must be 0. For example, you cannot have b = 2. If b = 2, then the dimension of the kernel of this matrix is only $1$, so the eigenvalue $1$ has $gemu(1) < almu(1)$, so this matrix would not be diagonalizable. Is there anything wrong with my reasoning?






      linear-algebra matrices diagonalization







      share|cite|improve this question















      I have this matrix A



      $$ begin{bmatrix}
      1 & c & 0 & 0 & 0 \
      0 & 1 & c & 0 & 0 \
      0 & 0 & 1 & a & 0\
      0 & 0 & 0 & b & c\
      0 & 0 & 0 & 0 & b
      end{bmatrix}
      $$




      (d) (4 points) Find all values of a, b, c, if any, such that A is
      diagonalizable but not orthogonally diagonalizable. Justify your
      answer.



      Solution: By spectral theorem, we know that A is orthogonally
      diagonalizable iff A is symmetric. So we know that at least one of a
      or c must be nonzero so that A is not symmetric. If b !=1, we know
      by part (c) that c=0 for the matrix to be diagonalizable. Therefore, a != 0.
      If b = 1, then the matrix is only diagonalizable iff a = c = 0 which
      is a contradiction as then A will be symmetric. Therefore, the
      requirements are the b != 1, c = 0, and a != 0.




      However, b must be 0. For example, you cannot have b = 2. If b = 2, then the dimension of the kernel of this matrix is only $1$, so the eigenvalue $1$ has $gemu(1) < almu(1)$, so this matrix would not be diagonalizable. Is there anything wrong with my reasoning?






      linear-algebra matrices diagonalization




      linear-algebra matrices diagonalization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 at 9:25









      José Carlos Santos

      148k22117218




      148k22117218










      asked Apr 18 at 23:13









      Goldname

      1,450928




      1,450928






















          2 Answers
          2






          active

          oldest

          votes


















          1














          Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.






          share|cite|improve this answer





















          • Ah I made mistake. I forgot c = 0.
            – Goldname
            Apr 19 at 1:11



















          1














          With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.






            share|cite|improve this answer





















            • Ah I made mistake. I forgot c = 0.
              – Goldname
              Apr 19 at 1:11
















            1














            Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.






            share|cite|improve this answer





















            • Ah I made mistake. I forgot c = 0.
              – Goldname
              Apr 19 at 1:11














            1












            1








            1






            Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.






            share|cite|improve this answer












            Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 18 at 23:34









            José Carlos Santos

            148k22117218




            148k22117218












            • Ah I made mistake. I forgot c = 0.
              – Goldname
              Apr 19 at 1:11


















            • Ah I made mistake. I forgot c = 0.
              – Goldname
              Apr 19 at 1:11
















            Ah I made mistake. I forgot c = 0.
            – Goldname
            Apr 19 at 1:11




            Ah I made mistake. I forgot c = 0.
            – Goldname
            Apr 19 at 1:11











            1














            With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.






            share|cite|improve this answer


























              1














              With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.






              share|cite|improve this answer
























                1












                1








                1






                With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.






                share|cite|improve this answer












                With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 18 at 23:50









                amd

                29k21050




                29k21050






























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