characterising the square 2x2 matrices on types of eigenvalues












0














If A is a 2 X 2 matrix with complex entries, then A is similar over $C$ to a matrix of one of the two types :-




  1. begin{bmatrix}
    a & 0 \
    0 & b
    end{bmatrix}


  2. begin{bmatrix}
    a & 0 \
    1 & a
    end{bmatrix}
    We know we get 2 eigenvalues since the matrix is over complex numbers.
    If the Eigenvalues are distinct, or if they are equal with geometric multiplicity 2 then it's similar to type 1. If they are equal but the geometric multiplicity is 1 then we get one eigenvector, how do I find out the other basis element?






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question





























    0














    If A is a 2 X 2 matrix with complex entries, then A is similar over $C$ to a matrix of one of the two types :-




    1. begin{bmatrix}
      a & 0 \
      0 & b
      end{bmatrix}


    2. begin{bmatrix}
      a & 0 \
      1 & a
      end{bmatrix}
      We know we get 2 eigenvalues since the matrix is over complex numbers.
      If the Eigenvalues are distinct, or if they are equal with geometric multiplicity 2 then it's similar to type 1. If they are equal but the geometric multiplicity is 1 then we get one eigenvector, how do I find out the other basis element?






    linear-algebra matrices eigenvalues-eigenvectors







    share|cite|improve this question



























      0












      0








      0







      If A is a 2 X 2 matrix with complex entries, then A is similar over $C$ to a matrix of one of the two types :-




      1. begin{bmatrix}
        a & 0 \
        0 & b
        end{bmatrix}


      2. begin{bmatrix}
        a & 0 \
        1 & a
        end{bmatrix}
        We know we get 2 eigenvalues since the matrix is over complex numbers.
        If the Eigenvalues are distinct, or if they are equal with geometric multiplicity 2 then it's similar to type 1. If they are equal but the geometric multiplicity is 1 then we get one eigenvector, how do I find out the other basis element?






      linear-algebra matrices eigenvalues-eigenvectors







      share|cite|improve this question















      If A is a 2 X 2 matrix with complex entries, then A is similar over $C$ to a matrix of one of the two types :-




      1. begin{bmatrix}
        a & 0 \
        0 & b
        end{bmatrix}


      2. begin{bmatrix}
        a & 0 \
        1 & a
        end{bmatrix}
        We know we get 2 eigenvalues since the matrix is over complex numbers.
        If the Eigenvalues are distinct, or if they are equal with geometric multiplicity 2 then it's similar to type 1. If they are equal but the geometric multiplicity is 1 then we get one eigenvector, how do I find out the other basis element?






      linear-algebra matrices eigenvalues-eigenvectors




      linear-algebra matrices eigenvalues-eigenvectors






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 24 at 12:10









      José Carlos Santos

      148k22117218




      148k22117218










      asked Nov 24 at 11:44









      Rohan Rajagopal

      5811




      5811






















          1 Answer
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          If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.






          share|cite|improve this answer





















          • I know, I got that. But how are we sure to get such a vector?
            – Rohan Rajagopal
            Nov 24 at 12:09








          • 1




            Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
            – José Carlos Santos
            Nov 24 at 12:17












          • Thank you very much. Deeply appreciate it.
            – Rohan Rajagopal
            Nov 24 at 13:01











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          1 Answer
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          1 Answer
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          active

          oldest

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          1














          If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.






          share|cite|improve this answer





















          • I know, I got that. But how are we sure to get such a vector?
            – Rohan Rajagopal
            Nov 24 at 12:09








          • 1




            Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
            – José Carlos Santos
            Nov 24 at 12:17












          • Thank you very much. Deeply appreciate it.
            – Rohan Rajagopal
            Nov 24 at 13:01
















          1














          If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.






          share|cite|improve this answer





















          • I know, I got that. But how are we sure to get such a vector?
            – Rohan Rajagopal
            Nov 24 at 12:09








          • 1




            Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
            – José Carlos Santos
            Nov 24 at 12:17












          • Thank you very much. Deeply appreciate it.
            – Rohan Rajagopal
            Nov 24 at 13:01














          1












          1








          1






          If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.






          share|cite|improve this answer












          If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 12:01









          José Carlos Santos

          148k22117218




          148k22117218












          • I know, I got that. But how are we sure to get such a vector?
            – Rohan Rajagopal
            Nov 24 at 12:09








          • 1




            Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
            – José Carlos Santos
            Nov 24 at 12:17












          • Thank you very much. Deeply appreciate it.
            – Rohan Rajagopal
            Nov 24 at 13:01


















          • I know, I got that. But how are we sure to get such a vector?
            – Rohan Rajagopal
            Nov 24 at 12:09








          • 1




            Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
            – José Carlos Santos
            Nov 24 at 12:17












          • Thank you very much. Deeply appreciate it.
            – Rohan Rajagopal
            Nov 24 at 13:01
















          I know, I got that. But how are we sure to get such a vector?
          – Rohan Rajagopal
          Nov 24 at 12:09






          I know, I got that. But how are we sure to get such a vector?
          – Rohan Rajagopal
          Nov 24 at 12:09






          1




          1




          Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
          – José Carlos Santos
          Nov 24 at 12:17






          Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
          – José Carlos Santos
          Nov 24 at 12:17














          Thank you very much. Deeply appreciate it.
          – Rohan Rajagopal
          Nov 24 at 13:01




          Thank you very much. Deeply appreciate it.
          – Rohan Rajagopal
          Nov 24 at 13:01


















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