Determine the principal strain of a 2x2 matrix












0












$begingroup$


For a 2D problem the strain matrix is given by



$$
begin{bmatrix}
varepsilon_{xx} & varepsilon_{xy} \
varepsilon_{xy} & varepsilon_{yy} \
end{bmatrix}
=
begin{bmatrix}
0 & 0.1 \
0.1 & 0 \
end{bmatrix}
$$



Determine the principal strains



Solution:



First the eigenvalues have to be computed:



begin{align}
det{textbf{A} - lambda mathbb{I}} &= det begin{bmatrix}
- lambda & 0.1 \
0.1 & -lambda
end{bmatrix}
= lambda^2 - 0.1^2 = 0\
lambda_{1,2} &= pm 0.1
end{align}



Then the first eignevector $vec{v}_1$ is determined by



begin{align}
begin{bmatrix}-0.1 & 0.1 \ 0.1 & -0.1 \
end{bmatrix}
begin{pmatrix} x \ y end{pmatrix} &=
begin{pmatrix} 0 \ 0 end{pmatrix} \
-0.1x + 0.1y &= 0 label{eq1} tag{1}\
vec{v}_1 & = frac{1}{sqrt{1^2 + 1^2}} begin{pmatrix} 1 \ 1 end{pmatrix} label{eq2} tag{2}
end{align}



and the second eigenvector is given by



$$
vec{v}_2
begin{pmatrix} -1/sqrt{2} \ 1/sqrt{2} end{pmatrix}
$$



My question



This is certainly a basic algebra question, but how to you get from (1) to (2)?






eigenvalues-eigenvectors matrix-equations deformation-theory







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    For a 2D problem the strain matrix is given by



    $$
    begin{bmatrix}
    varepsilon_{xx} & varepsilon_{xy} \
    varepsilon_{xy} & varepsilon_{yy} \
    end{bmatrix}
    =
    begin{bmatrix}
    0 & 0.1 \
    0.1 & 0 \
    end{bmatrix}
    $$



    Determine the principal strains



    Solution:



    First the eigenvalues have to be computed:



    begin{align}
    det{textbf{A} - lambda mathbb{I}} &= det begin{bmatrix}
    - lambda & 0.1 \
    0.1 & -lambda
    end{bmatrix}
    = lambda^2 - 0.1^2 = 0\
    lambda_{1,2} &= pm 0.1
    end{align}



    Then the first eignevector $vec{v}_1$ is determined by



    begin{align}
    begin{bmatrix}-0.1 & 0.1 \ 0.1 & -0.1 \
    end{bmatrix}
    begin{pmatrix} x \ y end{pmatrix} &=
    begin{pmatrix} 0 \ 0 end{pmatrix} \
    -0.1x + 0.1y &= 0 label{eq1} tag{1}\
    vec{v}_1 & = frac{1}{sqrt{1^2 + 1^2}} begin{pmatrix} 1 \ 1 end{pmatrix} label{eq2} tag{2}
    end{align}



    and the second eigenvector is given by



    $$
    vec{v}_2
    begin{pmatrix} -1/sqrt{2} \ 1/sqrt{2} end{pmatrix}
    $$



    My question



    This is certainly a basic algebra question, but how to you get from (1) to (2)?






    eigenvalues-eigenvectors matrix-equations deformation-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      For a 2D problem the strain matrix is given by



      $$
      begin{bmatrix}
      varepsilon_{xx} & varepsilon_{xy} \
      varepsilon_{xy} & varepsilon_{yy} \
      end{bmatrix}
      =
      begin{bmatrix}
      0 & 0.1 \
      0.1 & 0 \
      end{bmatrix}
      $$



      Determine the principal strains



      Solution:



      First the eigenvalues have to be computed:



      begin{align}
      det{textbf{A} - lambda mathbb{I}} &= det begin{bmatrix}
      - lambda & 0.1 \
      0.1 & -lambda
      end{bmatrix}
      = lambda^2 - 0.1^2 = 0\
      lambda_{1,2} &= pm 0.1
      end{align}



      Then the first eignevector $vec{v}_1$ is determined by



      begin{align}
      begin{bmatrix}-0.1 & 0.1 \ 0.1 & -0.1 \
      end{bmatrix}
      begin{pmatrix} x \ y end{pmatrix} &=
      begin{pmatrix} 0 \ 0 end{pmatrix} \
      -0.1x + 0.1y &= 0 label{eq1} tag{1}\
      vec{v}_1 & = frac{1}{sqrt{1^2 + 1^2}} begin{pmatrix} 1 \ 1 end{pmatrix} label{eq2} tag{2}
      end{align}



      and the second eigenvector is given by



      $$
      vec{v}_2
      begin{pmatrix} -1/sqrt{2} \ 1/sqrt{2} end{pmatrix}
      $$



      My question



      This is certainly a basic algebra question, but how to you get from (1) to (2)?






      eigenvalues-eigenvectors matrix-equations deformation-theory







      share|cite|improve this question









      $endgroup$




      For a 2D problem the strain matrix is given by



      $$
      begin{bmatrix}
      varepsilon_{xx} & varepsilon_{xy} \
      varepsilon_{xy} & varepsilon_{yy} \
      end{bmatrix}
      =
      begin{bmatrix}
      0 & 0.1 \
      0.1 & 0 \
      end{bmatrix}
      $$



      Determine the principal strains



      Solution:



      First the eigenvalues have to be computed:



      begin{align}
      det{textbf{A} - lambda mathbb{I}} &= det begin{bmatrix}
      - lambda & 0.1 \
      0.1 & -lambda
      end{bmatrix}
      = lambda^2 - 0.1^2 = 0\
      lambda_{1,2} &= pm 0.1
      end{align}



      Then the first eignevector $vec{v}_1$ is determined by



      begin{align}
      begin{bmatrix}-0.1 & 0.1 \ 0.1 & -0.1 \
      end{bmatrix}
      begin{pmatrix} x \ y end{pmatrix} &=
      begin{pmatrix} 0 \ 0 end{pmatrix} \
      -0.1x + 0.1y &= 0 label{eq1} tag{1}\
      vec{v}_1 & = frac{1}{sqrt{1^2 + 1^2}} begin{pmatrix} 1 \ 1 end{pmatrix} label{eq2} tag{2}
      end{align}



      and the second eigenvector is given by



      $$
      vec{v}_2
      begin{pmatrix} -1/sqrt{2} \ 1/sqrt{2} end{pmatrix}
      $$



      My question



      This is certainly a basic algebra question, but how to you get from (1) to (2)?






      eigenvalues-eigenvectors matrix-equations deformation-theory




      eigenvalues-eigenvectors matrix-equations deformation-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 23 '18 at 14:33









      ecjbecjb

      2668




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          $begingroup$

          You have $0.1y-0.1x=0implies x=y$. The eigenvector $begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}x\xend{bmatrix},xinBbb R-{0}$. So one possibility for the eigenvector is when $x=1/sqrt2$, that is,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer @Shubham Johri. Now where does the $frac{1}{sqrt{1^2+1^2}}$ come from?
            $endgroup$
            – ecjb
            Dec 23 '18 at 14:42












          • $begingroup$
            Note that if $bf v$ is an eigenvector of $A$ corresponding to eigenvalue $lambda$, $kbf v$$(kne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $Amathbf v=lambdamathbf vimplies A(kmathbf v)=lambda(kmathbf v)$.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:46










          • $begingroup$
            Take $x=1$ to get the eigenvector $mathbf v=(1,1)^T$. The norm of $mathbf v$ is $|mathbf v|=sqrt{1^2+1^2}$, so if you divide $bf v$ by $|mathbf v|=sqrt2$, you will get the unit vector $frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:49













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          active

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          $begingroup$

          You have $0.1y-0.1x=0implies x=y$. The eigenvector $begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}x\xend{bmatrix},xinBbb R-{0}$. So one possibility for the eigenvector is when $x=1/sqrt2$, that is,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer @Shubham Johri. Now where does the $frac{1}{sqrt{1^2+1^2}}$ come from?
            $endgroup$
            – ecjb
            Dec 23 '18 at 14:42












          • $begingroup$
            Note that if $bf v$ is an eigenvector of $A$ corresponding to eigenvalue $lambda$, $kbf v$$(kne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $Amathbf v=lambdamathbf vimplies A(kmathbf v)=lambda(kmathbf v)$.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:46










          • $begingroup$
            Take $x=1$ to get the eigenvector $mathbf v=(1,1)^T$. The norm of $mathbf v$ is $|mathbf v|=sqrt{1^2+1^2}$, so if you divide $bf v$ by $|mathbf v|=sqrt2$, you will get the unit vector $frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:49


















          1












          $begingroup$

          You have $0.1y-0.1x=0implies x=y$. The eigenvector $begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}x\xend{bmatrix},xinBbb R-{0}$. So one possibility for the eigenvector is when $x=1/sqrt2$, that is,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer @Shubham Johri. Now where does the $frac{1}{sqrt{1^2+1^2}}$ come from?
            $endgroup$
            – ecjb
            Dec 23 '18 at 14:42












          • $begingroup$
            Note that if $bf v$ is an eigenvector of $A$ corresponding to eigenvalue $lambda$, $kbf v$$(kne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $Amathbf v=lambdamathbf vimplies A(kmathbf v)=lambda(kmathbf v)$.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:46










          • $begingroup$
            Take $x=1$ to get the eigenvector $mathbf v=(1,1)^T$. The norm of $mathbf v$ is $|mathbf v|=sqrt{1^2+1^2}$, so if you divide $bf v$ by $|mathbf v|=sqrt2$, you will get the unit vector $frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:49
















          1












          1








          1





          $begingroup$

          You have $0.1y-0.1x=0implies x=y$. The eigenvector $begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}x\xend{bmatrix},xinBbb R-{0}$. So one possibility for the eigenvector is when $x=1/sqrt2$, that is,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.






          share|cite|improve this answer











          $endgroup$



          You have $0.1y-0.1x=0implies x=y$. The eigenvector $begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}x\xend{bmatrix},xinBbb R-{0}$. So one possibility for the eigenvector is when $x=1/sqrt2$, that is,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 23 '18 at 14:42

























          answered Dec 23 '18 at 14:39









          Shubham JohriShubham Johri

          5,102717




          5,102717












          • $begingroup$
            Thank you for your answer @Shubham Johri. Now where does the $frac{1}{sqrt{1^2+1^2}}$ come from?
            $endgroup$
            – ecjb
            Dec 23 '18 at 14:42












          • $begingroup$
            Note that if $bf v$ is an eigenvector of $A$ corresponding to eigenvalue $lambda$, $kbf v$$(kne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $Amathbf v=lambdamathbf vimplies A(kmathbf v)=lambda(kmathbf v)$.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:46










          • $begingroup$
            Take $x=1$ to get the eigenvector $mathbf v=(1,1)^T$. The norm of $mathbf v$ is $|mathbf v|=sqrt{1^2+1^2}$, so if you divide $bf v$ by $|mathbf v|=sqrt2$, you will get the unit vector $frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:49




















          • $begingroup$
            Thank you for your answer @Shubham Johri. Now where does the $frac{1}{sqrt{1^2+1^2}}$ come from?
            $endgroup$
            – ecjb
            Dec 23 '18 at 14:42












          • $begingroup$
            Note that if $bf v$ is an eigenvector of $A$ corresponding to eigenvalue $lambda$, $kbf v$$(kne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $Amathbf v=lambdamathbf vimplies A(kmathbf v)=lambda(kmathbf v)$.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:46










          • $begingroup$
            Take $x=1$ to get the eigenvector $mathbf v=(1,1)^T$. The norm of $mathbf v$ is $|mathbf v|=sqrt{1^2+1^2}$, so if you divide $bf v$ by $|mathbf v|=sqrt2$, you will get the unit vector $frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue.
            $endgroup$
            – Shubham Johri
            Dec 23 '18 at 14:49


















          $begingroup$
          Thank you for your answer @Shubham Johri. Now where does the $frac{1}{sqrt{1^2+1^2}}$ come from?
          $endgroup$
          – ecjb
          Dec 23 '18 at 14:42






          $begingroup$
          Thank you for your answer @Shubham Johri. Now where does the $frac{1}{sqrt{1^2+1^2}}$ come from?
          $endgroup$
          – ecjb
          Dec 23 '18 at 14:42














          $begingroup$
          Note that if $bf v$ is an eigenvector of $A$ corresponding to eigenvalue $lambda$, $kbf v$$(kne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $Amathbf v=lambdamathbf vimplies A(kmathbf v)=lambda(kmathbf v)$.
          $endgroup$
          – Shubham Johri
          Dec 23 '18 at 14:46




          $begingroup$
          Note that if $bf v$ is an eigenvector of $A$ corresponding to eigenvalue $lambda$, $kbf v$$(kne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $Amathbf v=lambdamathbf vimplies A(kmathbf v)=lambda(kmathbf v)$.
          $endgroup$
          – Shubham Johri
          Dec 23 '18 at 14:46












          $begingroup$
          Take $x=1$ to get the eigenvector $mathbf v=(1,1)^T$. The norm of $mathbf v$ is $|mathbf v|=sqrt{1^2+1^2}$, so if you divide $bf v$ by $|mathbf v|=sqrt2$, you will get the unit vector $frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue.
          $endgroup$
          – Shubham Johri
          Dec 23 '18 at 14:49






          $begingroup$
          Take $x=1$ to get the eigenvector $mathbf v=(1,1)^T$. The norm of $mathbf v$ is $|mathbf v|=sqrt{1^2+1^2}$, so if you divide $bf v$ by $|mathbf v|=sqrt2$, you will get the unit vector $frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue.
          $endgroup$
          – Shubham Johri
          Dec 23 '18 at 14:49




















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