Determine the principal strain of a 2x2 matrix
$begingroup$
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
asked Dec 23 '18 at 14:33
ecjbecjb
2668
$endgroup$
add a comment |
$begingroup$
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
asked Dec 23 '18 at 14:33
ecjbecjb
2668
$endgroup$
add a comment |
$begingroup$
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
asked Dec 23 '18 at 14:33
ecjbecjb
2668
$endgroup$
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
asked Dec 23 '18 at 14:33
ecjbecjb
2668
asked Dec 23 '18 at 14:33
ecjbecjb
2668
asked Dec 23 '18 at 14:33
ecjbecjb
2668
asked Dec 23 '18 at 14:33
ecjbecjb
2668
asked Dec 23 '18 at 14:33
ecjbecjb
2668
2668
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.
answered Dec 23 '18 at 14:39
Shubham JohriShubham Johri
5,102717
$endgroup$
$begingroup$
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$$(kne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $Amathbf v=lambdamathbf vimplies A(kmathbf v)=lambda(kmathbf v)$.
$endgroup$
– Shubham Johri
Dec 23 '18 at 14:46
$begingroup$
Take $x=1$ to get the eigenvector $mathbf v=(1,1)^T$. The norm of $mathbf v$ is $|mathbf v|=sqrt{1^2+1^2}$, so if you divide $bf v$ by $|mathbf v|=sqrt2$, you will get the unit vector $frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue.
$endgroup$
– Shubham Johri
Dec 23 '18 at 14:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You have $0.1y-0.1x=0implies x=y$. The eigenvector $begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}x\xend{bmatrix},xinBbb R-{0}$. So one possibility for the eigenvector is when $x=1/sqrt2$, that is,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.
answered Dec 23 '18 at 14:39
Shubham JohriShubham Johri
5,102717
$endgroup$
$begingroup$
Thank you for your answer @Shubham Johri. Now where does the $frac{1}{sqrt{1^2+1^2}}$ come from?
$endgroup$
– ecjb
Dec 23 '18 at 14:42
$begingroup$
Note that if $bf v$ is an eigenvector of $A$ corresponding to eigenvalue $lambda$, $kbf v$$(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
add a comment |
$begingroup$
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,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.
answered Dec 23 '18 at 14:39
Shubham JohriShubham Johri
5,102717
$endgroup$
$begingroup$
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$$(kne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $Amathbf v=lambdamathbf vimplies A(kmathbf v)=lambda(kmathbf v)$.
$endgroup$
– Shubham Johri
Dec 23 '18 at 14:46
$begingroup$
Take $x=1$ to get the eigenvector $mathbf v=(1,1)^T$. The norm of $mathbf v$ is $|mathbf v|=sqrt{1^2+1^2}$, so if you divide $bf v$ by $|mathbf v|=sqrt2$, you will get the unit vector $frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue.
$endgroup$
– Shubham Johri
Dec 23 '18 at 14:49
add a comment |
$begingroup$
You have $0.1y-0.1x=0implies x=y$. The eigenvector $begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}x\xend{bmatrix},xinBbb R-{0}$. So one possibility for the eigenvector is when $x=1/sqrt2$, that is,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.
answered Dec 23 '18 at 14:39
Shubham JohriShubham Johri
5,102717
$endgroup$
You have $0.1y-0.1x=0implies x=y$. The eigenvector $begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}x\xend{bmatrix},xinBbb R-{0}$. So one possibility for the eigenvector is when $x=1/sqrt2$, that is,$begin{bmatrix}1/sqrt2\1/sqrt2end{bmatrix}=frac1{sqrt2}begin{bmatrix}1\1end{bmatrix}$.
answered Dec 23 '18 at 14:39
Shubham JohriShubham Johri
5,102717
edited Dec 23 '18 at 14:42
answered Dec 23 '18 at 14:39
Shubham JohriShubham Johri
5,102717
answered Dec 23 '18 at 14:39
Shubham JohriShubham Johri
5,102717
answered Dec 23 '18 at 14:39
Shubham JohriShubham Johri
5,102717
5,102717
$begingroup$
Thank you for your answer @Shubham Johri. Now where does the $frac{1}{sqrt{1^2+1^2}}$ come from?
$endgroup$
– ecjb
Dec 23 '18 at 14:42
$begingroup$
Note that if $bf v$ is an eigenvector of $A$ corresponding to eigenvalue $lambda$, $kbf v$$(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
add a comment |
$begingroup$
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
add a comment |
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