Finding the sec. Eigenvector, when knowing the first Eigenvector and Eigenvalue
$begingroup$
So here is my problem,
Let A be a symmetric Matrix (2x2) with EV=(-2, 3) and the Eigenvalue being -5.
Find the 2. Eigenvector to the second Eigenvalue.
The only info i can think of is that the EV of a symmetric matrix have to be orthogonal, so the dot product has to be 0.
So -2x1 +3x2 =0
I then can fix x1=a for example and then calculate, x2=2/3
$$ (1, 2/3)^t *a, $$
for all a Elements of R without 0
Is this the correct answer it seems a bit too simple....
linear-algebra eigenvalues-eigenvectors
asked Dec 9 '18 at 19:50
LillysLillys
778
$endgroup$
add a comment |
$begingroup$
So here is my problem,
Let A be a symmetric Matrix (2x2) with EV=(-2, 3) and the Eigenvalue being -5.
Find the 2. Eigenvector to the second Eigenvalue.
The only info i can think of is that the EV of a symmetric matrix have to be orthogonal, so the dot product has to be 0.
So -2x1 +3x2 =0
I then can fix x1=a for example and then calculate, x2=2/3
$$ (1, 2/3)^t *a, $$
for all a Elements of R without 0
Is this the correct answer it seems a bit too simple....
linear-algebra eigenvalues-eigenvectors
asked Dec 9 '18 at 19:50
LillysLillys
778
$endgroup$
add a comment |
$begingroup$
So here is my problem,
Let A be a symmetric Matrix (2x2) with EV=(-2, 3) and the Eigenvalue being -5.
Find the 2. Eigenvector to the second Eigenvalue.
The only info i can think of is that the EV of a symmetric matrix have to be orthogonal, so the dot product has to be 0.
So -2x1 +3x2 =0
I then can fix x1=a for example and then calculate, x2=2/3
$$ (1, 2/3)^t *a, $$
for all a Elements of R without 0
Is this the correct answer it seems a bit too simple....
linear-algebra eigenvalues-eigenvectors
asked Dec 9 '18 at 19:50
LillysLillys
778
$endgroup$
So here is my problem,
Let A be a symmetric Matrix (2x2) with EV=(-2, 3) and the Eigenvalue being -5.
Find the 2. Eigenvector to the second Eigenvalue.
The only info i can think of is that the EV of a symmetric matrix have to be orthogonal, so the dot product has to be 0.
So -2x1 +3x2 =0
I then can fix x1=a for example and then calculate, x2=2/3
$$ (1, 2/3)^t *a, $$
for all a Elements of R without 0
Is this the correct answer it seems a bit too simple....
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Dec 9 '18 at 19:50
LillysLillys
778
asked Dec 9 '18 at 19:50
LillysLillys
778
asked Dec 9 '18 at 19:50
LillysLillys
778
asked Dec 9 '18 at 19:50
LillysLillys
778
asked Dec 9 '18 at 19:50
LillysLillys
778
778
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You know that $begin{pmatrix}-2\3end{pmatrix}$ is an eigenvector relative to $-5$. You correctly argue that the eigenvectors relative to the other eigenvalues are orthogonal to this one, so we find $-2x_1+3x_2=0$. Any such vector is good, so we can take $begin{pmatrix}3\2end{pmatrix}$.
You are correct.
The spectral theorem says that
$$
A=-5vv^T+lambda ww^T
$$
where $v$ and $w$ are the normalization of the two eigenvectors. There is no way to determine the second eigenvalue from the given data. You have
$$
A=-frac{5}{13}begin{pmatrix} 4 & -6 \ -6 & 9end{pmatrix}
+frac{lambda}{13}begin{pmatrix} 9 & 6 \ 6 & 4end{pmatrix}
$$
answered Dec 9 '18 at 21:07
egregegreg
181k1485202
$endgroup$
add a comment |
$begingroup$
Your solution and reasoning are correct. Every $2times2$ symmetric matrix has a pair of orthogonal eigenvectors. Thus, having one of the two eigenvectors specifies the other, as in two dimensions there are only two orthogonal directions.
If you want to see this algebraically, it follows from
begin{multline}
left[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} x \ yend{matrix}right] = lambda_1 left[begin{matrix} x \ yend{matrix}right]Longrightarrowbegin{matrix}a x + b y = lambda_1 x \ b x + c y = lambda_1 yend{matrix}Longrightarrowbegin{matrix} (c-lambda_1)y + b x =0 \ -b y +(lambda_1 - a)x = 0end{matrix} \ Longrightarrowbegin{matrix} -(a - a - c + lambda_1)y + b x =0 \ -b y +(c - c + lambda_1 - a)x = 0end{matrix}Longrightarrowbegin{matrix} -ay + b x = -(a+c-lambda_1)y \ -b y +c x = (a + c -lambda_1) x end{matrix}\Longrightarrowleft[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} -y \ xend{matrix}right]=(a + c - lambda_1) left[begin{matrix} -y \ xend{matrix}right].
end{multline}
answered Dec 9 '18 at 21:00
eyeballfrogeyeballfrog
6,103629
$endgroup$
add a comment |
$begingroup$
If $$M=begin{pmatrix}
a & b \ b & c
end{pmatrix}.$$
Then $lambda+ lambda ' = a+c$ and $lambda lambda ' = ac-b^2$ and $$-2a+3b =10$$ $$-2b+3c =-15$$
Let $b=6t$ and $lambda ' = -5$ then $$a= 9t-5$$ $$c= 4t-5$$
so we have $lambda = 13t-5$. Now we see that your matrix is not uniqely determined: $$M=begin{pmatrix}
9t-5 & 6t \ 6t & 4t-5
end{pmatrix}.$$
answered Dec 9 '18 at 19:58
greedoidgreedoid
40.1k114799
$endgroup$
$begingroup$
Sorry I still don’t get it. U used the trace, and than that the product of the eigenvalues =the determinant?? I have never heard of it. How did you get to the equations and how do i solve them for an Eigenvector, i have 3 variables and two equations.
$endgroup$
– Lillys
Dec 9 '18 at 20:06
add a comment |
Your Answer
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You know that $begin{pmatrix}-2\3end{pmatrix}$ is an eigenvector relative to $-5$. You correctly argue that the eigenvectors relative to the other eigenvalues are orthogonal to this one, so we find $-2x_1+3x_2=0$. Any such vector is good, so we can take $begin{pmatrix}3\2end{pmatrix}$.
You are correct.
The spectral theorem says that
$$
A=-5vv^T+lambda ww^T
$$
where $v$ and $w$ are the normalization of the two eigenvectors. There is no way to determine the second eigenvalue from the given data. You have
$$
A=-frac{5}{13}begin{pmatrix} 4 & -6 \ -6 & 9end{pmatrix}
+frac{lambda}{13}begin{pmatrix} 9 & 6 \ 6 & 4end{pmatrix}
$$
answered Dec 9 '18 at 21:07
egregegreg
181k1485202
$endgroup$
add a comment |
$begingroup$
You know that $begin{pmatrix}-2\3end{pmatrix}$ is an eigenvector relative to $-5$. You correctly argue that the eigenvectors relative to the other eigenvalues are orthogonal to this one, so we find $-2x_1+3x_2=0$. Any such vector is good, so we can take $begin{pmatrix}3\2end{pmatrix}$.
You are correct.
The spectral theorem says that
$$
A=-5vv^T+lambda ww^T
$$
where $v$ and $w$ are the normalization of the two eigenvectors. There is no way to determine the second eigenvalue from the given data. You have
$$
A=-frac{5}{13}begin{pmatrix} 4 & -6 \ -6 & 9end{pmatrix}
+frac{lambda}{13}begin{pmatrix} 9 & 6 \ 6 & 4end{pmatrix}
$$
answered Dec 9 '18 at 21:07
egregegreg
181k1485202
$endgroup$
add a comment |
$begingroup$
You know that $begin{pmatrix}-2\3end{pmatrix}$ is an eigenvector relative to $-5$. You correctly argue that the eigenvectors relative to the other eigenvalues are orthogonal to this one, so we find $-2x_1+3x_2=0$. Any such vector is good, so we can take $begin{pmatrix}3\2end{pmatrix}$.
You are correct.
The spectral theorem says that
$$
A=-5vv^T+lambda ww^T
$$
where $v$ and $w$ are the normalization of the two eigenvectors. There is no way to determine the second eigenvalue from the given data. You have
$$
A=-frac{5}{13}begin{pmatrix} 4 & -6 \ -6 & 9end{pmatrix}
+frac{lambda}{13}begin{pmatrix} 9 & 6 \ 6 & 4end{pmatrix}
$$
answered Dec 9 '18 at 21:07
egregegreg
181k1485202
$endgroup$
You know that $begin{pmatrix}-2\3end{pmatrix}$ is an eigenvector relative to $-5$. You correctly argue that the eigenvectors relative to the other eigenvalues are orthogonal to this one, so we find $-2x_1+3x_2=0$. Any such vector is good, so we can take $begin{pmatrix}3\2end{pmatrix}$.
You are correct.
The spectral theorem says that
$$
A=-5vv^T+lambda ww^T
$$
where $v$ and $w$ are the normalization of the two eigenvectors. There is no way to determine the second eigenvalue from the given data. You have
$$
A=-frac{5}{13}begin{pmatrix} 4 & -6 \ -6 & 9end{pmatrix}
+frac{lambda}{13}begin{pmatrix} 9 & 6 \ 6 & 4end{pmatrix}
$$
answered Dec 9 '18 at 21:07
egregegreg
181k1485202
answered Dec 9 '18 at 21:07
egregegreg
181k1485202
answered Dec 9 '18 at 21:07
egregegreg
181k1485202
answered Dec 9 '18 at 21:07
egregegreg
181k1485202
181k1485202
add a comment |
add a comment |
$begingroup$
Your solution and reasoning are correct. Every $2times2$ symmetric matrix has a pair of orthogonal eigenvectors. Thus, having one of the two eigenvectors specifies the other, as in two dimensions there are only two orthogonal directions.
If you want to see this algebraically, it follows from
begin{multline}
left[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} x \ yend{matrix}right] = lambda_1 left[begin{matrix} x \ yend{matrix}right]Longrightarrowbegin{matrix}a x + b y = lambda_1 x \ b x + c y = lambda_1 yend{matrix}Longrightarrowbegin{matrix} (c-lambda_1)y + b x =0 \ -b y +(lambda_1 - a)x = 0end{matrix} \ Longrightarrowbegin{matrix} -(a - a - c + lambda_1)y + b x =0 \ -b y +(c - c + lambda_1 - a)x = 0end{matrix}Longrightarrowbegin{matrix} -ay + b x = -(a+c-lambda_1)y \ -b y +c x = (a + c -lambda_1) x end{matrix}\Longrightarrowleft[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} -y \ xend{matrix}right]=(a + c - lambda_1) left[begin{matrix} -y \ xend{matrix}right].
end{multline}
answered Dec 9 '18 at 21:00
eyeballfrogeyeballfrog
6,103629
$endgroup$
add a comment |
$begingroup$
Your solution and reasoning are correct. Every $2times2$ symmetric matrix has a pair of orthogonal eigenvectors. Thus, having one of the two eigenvectors specifies the other, as in two dimensions there are only two orthogonal directions.
If you want to see this algebraically, it follows from
begin{multline}
left[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} x \ yend{matrix}right] = lambda_1 left[begin{matrix} x \ yend{matrix}right]Longrightarrowbegin{matrix}a x + b y = lambda_1 x \ b x + c y = lambda_1 yend{matrix}Longrightarrowbegin{matrix} (c-lambda_1)y + b x =0 \ -b y +(lambda_1 - a)x = 0end{matrix} \ Longrightarrowbegin{matrix} -(a - a - c + lambda_1)y + b x =0 \ -b y +(c - c + lambda_1 - a)x = 0end{matrix}Longrightarrowbegin{matrix} -ay + b x = -(a+c-lambda_1)y \ -b y +c x = (a + c -lambda_1) x end{matrix}\Longrightarrowleft[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} -y \ xend{matrix}right]=(a + c - lambda_1) left[begin{matrix} -y \ xend{matrix}right].
end{multline}
answered Dec 9 '18 at 21:00
eyeballfrogeyeballfrog
6,103629
$endgroup$
add a comment |
$begingroup$
Your solution and reasoning are correct. Every $2times2$ symmetric matrix has a pair of orthogonal eigenvectors. Thus, having one of the two eigenvectors specifies the other, as in two dimensions there are only two orthogonal directions.
If you want to see this algebraically, it follows from
begin{multline}
left[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} x \ yend{matrix}right] = lambda_1 left[begin{matrix} x \ yend{matrix}right]Longrightarrowbegin{matrix}a x + b y = lambda_1 x \ b x + c y = lambda_1 yend{matrix}Longrightarrowbegin{matrix} (c-lambda_1)y + b x =0 \ -b y +(lambda_1 - a)x = 0end{matrix} \ Longrightarrowbegin{matrix} -(a - a - c + lambda_1)y + b x =0 \ -b y +(c - c + lambda_1 - a)x = 0end{matrix}Longrightarrowbegin{matrix} -ay + b x = -(a+c-lambda_1)y \ -b y +c x = (a + c -lambda_1) x end{matrix}\Longrightarrowleft[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} -y \ xend{matrix}right]=(a + c - lambda_1) left[begin{matrix} -y \ xend{matrix}right].
end{multline}
answered Dec 9 '18 at 21:00
eyeballfrogeyeballfrog
6,103629
$endgroup$
Your solution and reasoning are correct. Every $2times2$ symmetric matrix has a pair of orthogonal eigenvectors. Thus, having one of the two eigenvectors specifies the other, as in two dimensions there are only two orthogonal directions.
If you want to see this algebraically, it follows from
begin{multline}
left[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} x \ yend{matrix}right] = lambda_1 left[begin{matrix} x \ yend{matrix}right]Longrightarrowbegin{matrix}a x + b y = lambda_1 x \ b x + c y = lambda_1 yend{matrix}Longrightarrowbegin{matrix} (c-lambda_1)y + b x =0 \ -b y +(lambda_1 - a)x = 0end{matrix} \ Longrightarrowbegin{matrix} -(a - a - c + lambda_1)y + b x =0 \ -b y +(c - c + lambda_1 - a)x = 0end{matrix}Longrightarrowbegin{matrix} -ay + b x = -(a+c-lambda_1)y \ -b y +c x = (a + c -lambda_1) x end{matrix}\Longrightarrowleft[begin{matrix} a & b \ b & cend{matrix}right]left[begin{matrix} -y \ xend{matrix}right]=(a + c - lambda_1) left[begin{matrix} -y \ xend{matrix}right].
end{multline}
answered Dec 9 '18 at 21:00
eyeballfrogeyeballfrog
6,103629
answered Dec 9 '18 at 21:00
eyeballfrogeyeballfrog
6,103629
answered Dec 9 '18 at 21:00
eyeballfrogeyeballfrog
6,103629
answered Dec 9 '18 at 21:00
eyeballfrogeyeballfrog
6,103629
6,103629
add a comment |
add a comment |
$begingroup$
If $$M=begin{pmatrix}
a & b \ b & c
end{pmatrix}.$$
Then $lambda+ lambda ' = a+c$ and $lambda lambda ' = ac-b^2$ and $$-2a+3b =10$$ $$-2b+3c =-15$$
Let $b=6t$ and $lambda ' = -5$ then $$a= 9t-5$$ $$c= 4t-5$$
so we have $lambda = 13t-5$. Now we see that your matrix is not uniqely determined: $$M=begin{pmatrix}
9t-5 & 6t \ 6t & 4t-5
end{pmatrix}.$$
answered Dec 9 '18 at 19:58
greedoidgreedoid
40.1k114799
$endgroup$
$begingroup$
Sorry I still don’t get it. U used the trace, and than that the product of the eigenvalues =the determinant?? I have never heard of it. How did you get to the equations and how do i solve them for an Eigenvector, i have 3 variables and two equations.
$endgroup$
– Lillys
Dec 9 '18 at 20:06
add a comment |
$begingroup$
If $$M=begin{pmatrix}
a & b \ b & c
end{pmatrix}.$$
Then $lambda+ lambda ' = a+c$ and $lambda lambda ' = ac-b^2$ and $$-2a+3b =10$$ $$-2b+3c =-15$$
Let $b=6t$ and $lambda ' = -5$ then $$a= 9t-5$$ $$c= 4t-5$$
so we have $lambda = 13t-5$. Now we see that your matrix is not uniqely determined: $$M=begin{pmatrix}
9t-5 & 6t \ 6t & 4t-5
end{pmatrix}.$$
answered Dec 9 '18 at 19:58
greedoidgreedoid
40.1k114799
$endgroup$
$begingroup$
Sorry I still don’t get it. U used the trace, and than that the product of the eigenvalues =the determinant?? I have never heard of it. How did you get to the equations and how do i solve them for an Eigenvector, i have 3 variables and two equations.
$endgroup$
– Lillys
Dec 9 '18 at 20:06
add a comment |
$begingroup$
If $$M=begin{pmatrix}
a & b \ b & c
end{pmatrix}.$$
Then $lambda+ lambda ' = a+c$ and $lambda lambda ' = ac-b^2$ and $$-2a+3b =10$$ $$-2b+3c =-15$$
Let $b=6t$ and $lambda ' = -5$ then $$a= 9t-5$$ $$c= 4t-5$$
so we have $lambda = 13t-5$. Now we see that your matrix is not uniqely determined: $$M=begin{pmatrix}
9t-5 & 6t \ 6t & 4t-5
end{pmatrix}.$$
answered Dec 9 '18 at 19:58
greedoidgreedoid
40.1k114799
$endgroup$
If $$M=begin{pmatrix}
a & b \ b & c
end{pmatrix}.$$
Then $lambda+ lambda ' = a+c$ and $lambda lambda ' = ac-b^2$ and $$-2a+3b =10$$ $$-2b+3c =-15$$
Let $b=6t$ and $lambda ' = -5$ then $$a= 9t-5$$ $$c= 4t-5$$
so we have $lambda = 13t-5$. Now we see that your matrix is not uniqely determined: $$M=begin{pmatrix}
9t-5 & 6t \ 6t & 4t-5
end{pmatrix}.$$
answered Dec 9 '18 at 19:58
greedoidgreedoid
40.1k114799
edited Dec 9 '18 at 20:15
answered Dec 9 '18 at 19:58
greedoidgreedoid
40.1k114799
answered Dec 9 '18 at 19:58
greedoidgreedoid
40.1k114799
answered Dec 9 '18 at 19:58
greedoidgreedoid
40.1k114799
40.1k114799
$begingroup$
Sorry I still don’t get it. U used the trace, and than that the product of the eigenvalues =the determinant?? I have never heard of it. How did you get to the equations and how do i solve them for an Eigenvector, i have 3 variables and two equations.
$endgroup$
– Lillys
Dec 9 '18 at 20:06
add a comment |
$begingroup$
Sorry I still don’t get it. U used the trace, and than that the product of the eigenvalues =the determinant?? I have never heard of it. How did you get to the equations and how do i solve them for an Eigenvector, i have 3 variables and two equations.
$endgroup$
– Lillys
Dec 9 '18 at 20:06
$begingroup$
Sorry I still don’t get it. U used the trace, and than that the product of the eigenvalues =the determinant?? I have never heard of it. How did you get to the equations and how do i solve them for an Eigenvector, i have 3 variables and two equations.
$endgroup$
– Lillys
Dec 9 '18 at 20:06
$begingroup$
Sorry I still don’t get it. U used the trace, and than that the product of the eigenvalues =the determinant?? I have never heard of it. How did you get to the equations and how do i solve them for an Eigenvector, i have 3 variables and two equations.
$endgroup$
– Lillys
Dec 9 '18 at 20:06
add a comment |
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