Find locus of points by finding eigenvalues
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Let $boldsymbol{x}=left(begin{matrix}x\ yend{matrix}right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $boldsymbol{M}$, sketch the locus of points $boldsymbol{x}$ that satisfy $$ boldsymbol{x^TMx}=4$$
given that
$$boldsymbol{M}=left(begin{matrix}&5 &sqrt{3}\ &sqrt{3} &3end{matrix}right). $$
I found two eigenvalues to be $lambda_1 = 6$ and $lambda_2=2$, and the corresponding eigenvectors are
$$ boldsymbol{v}_1=left(begin{matrix}sqrt{3}\ 1end{matrix}right)quadtext{ and }quad boldsymbol{v}_2=left(begin{matrix}1\ -sqrt{3}end{matrix}right)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $boldsymbol{x^TMx}=4$.
Any hints?
linear-algebra vector-spaces eigenvalues-eigenvectors
edited 38 secs ago
bubba
29.6k32985
asked 1 hour ago
VanDerWarden
577629
add a comment |
up vote
4
down vote
favorite
Let $boldsymbol{x}=left(begin{matrix}x\ yend{matrix}right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $boldsymbol{M}$, sketch the locus of points $boldsymbol{x}$ that satisfy $$ boldsymbol{x^TMx}=4$$
given that
$$boldsymbol{M}=left(begin{matrix}&5 &sqrt{3}\ &sqrt{3} &3end{matrix}right). $$
I found two eigenvalues to be $lambda_1 = 6$ and $lambda_2=2$, and the corresponding eigenvectors are
$$ boldsymbol{v}_1=left(begin{matrix}sqrt{3}\ 1end{matrix}right)quadtext{ and }quad boldsymbol{v}_2=left(begin{matrix}1\ -sqrt{3}end{matrix}right)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $boldsymbol{x^TMx}=4$.
Any hints?
linear-algebra vector-spaces eigenvalues-eigenvectors
edited 38 secs ago
bubba
29.6k32985
asked 1 hour ago
VanDerWarden
577629
The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
– Shubham Johri
1 hour ago
@ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
– VanDerWarden
1 hour ago
Yes, now the eigenvalues and eigenvectors are fine
– Shubham Johri
1 hour ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $boldsymbol{x}=left(begin{matrix}x\ yend{matrix}right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $boldsymbol{M}$, sketch the locus of points $boldsymbol{x}$ that satisfy $$ boldsymbol{x^TMx}=4$$
given that
$$boldsymbol{M}=left(begin{matrix}&5 &sqrt{3}\ &sqrt{3} &3end{matrix}right). $$
I found two eigenvalues to be $lambda_1 = 6$ and $lambda_2=2$, and the corresponding eigenvectors are
$$ boldsymbol{v}_1=left(begin{matrix}sqrt{3}\ 1end{matrix}right)quadtext{ and }quad boldsymbol{v}_2=left(begin{matrix}1\ -sqrt{3}end{matrix}right)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $boldsymbol{x^TMx}=4$.
Any hints?
linear-algebra vector-spaces eigenvalues-eigenvectors
edited 38 secs ago
bubba
29.6k32985
asked 1 hour ago
VanDerWarden
577629
Let $boldsymbol{x}=left(begin{matrix}x\ yend{matrix}right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $boldsymbol{M}$, sketch the locus of points $boldsymbol{x}$ that satisfy $$ boldsymbol{x^TMx}=4$$
given that
$$boldsymbol{M}=left(begin{matrix}&5 &sqrt{3}\ &sqrt{3} &3end{matrix}right). $$
I found two eigenvalues to be $lambda_1 = 6$ and $lambda_2=2$, and the corresponding eigenvectors are
$$ boldsymbol{v}_1=left(begin{matrix}sqrt{3}\ 1end{matrix}right)quadtext{ and }quad boldsymbol{v}_2=left(begin{matrix}1\ -sqrt{3}end{matrix}right)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $boldsymbol{x^TMx}=4$.
Any hints?
linear-algebra vector-spaces eigenvalues-eigenvectors
linear-algebra vector-spaces eigenvalues-eigenvectors
edited 38 secs ago
bubba
29.6k32985
asked 1 hour ago
VanDerWarden
577629
edited 38 secs ago
bubba
29.6k32985
asked 1 hour ago
VanDerWarden
577629
edited 38 secs ago
bubba
29.6k32985
edited 38 secs ago
bubba
29.6k32985
edited 38 secs ago
bubba
29.6k32985
29.6k32985
asked 1 hour ago
VanDerWarden
577629
asked 1 hour ago
VanDerWarden
577629
asked 1 hour ago
VanDerWarden
577629
577629
The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
– Shubham Johri
1 hour ago
@ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
– VanDerWarden
1 hour ago
Yes, now the eigenvalues and eigenvectors are fine
– Shubham Johri
1 hour ago
add a comment |
The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
– Shubham Johri
1 hour ago
@ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
– VanDerWarden
1 hour ago
Yes, now the eigenvalues and eigenvectors are fine
– Shubham Johri
1 hour ago
The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
– Shubham Johri
1 hour ago
The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
– Shubham Johri
1 hour ago
@ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
– VanDerWarden
1 hour ago
@ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
– VanDerWarden
1 hour ago
Yes, now the eigenvalues and eigenvectors are fine
– Shubham Johri
1 hour ago
Yes, now the eigenvalues and eigenvectors are fine
– Shubham Johri
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.
$mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$
$implies6c_1^2+2c_2^2=1$
answered 1 hour ago
Shubham Johri
2,648413
add a comment |
up vote
5
down vote
Some hints (as you requested):
What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).
After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.
The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates
In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.
answered 1 hour ago
bubba
29.6k32985
Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
– Anvit
1 hour ago
@Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
– bubba
1 hour ago
@bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
– AmbretteOrrisey
57 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.
$mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$
$implies6c_1^2+2c_2^2=1$
answered 1 hour ago
Shubham Johri
2,648413
add a comment |
up vote
3
down vote
accepted
The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.
$mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$
$implies6c_1^2+2c_2^2=1$
answered 1 hour ago
Shubham Johri
2,648413
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.
$mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$
$implies6c_1^2+2c_2^2=1$
answered 1 hour ago
Shubham Johri
2,648413
The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=begin{bmatrix}x\yend{bmatrix}=c_1mathbf x_1+c_2mathbf x_2$.
$mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$
$implies6c_1^2+2c_2^2=1$
answered 1 hour ago
Shubham Johri
2,648413
edited 1 hour ago
answered 1 hour ago
Shubham Johri
2,648413
answered 1 hour ago
Shubham Johri
2,648413
answered 1 hour ago
Shubham Johri
2,648413
2,648413
add a comment |
add a comment |
up vote
5
down vote
Some hints (as you requested):
What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).
After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.
The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates
In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.
answered 1 hour ago
bubba
29.6k32985
Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
– Anvit
1 hour ago
@Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
– bubba
1 hour ago
@bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
– AmbretteOrrisey
57 mins ago
add a comment |
up vote
5
down vote
Some hints (as you requested):
What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).
After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.
The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates
In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.
answered 1 hour ago
bubba
29.6k32985
Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
– Anvit
1 hour ago
@Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
– bubba
1 hour ago
@bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
– AmbretteOrrisey
57 mins ago
add a comment |
up vote
5
down vote
up vote
5
down vote
Some hints (as you requested):
What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).
After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.
The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates
In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.
answered 1 hour ago
bubba
29.6k32985
Some hints (as you requested):
What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).
After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $mathbf{M}$.
The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates
In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.
answered 1 hour ago
bubba
29.6k32985
edited 1 min ago
answered 1 hour ago
bubba
29.6k32985
answered 1 hour ago
bubba
29.6k32985
answered 1 hour ago
bubba
29.6k32985
29.6k32985
Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
– Anvit
1 hour ago
@Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
– bubba
1 hour ago
@bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
– AmbretteOrrisey
57 mins ago
add a comment |
Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
– Anvit
1 hour ago
@Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
– bubba
1 hour ago
@bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
– AmbretteOrrisey
57 mins ago
Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
– Anvit
1 hour ago
Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct
– Anvit
1 hour ago
@Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
– bubba
1 hour ago
@Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $mathbf v$ is an eigenvector, then so is $kmathbf v$, for any $k ne 0$, so the eigenvectors can't tell you anything about scale.
– bubba
1 hour ago
@bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
– AmbretteOrrisey
57 mins ago
@bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall.
– AmbretteOrrisey
57 mins ago
add a comment |
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The eigenvalues are $5pmsqrt3$, not $6,2$ because $6+2ne5+5$
– Shubham Johri
1 hour ago
@ShubhamJohri see the edited question. I wrongly typed one entry of $M$.
– VanDerWarden
1 hour ago
Yes, now the eigenvalues and eigenvectors are fine
– Shubham Johri
1 hour ago