Mixing
Why are the axis of an ellipsoid eigenvectors?
$begingroup$
Consider an ellipsoid ${x| x^TAx = 1}$.
Let $A$ be a real symmetric matrix, and consider the eigen-decomposition
$A=PLambda P^{-1} =PLambda P^T$,
where the matrix $P$ is orthogonal because $A$ is symmetric, and the diagonal matrix $Lambda$ contains the eigenvalues of $A$.
Then the ellipsoid can be rewritten as:
$$x^TPLambda P^Tx = (P^Tx)^TLambda(P^Tx) = y^TLambda y = 1$$
Since the matrix is diagonal, the quadratic form gives:
$$y^TLambda y =lambda_1y_1^2+...+lambda_ny_n^2 = 1$$
This is clearly the equation of ellipsoid.
My question is, why are the axes of this ellipsoid the eigenvectors of $A$?
linear-algebra algebra-precalculus geometry convex-optimization conic-sections
asked Jan 23 at 6:55
The man of your dreamThe man of your dream
23110
$endgroup$
add a comment |
$begingroup$
Consider an ellipsoid ${x| x^TAx = 1}$.
Let $A$ be a real symmetric matrix, and consider the eigen-decomposition
$A=PLambda P^{-1} =PLambda P^T$,
where the matrix $P$ is orthogonal because $A$ is symmetric, and the diagonal matrix $Lambda$ contains the eigenvalues of $A$.
Then the ellipsoid can be rewritten as:
$$x^TPLambda P^Tx = (P^Tx)^TLambda(P^Tx) = y^TLambda y = 1$$
Since the matrix is diagonal, the quadratic form gives:
$$y^TLambda y =lambda_1y_1^2+...+lambda_ny_n^2 = 1$$
This is clearly the equation of ellipsoid.
My question is, why are the axes of this ellipsoid the eigenvectors of $A$?
linear-algebra algebra-precalculus geometry convex-optimization conic-sections
asked Jan 23 at 6:55
The man of your dreamThe man of your dream
23110
$endgroup$
add a comment |
$begingroup$
Consider an ellipsoid ${x| x^TAx = 1}$.
Let $A$ be a real symmetric matrix, and consider the eigen-decomposition
$A=PLambda P^{-1} =PLambda P^T$,
where the matrix $P$ is orthogonal because $A$ is symmetric, and the diagonal matrix $Lambda$ contains the eigenvalues of $A$.
Then the ellipsoid can be rewritten as:
$$x^TPLambda P^Tx = (P^Tx)^TLambda(P^Tx) = y^TLambda y = 1$$
Since the matrix is diagonal, the quadratic form gives:
$$y^TLambda y =lambda_1y_1^2+...+lambda_ny_n^2 = 1$$
This is clearly the equation of ellipsoid.
My question is, why are the axes of this ellipsoid the eigenvectors of $A$?
linear-algebra algebra-precalculus geometry convex-optimization conic-sections
asked Jan 23 at 6:55
The man of your dreamThe man of your dream
23110
$endgroup$
Consider an ellipsoid ${x| x^TAx = 1}$.
Let $A$ be a real symmetric matrix, and consider the eigen-decomposition
$A=PLambda P^{-1} =PLambda P^T$,
where the matrix $P$ is orthogonal because $A$ is symmetric, and the diagonal matrix $Lambda$ contains the eigenvalues of $A$.
Then the ellipsoid can be rewritten as:
$$x^TPLambda P^Tx = (P^Tx)^TLambda(P^Tx) = y^TLambda y = 1$$
Since the matrix is diagonal, the quadratic form gives:
$$y^TLambda y =lambda_1y_1^2+...+lambda_ny_n^2 = 1$$
This is clearly the equation of ellipsoid.
My question is, why are the axes of this ellipsoid the eigenvectors of $A$?
linear-algebra algebra-precalculus geometry convex-optimization conic-sections
linear-algebra algebra-precalculus geometry convex-optimization conic-sections
asked Jan 23 at 6:55
The man of your dreamThe man of your dream
23110
asked Jan 23 at 6:55
The man of your dreamThe man of your dream
23110
edited Jan 24 at 1:32
The man of your dream
asked Jan 23 at 6:55
The man of your dreamThe man of your dream
23110
asked Jan 23 at 6:55
The man of your dreamThe man of your dream
23110
asked Jan 23 at 6:55
The man of your dreamThe man of your dream
23110
23110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The axes of this ellipsoid are (the multiples of) $e_1=(1,0,0,ldots,0)$, $e_2=(0,1,0,ldots,0)$, …, $e_n=(0,0,0,ldots,1)$ (and the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). But $Lambda=P^{-1}AP$ and $P$ is a change-of-bases matrix. So, the columns of $P$ are eigenvectors of $A$ (and, again, the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). This was just a change of basis, so, geometrically, the columns of $P$ are the still the vectors $e_1,ldots,e_n$.
answered Jan 23 at 7:28
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
I still don't quite understand..I'm using these slide ee263.stanford.edu/lectures/ellipsoids.pdf by the way
$endgroup$
– The man of your dream
Jan 23 at 7:37
$begingroup$
The axis of the ellipsoif are $s_1$ and $s_2$. When you use that orthogonal matrix $P$, what you actually do is to rotate the ellipsoid so that $s_1$ and $s_2$ become $e_1$ and $e_2$ (perhaps with a change of order); this can be done, since thos axes are always orthogonal.
$endgroup$
– José Carlos Santos
Jan 23 at 8:37
add a comment |
$begingroup$
As you mentioned
$$left(frac{y_1}{sqrt{lambda_1}}right)^2 + left(frac{y_2}{sqrt{lambda_2}}right)^2 + cdots + left(frac{y_n}{sqrt{lambda_n}}right)^2 = 1$$
is the equation of an ellipse with semiaxis lengths $sqrt{lambda_1},sqrt{lambda_2},ldots,sqrt{lambda_n}$ (each $lambda_i > 0$ because the original matrix $A$ is positive semidefinite). Specifically, this is an axis aligned ellipse in the y-coordinate space. Set any $y_i$ to be $sqrt{lambda_i}$ and the rest to $0$ and plug into the equality to convince yourself this is so.
From your question statement we have $y = P^Tx = P^{-1}x$ or equivalently $x = Py$. We know the axes of the ellipse in y-coordinate space are just the standard basis vectors $mathbf{e}_1, mathbf{e}_2, ldots, mathbf{e}_n$.
The axes in x-coordinate space then must be $Pmathbf{e}_1, Pmathbf{e}_2, ldots, Pmathbf{e}_n$. That is to say, the ellipse axes are just the columns of $P$. The columns of $P$ are of course the eigenvectors of $A$ because of how $P$ was defined to be the matrix diagonalizing $A$.
answered Feb 21 at 11:05
CaseyCasey
1676
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The axes of this ellipsoid are (the multiples of) $e_1=(1,0,0,ldots,0)$, $e_2=(0,1,0,ldots,0)$, …, $e_n=(0,0,0,ldots,1)$ (and the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). But $Lambda=P^{-1}AP$ and $P$ is a change-of-bases matrix. So, the columns of $P$ are eigenvectors of $A$ (and, again, the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). This was just a change of basis, so, geometrically, the columns of $P$ are the still the vectors $e_1,ldots,e_n$.
answered Jan 23 at 7:28
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
I still don't quite understand..I'm using these slide ee263.stanford.edu/lectures/ellipsoids.pdf by the way
$endgroup$
– The man of your dream
Jan 23 at 7:37
$begingroup$
The axis of the ellipsoif are $s_1$ and $s_2$. When you use that orthogonal matrix $P$, what you actually do is to rotate the ellipsoid so that $s_1$ and $s_2$ become $e_1$ and $e_2$ (perhaps with a change of order); this can be done, since thos axes are always orthogonal.
$endgroup$
– José Carlos Santos
Jan 23 at 8:37
add a comment |
$begingroup$
The axes of this ellipsoid are (the multiples of) $e_1=(1,0,0,ldots,0)$, $e_2=(0,1,0,ldots,0)$, …, $e_n=(0,0,0,ldots,1)$ (and the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). But $Lambda=P^{-1}AP$ and $P$ is a change-of-bases matrix. So, the columns of $P$ are eigenvectors of $A$ (and, again, the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). This was just a change of basis, so, geometrically, the columns of $P$ are the still the vectors $e_1,ldots,e_n$.
answered Jan 23 at 7:28
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
I still don't quite understand..I'm using these slide ee263.stanford.edu/lectures/ellipsoids.pdf by the way
$endgroup$
– The man of your dream
Jan 23 at 7:37
$begingroup$
The axis of the ellipsoif are $s_1$ and $s_2$. When you use that orthogonal matrix $P$, what you actually do is to rotate the ellipsoid so that $s_1$ and $s_2$ become $e_1$ and $e_2$ (perhaps with a change of order); this can be done, since thos axes are always orthogonal.
$endgroup$
– José Carlos Santos
Jan 23 at 8:37
add a comment |
$begingroup$
The axes of this ellipsoid are (the multiples of) $e_1=(1,0,0,ldots,0)$, $e_2=(0,1,0,ldots,0)$, …, $e_n=(0,0,0,ldots,1)$ (and the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). But $Lambda=P^{-1}AP$ and $P$ is a change-of-bases matrix. So, the columns of $P$ are eigenvectors of $A$ (and, again, the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). This was just a change of basis, so, geometrically, the columns of $P$ are the still the vectors $e_1,ldots,e_n$.
answered Jan 23 at 7:28
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
The axes of this ellipsoid are (the multiples of) $e_1=(1,0,0,ldots,0)$, $e_2=(0,1,0,ldots,0)$, …, $e_n=(0,0,0,ldots,1)$ (and the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). But $Lambda=P^{-1}AP$ and $P$ is a change-of-bases matrix. So, the columns of $P$ are eigenvectors of $A$ (and, again, the corresponding eigenvalues are $lambda_1,lambda_2,ldots,lambda_n$). This was just a change of basis, so, geometrically, the columns of $P$ are the still the vectors $e_1,ldots,e_n$.
answered Jan 23 at 7:28
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 7:28
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 7:28
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 7:28
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
$begingroup$
I still don't quite understand..I'm using these slide ee263.stanford.edu/lectures/ellipsoids.pdf by the way
$endgroup$
– The man of your dream
Jan 23 at 7:37
$begingroup$
The axis of the ellipsoif are $s_1$ and $s_2$. When you use that orthogonal matrix $P$, what you actually do is to rotate the ellipsoid so that $s_1$ and $s_2$ become $e_1$ and $e_2$ (perhaps with a change of order); this can be done, since thos axes are always orthogonal.
$endgroup$
– José Carlos Santos
Jan 23 at 8:37
add a comment |
$begingroup$
I still don't quite understand..I'm using these slide ee263.stanford.edu/lectures/ellipsoids.pdf by the way
$endgroup$
– The man of your dream
Jan 23 at 7:37
$begingroup$
The axis of the ellipsoif are $s_1$ and $s_2$. When you use that orthogonal matrix $P$, what you actually do is to rotate the ellipsoid so that $s_1$ and $s_2$ become $e_1$ and $e_2$ (perhaps with a change of order); this can be done, since thos axes are always orthogonal.
$endgroup$
– José Carlos Santos
Jan 23 at 8:37
$begingroup$
I still don't quite understand..I'm using these slide ee263.stanford.edu/lectures/ellipsoids.pdf by the way
$endgroup$
– The man of your dream
Jan 23 at 7:37
$begingroup$
I still don't quite understand..I'm using these slide ee263.stanford.edu/lectures/ellipsoids.pdf by the way
$endgroup$
– The man of your dream
Jan 23 at 7:37
$begingroup$
The axis of the ellipsoif are $s_1$ and $s_2$. When you use that orthogonal matrix $P$, what you actually do is to rotate the ellipsoid so that $s_1$ and $s_2$ become $e_1$ and $e_2$ (perhaps with a change of order); this can be done, since thos axes are always orthogonal.
$endgroup$
– José Carlos Santos
Jan 23 at 8:37
$begingroup$
The axis of the ellipsoif are $s_1$ and $s_2$. When you use that orthogonal matrix $P$, what you actually do is to rotate the ellipsoid so that $s_1$ and $s_2$ become $e_1$ and $e_2$ (perhaps with a change of order); this can be done, since thos axes are always orthogonal.
$endgroup$
– José Carlos Santos
Jan 23 at 8:37
add a comment |
$begingroup$
As you mentioned
$$left(frac{y_1}{sqrt{lambda_1}}right)^2 + left(frac{y_2}{sqrt{lambda_2}}right)^2 + cdots + left(frac{y_n}{sqrt{lambda_n}}right)^2 = 1$$
is the equation of an ellipse with semiaxis lengths $sqrt{lambda_1},sqrt{lambda_2},ldots,sqrt{lambda_n}$ (each $lambda_i > 0$ because the original matrix $A$ is positive semidefinite). Specifically, this is an axis aligned ellipse in the y-coordinate space. Set any $y_i$ to be $sqrt{lambda_i}$ and the rest to $0$ and plug into the equality to convince yourself this is so.
From your question statement we have $y = P^Tx = P^{-1}x$ or equivalently $x = Py$. We know the axes of the ellipse in y-coordinate space are just the standard basis vectors $mathbf{e}_1, mathbf{e}_2, ldots, mathbf{e}_n$.
The axes in x-coordinate space then must be $Pmathbf{e}_1, Pmathbf{e}_2, ldots, Pmathbf{e}_n$. That is to say, the ellipse axes are just the columns of $P$. The columns of $P$ are of course the eigenvectors of $A$ because of how $P$ was defined to be the matrix diagonalizing $A$.
answered Feb 21 at 11:05
CaseyCasey
1676
$endgroup$
add a comment |
$begingroup$
As you mentioned
$$left(frac{y_1}{sqrt{lambda_1}}right)^2 + left(frac{y_2}{sqrt{lambda_2}}right)^2 + cdots + left(frac{y_n}{sqrt{lambda_n}}right)^2 = 1$$
is the equation of an ellipse with semiaxis lengths $sqrt{lambda_1},sqrt{lambda_2},ldots,sqrt{lambda_n}$ (each $lambda_i > 0$ because the original matrix $A$ is positive semidefinite). Specifically, this is an axis aligned ellipse in the y-coordinate space. Set any $y_i$ to be $sqrt{lambda_i}$ and the rest to $0$ and plug into the equality to convince yourself this is so.
From your question statement we have $y = P^Tx = P^{-1}x$ or equivalently $x = Py$. We know the axes of the ellipse in y-coordinate space are just the standard basis vectors $mathbf{e}_1, mathbf{e}_2, ldots, mathbf{e}_n$.
The axes in x-coordinate space then must be $Pmathbf{e}_1, Pmathbf{e}_2, ldots, Pmathbf{e}_n$. That is to say, the ellipse axes are just the columns of $P$. The columns of $P$ are of course the eigenvectors of $A$ because of how $P$ was defined to be the matrix diagonalizing $A$.
answered Feb 21 at 11:05
CaseyCasey
1676
$endgroup$
add a comment |
$begingroup$
As you mentioned
$$left(frac{y_1}{sqrt{lambda_1}}right)^2 + left(frac{y_2}{sqrt{lambda_2}}right)^2 + cdots + left(frac{y_n}{sqrt{lambda_n}}right)^2 = 1$$
is the equation of an ellipse with semiaxis lengths $sqrt{lambda_1},sqrt{lambda_2},ldots,sqrt{lambda_n}$ (each $lambda_i > 0$ because the original matrix $A$ is positive semidefinite). Specifically, this is an axis aligned ellipse in the y-coordinate space. Set any $y_i$ to be $sqrt{lambda_i}$ and the rest to $0$ and plug into the equality to convince yourself this is so.
From your question statement we have $y = P^Tx = P^{-1}x$ or equivalently $x = Py$. We know the axes of the ellipse in y-coordinate space are just the standard basis vectors $mathbf{e}_1, mathbf{e}_2, ldots, mathbf{e}_n$.
The axes in x-coordinate space then must be $Pmathbf{e}_1, Pmathbf{e}_2, ldots, Pmathbf{e}_n$. That is to say, the ellipse axes are just the columns of $P$. The columns of $P$ are of course the eigenvectors of $A$ because of how $P$ was defined to be the matrix diagonalizing $A$.
answered Feb 21 at 11:05
CaseyCasey
1676
$endgroup$
As you mentioned
$$left(frac{y_1}{sqrt{lambda_1}}right)^2 + left(frac{y_2}{sqrt{lambda_2}}right)^2 + cdots + left(frac{y_n}{sqrt{lambda_n}}right)^2 = 1$$
is the equation of an ellipse with semiaxis lengths $sqrt{lambda_1},sqrt{lambda_2},ldots,sqrt{lambda_n}$ (each $lambda_i > 0$ because the original matrix $A$ is positive semidefinite). Specifically, this is an axis aligned ellipse in the y-coordinate space. Set any $y_i$ to be $sqrt{lambda_i}$ and the rest to $0$ and plug into the equality to convince yourself this is so.
From your question statement we have $y = P^Tx = P^{-1}x$ or equivalently $x = Py$. We know the axes of the ellipse in y-coordinate space are just the standard basis vectors $mathbf{e}_1, mathbf{e}_2, ldots, mathbf{e}_n$.
The axes in x-coordinate space then must be $Pmathbf{e}_1, Pmathbf{e}_2, ldots, Pmathbf{e}_n$. That is to say, the ellipse axes are just the columns of $P$. The columns of $P$ are of course the eigenvectors of $A$ because of how $P$ was defined to be the matrix diagonalizing $A$.
answered Feb 21 at 11:05
CaseyCasey
1676
answered Feb 21 at 11:05
CaseyCasey
1676
answered Feb 21 at 11:05
CaseyCasey
1676
answered Feb 21 at 11:05
CaseyCasey
1676
1676
add a comment |
add a comment |
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