Signature of matrix associated with q
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For $alphainmathbb{R}$, let $q(x_1, x_2) = x_1^2
+ 2alpha x_1x_2 + dfrac{1}{2}2x_2^2$, for $(x_1, x_2) in mathbb{R^2}$.Find all values of $alpha$ for which the signature of $q$ is 1.
The matrix associated with the quadratic form is $begin{bmatrix} 1& alpha\alpha½end{bmatrix}$. To find the signature we reduce it to a diagonal form by simultaneous row and column transformations as $$begin{bmatrix} 1& alpha\alpha½end{bmatrix}xrightarrow[C_2rightarrow C_2-alpha C_1]{R_2rightarrow R_2-alpha R_1}begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}.$$
Signature of a matrix is defined as number of positive entries - total no of negative entries.
Signature of matrix of $q$ is $1$ if ${displaystylealphageqfrac{1}{2}}cup{alphaleqfrac{-1}{2}} $. But there is also another formula so to say viz $2p-r$ where $p$ is the number of positive entries and $r$ is the rank of the diagonal form. For $2p-r=1$ we require $p=1$ and $r=1$. Thus the values of $alpha=pmdfrac{1}{2}$. Is it weird that we are getting two ranges for $alpha$ for the same quadratic form? I am sure I have made a mistake. Can anyone please clarify?
linear-algebra proof-explanation quadratic-forms
edited Nov 15 at 9:28
pointguard0
1,310821
asked Nov 15 at 9:06
Yadati Kiran
964316
add a comment |
up vote
1
down vote
favorite
For $alphainmathbb{R}$, let $q(x_1, x_2) = x_1^2
+ 2alpha x_1x_2 + dfrac{1}{2}2x_2^2$, for $(x_1, x_2) in mathbb{R^2}$.Find all values of $alpha$ for which the signature of $q$ is 1.
The matrix associated with the quadratic form is $begin{bmatrix} 1& alpha\alpha½end{bmatrix}$. To find the signature we reduce it to a diagonal form by simultaneous row and column transformations as $$begin{bmatrix} 1& alpha\alpha½end{bmatrix}xrightarrow[C_2rightarrow C_2-alpha C_1]{R_2rightarrow R_2-alpha R_1}begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}.$$
Signature of a matrix is defined as number of positive entries - total no of negative entries.
Signature of matrix of $q$ is $1$ if ${displaystylealphageqfrac{1}{2}}cup{alphaleqfrac{-1}{2}} $. But there is also another formula so to say viz $2p-r$ where $p$ is the number of positive entries and $r$ is the rank of the diagonal form. For $2p-r=1$ we require $p=1$ and $r=1$. Thus the values of $alpha=pmdfrac{1}{2}$. Is it weird that we are getting two ranges for $alpha$ for the same quadratic form? I am sure I have made a mistake. Can anyone please clarify?
linear-algebra proof-explanation quadratic-forms
edited Nov 15 at 9:28
pointguard0
1,310821
asked Nov 15 at 9:06
Yadati Kiran
964316
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For $alphainmathbb{R}$, let $q(x_1, x_2) = x_1^2
+ 2alpha x_1x_2 + dfrac{1}{2}2x_2^2$, for $(x_1, x_2) in mathbb{R^2}$.Find all values of $alpha$ for which the signature of $q$ is 1.
The matrix associated with the quadratic form is $begin{bmatrix} 1& alpha\alpha½end{bmatrix}$. To find the signature we reduce it to a diagonal form by simultaneous row and column transformations as $$begin{bmatrix} 1& alpha\alpha½end{bmatrix}xrightarrow[C_2rightarrow C_2-alpha C_1]{R_2rightarrow R_2-alpha R_1}begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}.$$
Signature of a matrix is defined as number of positive entries - total no of negative entries.
Signature of matrix of $q$ is $1$ if ${displaystylealphageqfrac{1}{2}}cup{alphaleqfrac{-1}{2}} $. But there is also another formula so to say viz $2p-r$ where $p$ is the number of positive entries and $r$ is the rank of the diagonal form. For $2p-r=1$ we require $p=1$ and $r=1$. Thus the values of $alpha=pmdfrac{1}{2}$. Is it weird that we are getting two ranges for $alpha$ for the same quadratic form? I am sure I have made a mistake. Can anyone please clarify?
linear-algebra proof-explanation quadratic-forms
edited Nov 15 at 9:28
pointguard0
1,310821
asked Nov 15 at 9:06
Yadati Kiran
964316
For $alphainmathbb{R}$, let $q(x_1, x_2) = x_1^2
+ 2alpha x_1x_2 + dfrac{1}{2}2x_2^2$, for $(x_1, x_2) in mathbb{R^2}$.Find all values of $alpha$ for which the signature of $q$ is 1.
The matrix associated with the quadratic form is $begin{bmatrix} 1& alpha\alpha½end{bmatrix}$. To find the signature we reduce it to a diagonal form by simultaneous row and column transformations as $$begin{bmatrix} 1& alpha\alpha½end{bmatrix}xrightarrow[C_2rightarrow C_2-alpha C_1]{R_2rightarrow R_2-alpha R_1}begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}.$$
Signature of a matrix is defined as number of positive entries - total no of negative entries.
Signature of matrix of $q$ is $1$ if ${displaystylealphageqfrac{1}{2}}cup{alphaleqfrac{-1}{2}} $. But there is also another formula so to say viz $2p-r$ where $p$ is the number of positive entries and $r$ is the rank of the diagonal form. For $2p-r=1$ we require $p=1$ and $r=1$. Thus the values of $alpha=pmdfrac{1}{2}$. Is it weird that we are getting two ranges for $alpha$ for the same quadratic form? I am sure I have made a mistake. Can anyone please clarify?
linear-algebra proof-explanation quadratic-forms
linear-algebra proof-explanation quadratic-forms
edited Nov 15 at 9:28
pointguard0
1,310821
asked Nov 15 at 9:06
Yadati Kiran
964316
edited Nov 15 at 9:28
pointguard0
1,310821
asked Nov 15 at 9:06
Yadati Kiran
964316
edited Nov 15 at 9:28
pointguard0
1,310821
edited Nov 15 at 9:28
pointguard0
1,310821
edited Nov 15 at 9:28
pointguard0
1,310821
1,310821
asked Nov 15 at 9:06
Yadati Kiran
964316
asked Nov 15 at 9:06
Yadati Kiran
964316
asked Nov 15 at 9:06
Yadati Kiran
964316
964316
add a comment |
add a comment |
1 Answer
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votes
up vote
1
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$r$ isn't the rank of the matrix. It's the rank of a quadratic form. If the matrix is non-degenerate and $frac 1 2 - alpha^2 < 0$ then its signature is $1$.
For definitions see here.
answered Nov 15 at 10:29
pointguard0
1,310821
$r$ is the total number of square terms of the quadratic form viz the rank of $begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}$.
– Yadati Kiran
Nov 15 at 10:39
can we caluculate signature for degenerate quadratic form?
– Yadati Kiran
Nov 15 at 10:45
well, in case of non-degenerate matrices the signature is usually given by tuple $(p, q) equiv (p, r - p)$, while for degenerate matrices it is described by $(p, q, z)$, where $z$ indicates the number of non-zero $y_{t}$s such that they square to 0. For this case the quadratic form can be written as follows: $$ Q = y_1^2 + dots + y_p^2 - y_{p+1}^2 - dots - y_{p+q}^2 + y_{p+q+1}^2 + dots + y_{p+q+z}^2 $$
– pointguard0
Nov 15 at 11:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$r$ isn't the rank of the matrix. It's the rank of a quadratic form. If the matrix is non-degenerate and $frac 1 2 - alpha^2 < 0$ then its signature is $1$.
For definitions see here.
answered Nov 15 at 10:29
pointguard0
1,310821
$r$ is the total number of square terms of the quadratic form viz the rank of $begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}$.
– Yadati Kiran
Nov 15 at 10:39
can we caluculate signature for degenerate quadratic form?
– Yadati Kiran
Nov 15 at 10:45
well, in case of non-degenerate matrices the signature is usually given by tuple $(p, q) equiv (p, r - p)$, while for degenerate matrices it is described by $(p, q, z)$, where $z$ indicates the number of non-zero $y_{t}$s such that they square to 0. For this case the quadratic form can be written as follows: $$ Q = y_1^2 + dots + y_p^2 - y_{p+1}^2 - dots - y_{p+q}^2 + y_{p+q+1}^2 + dots + y_{p+q+z}^2 $$
– pointguard0
Nov 15 at 11:42
add a comment |
up vote
1
down vote
accepted
$r$ isn't the rank of the matrix. It's the rank of a quadratic form. If the matrix is non-degenerate and $frac 1 2 - alpha^2 < 0$ then its signature is $1$.
For definitions see here.
answered Nov 15 at 10:29
pointguard0
1,310821
$r$ is the total number of square terms of the quadratic form viz the rank of $begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}$.
– Yadati Kiran
Nov 15 at 10:39
can we caluculate signature for degenerate quadratic form?
– Yadati Kiran
Nov 15 at 10:45
well, in case of non-degenerate matrices the signature is usually given by tuple $(p, q) equiv (p, r - p)$, while for degenerate matrices it is described by $(p, q, z)$, where $z$ indicates the number of non-zero $y_{t}$s such that they square to 0. For this case the quadratic form can be written as follows: $$ Q = y_1^2 + dots + y_p^2 - y_{p+1}^2 - dots - y_{p+q}^2 + y_{p+q+1}^2 + dots + y_{p+q+z}^2 $$
– pointguard0
Nov 15 at 11:42
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$r$ isn't the rank of the matrix. It's the rank of a quadratic form. If the matrix is non-degenerate and $frac 1 2 - alpha^2 < 0$ then its signature is $1$.
For definitions see here.
answered Nov 15 at 10:29
pointguard0
1,310821
$r$ isn't the rank of the matrix. It's the rank of a quadratic form. If the matrix is non-degenerate and $frac 1 2 - alpha^2 < 0$ then its signature is $1$.
For definitions see here.
answered Nov 15 at 10:29
pointguard0
1,310821
answered Nov 15 at 10:29
pointguard0
1,310821
answered Nov 15 at 10:29
pointguard0
1,310821
answered Nov 15 at 10:29
pointguard0
1,310821
1,310821
$r$ is the total number of square terms of the quadratic form viz the rank of $begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}$.
– Yadati Kiran
Nov 15 at 10:39
can we caluculate signature for degenerate quadratic form?
– Yadati Kiran
Nov 15 at 10:45
well, in case of non-degenerate matrices the signature is usually given by tuple $(p, q) equiv (p, r - p)$, while for degenerate matrices it is described by $(p, q, z)$, where $z$ indicates the number of non-zero $y_{t}$s such that they square to 0. For this case the quadratic form can be written as follows: $$ Q = y_1^2 + dots + y_p^2 - y_{p+1}^2 - dots - y_{p+q}^2 + y_{p+q+1}^2 + dots + y_{p+q+z}^2 $$
– pointguard0
Nov 15 at 11:42
add a comment |
$r$ is the total number of square terms of the quadratic form viz the rank of $begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}$.
– Yadati Kiran
Nov 15 at 10:39
can we caluculate signature for degenerate quadratic form?
– Yadati Kiran
Nov 15 at 10:45
well, in case of non-degenerate matrices the signature is usually given by tuple $(p, q) equiv (p, r - p)$, while for degenerate matrices it is described by $(p, q, z)$, where $z$ indicates the number of non-zero $y_{t}$s such that they square to 0. For this case the quadratic form can be written as follows: $$ Q = y_1^2 + dots + y_p^2 - y_{p+1}^2 - dots - y_{p+q}^2 + y_{p+q+1}^2 + dots + y_{p+q+z}^2 $$
– pointguard0
Nov 15 at 11:42
$r$ is the total number of square terms of the quadratic form viz the rank of $begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}$.
– Yadati Kiran
Nov 15 at 10:39
$r$ is the total number of square terms of the quadratic form viz the rank of $begin{bmatrix} 1& 0\0½-alpha^2end{bmatrix}$.
– Yadati Kiran
Nov 15 at 10:39
can we caluculate signature for degenerate quadratic form?
– Yadati Kiran
Nov 15 at 10:45
can we caluculate signature for degenerate quadratic form?
– Yadati Kiran
Nov 15 at 10:45
well, in case of non-degenerate matrices the signature is usually given by tuple $(p, q) equiv (p, r - p)$, while for degenerate matrices it is described by $(p, q, z)$, where $z$ indicates the number of non-zero $y_{t}$s such that they square to 0. For this case the quadratic form can be written as follows: $$ Q = y_1^2 + dots + y_p^2 - y_{p+1}^2 - dots - y_{p+q}^2 + y_{p+q+1}^2 + dots + y_{p+q+z}^2 $$
– pointguard0
Nov 15 at 11:42
well, in case of non-degenerate matrices the signature is usually given by tuple $(p, q) equiv (p, r - p)$, while for degenerate matrices it is described by $(p, q, z)$, where $z$ indicates the number of non-zero $y_{t}$s such that they square to 0. For this case the quadratic form can be written as follows: $$ Q = y_1^2 + dots + y_p^2 - y_{p+1}^2 - dots - y_{p+q}^2 + y_{p+q+1}^2 + dots + y_{p+q+z}^2 $$
– pointguard0
Nov 15 at 11:42
add a comment |
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