Quadratic function of multiple variables - Revision history

Vipul: /* Case of general matrix= */

2014-05-26T16:12:59Z

Case of general matrix=

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	===Second-order partial derivatives and Hessian matrix===		===Second-order partial derivatives and Hessian matrix===

	===Case of general matrix====		====Case of general matrix====

	Recall that we had obtained (we replace the dummy variable <math>j</math> by <math>k</math> to facilitate differentiation with respect to <math>j</math> in the next step):		Recall that we had obtained (we replace the dummy variable <math>j</math> by <math>k</math> to facilitate differentiation with respect to <math>j</math> in the next step):

Vipul at 15:59, 26 May 2014

2014-05-26T15:59:51Z

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 * In the case that <math>A</math> is a symmetric positive-semidefinite matrix, i.e., all its nonzero eigenvalues are positive, the function attains a local minimum and also its absolute minimum at all its critical points. Note that, since the function is constant at this minimum value on the affine space, none of these is a point of ''strict'' local minimum.
 * In the case that <math>A</math> is a symmetric negative-semidefinite matrix, i.e., all its nonzero eigenvalues are negative, the function attains a local maximum and also its absolute maximum value at all its critical points. Note that, since the function is constant at this minimum value on the affine space, none of these is a point of ''strict'' local minimum.
 ==Alternate analysis of extreme values==

Vipul: /* Cases */

2014-05-26T15:50:40Z

Cases

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	* In the case that <math>A</math> is a symmetric negative-semidefinite matrix, i.e., all its nonzero eigenvalues are negative, the function attains a local maximum and also its absolute maximum value at all its critical points. Note that, since the function is constant at this minimum value on the affine space, none of these is a point of ''strict'' local minimum.		* In the case that <math>A</math> is a symmetric negative-semidefinite matrix, i.e., all its nonzero eigenvalues are negative, the function attains a local maximum and also its absolute maximum value at all its critical points. Note that, since the function is constant at this minimum value on the affine space, none of these is a point of ''strict'' local minimum.

	==~~Cases~~==		==Alternate analysis of extreme values==

	For the discussion of cases, assume that <math>A</math> is a symmetric matrix. If <math>A</math> is not symmetric, replace it by the symmetric matrix <math>(A + A^T)/2</math>.		For the discussion of cases, assume that <math>A</math> is a symmetric matrix. If <math>A</math> is not symmetric, replace it by the symmetric matrix <math>(A + A^T)/2</math>.

Vipul: /* Higher derivatives */

2014-05-26T15:50:16Z

Higher derivatives

← Older revision		Revision as of 15:50, 26 May 2014
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	Therefore, the higher derivative tensors (the higher-order analogues of the gradient vector and Hessian matrix) are also identically zero.		Therefore, the higher derivative tensors (the higher-order analogues of the gradient vector and Hessian matrix) are also identically zero.

			==Points and intervals of interest==

			For the discussion here, assume that <math>A</math> is symmetric. If it is not, replace <math>A</math> by the matrix <math>(A + A^T)/2</math>.

			===Critical points===

			====Case that the matrix <math>A</math> is invertible====

			To find the critical points, we need to set the gradient vector equal to zero. This gives the vector equation:

			<math>2A\vec{x} + \vec{b} = \vec{0}</math>

			In other words:

			<math>A \vec{x} = \frac{-1}{2}\vec{b}</math>

			Under the assumption that <math>A</math> is invertible (or equivalently, that all its eigenvalues are nonzero), we can left-multiply both sides by <math>A^{-1}</math> and obtain:

			<math>\vec{x} = \frac{-1}{2} A^{-1} \vec{b}</math>

			We thus have a ''unique'' critical point as described above.

			====Case that the matrix <math>A</math> is non-invertible====

			We still need to solve the same linear system:

			<math>2A \vec{x} + \vec{b}</math>

			However, since <math>A</math> is no longer invertible (i.e., it has zero as an eigenvalue, or equivalently, it has a nonzero kernel), two cases arise:

			* No solution exists. This happens if the vector <math>\vec{b}</math> is not in the image of the linear transformation defined by <math>A</math>. Conceptually, what is happening is that the linear part of <math>f</math> is not constrained by the quadratic part, and therefore the function is unbounded.
			* A solution exists, but it is not unique. This happens if the vector <math>\vec{b}</math> is in the image of the linear transformation defined by <math>A</math>. In fact, the solution set is an affine space of dimension equal to the nullity of <math>A</math>. Thus, we get an affine space's worth of critical points.

			===Determination of local extremum behavior at critical points===

			====Case that the matrix <math>A</math> is invertible====

			Recall that the [[Hessian matrix]] of <math>f</math> is <math>2A</math>. Therefore, by the [[second derivative test for a function of multiple variables]], we obtain the following:

			* If <math>A</math> is a [[linear:symmetric positive-definite matrix\|symmetric positive-definite matrix]], i.e., all its eigenvalues are positive, then the unique critical point <math>\frac{-1}{2}A^{-1}\vec{b}</math> is a point of local minimum and is the unique point of absolute minimum (an alternate derivation of this fact is later in the page).
			* If <math>A</math> is negative-definite matrix, i.e., all its eigenvalues are negative, then the unique critical point <math>\frac{-1}{2}A^{-1}\vec{b}</math> is a point of local maximum and is the unique point of absolute maximum.
			* If <math>A</math> has both positive and negative eigenvalues, then the unique critical point <math>\frac{-1}{2}A^{-1}\vec{b}</math> is neither a point of local minimum nor a point of local maximum. In fact, it gives a saddle point for the function. There is no point of local extremum and the range of the function is all of <math>\R</math>.

			====Case that the matrix <math>A</math> is non-invertible====

			In this case, the Hessian matrix, <math>2A</math>, is also non-invertible and in particular has zero as an eigenvalue.

			In the case that there are no critical points, there is nothing to say.

			In the case that there are critical points, we note that, through any critical point, there is a direction along which the second derivative is zero. In fact, what's happening geometrically is that the set of critical points form an affine subspace and the quadratic function <math>f</math> is constant on that affine subspace. How the behavior of that affine subspace compares with the values of the function elsewhere depends on the nonzero eigenvalues.

			* In the case that <math>A</math> is a symmetric positive-semidefinite matrix, i.e., all its nonzero eigenvalues are positive, the function attains a local minimum and also its absolute minimum at all its critical points. Note that, since the function is constant at this minimum value on the affine space, none of these is a point of ''strict'' local minimum.
			* In the case that <math>A</math> is a symmetric negative-semidefinite matrix, i.e., all its nonzero eigenvalues are negative, the function attains a local maximum and also its absolute maximum value at all its critical points. Note that, since the function is constant at this minimum value on the affine space, none of these is a point of ''strict'' local minimum.

	==Cases==		==Cases==

Vipul: /* Higher derivatives */

2014-05-26T15:22:21Z

Higher derivatives

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	===Higher derivatives===		===Higher derivatives===

	All the higher derivative tensors are zero.		All higher order partial derivatives (pure or mixed) are zero. This can be seen directly from the fact that the second-order partial derivatives are all constants, so differentiating them further (with respect to any variable) gives zero.

			Therefore, the higher derivative tensors (the higher-order analogues of the gradient vector and Hessian matrix) are also identically zero.

	==Cases==		==Cases==

Vipul: /* Differentiation */

2014-05-26T15:20:32Z

Differentiation

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 ===Partial derivatives and gradient vector===
 The [[partial derivative]] with respect to the variable <math>x_i</math>, and therefore also the <math>i^{th}</math> coordinate of the [[gradient vector]], is given by:
@@ Line 45: / Line 46: @@
 <math>(\nabla f)(\vec{x}) = (A + A^T)\vec{x} + \vec{b}</math>
-In the case that <math>A</math> is a symmetric matrix, this simplifies to:
+====Case of symmetric matrix====
 <math>(\nabla f)(\vec{x}) = 2A\vec{x} + \vec{b}</math>
-===Hessian matrix===
+====Case <math>n = 1</math>====
-The [[Hessian matrix]] of the quadratic function is the matrix <math>A + A^T</math>. In the case that <math>A</math> is symmetric, this simplifies to <math>2A</math>.
+This is indeed the case. The only second-order partial derivative is <math>f''(x) = 2a</math>. This agrees both with the formula for the second-order partial derivative and with the formula for the Hessian matrix.
 ===Higher derivatives===

Vipul: /* Key data */

2014-05-26T03:42:28Z

Key data

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	\| [[local minimum value]] and points of attainment \|\| If the matrix <math>A</math> is positive definite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is positive semidefinite but not positive definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The positive definite case corresponds to <math>a > 0</math>: Here, the local minimum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The negative definite case corresponds to <math>a < 0</math>, and there is no minimum in this case.		\| [[local minimum value]] and points of attainment \|\| If the matrix <math>A</math> is positive definite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is positive semidefinite but not positive definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The positive definite case corresponds to <math>a > 0</math>: Here, the local minimum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The negative definite case corresponds to <math>a < 0</math>, and there is no minimum in this case.
	\|-		\|-
	\| [[local maximum value]] and points of attainment \|\| If the matrix <math>A</math> is negative definite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is negative semidefinite but not negative definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not ~~positive~~ semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The negative definite case corresponds to <math>a < 0</math>: Here, the local maximum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The positive definite case corresponds to <math>a > 0</math>, and there is no maximum in this case.		\| [[local maximum value]] and points of attainment \|\| If the matrix <math>A</math> is negative definite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is negative semidefinite but not negative definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not negative semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The negative definite case corresponds to <math>a < 0</math>: Here, the local maximum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The positive definite case corresponds to <math>a > 0</math>, and there is no maximum in this case.
	\|-		\|-
	\| [[gradient vector]] function (analogous to the [[derivative]]) \|\| <math>\vec{x} \mapsto 2A\vec{x} + \vec{b}</math> \|\| the [[derivative]] is <math>x \mapsto 2ax + b</math> (consistent with the matrix formulation)		\| [[gradient vector]] function (analogous to the [[derivative]]) \|\| <math>\vec{x} \mapsto 2A\vec{x} + \vec{b}</math> \|\| the [[derivative]] is <math>x \mapsto 2ax + b</math> (consistent with the matrix formulation)

Vipul: /* Key data */

2014-05-26T03:41:49Z

Key data

← Older revision		Revision as of 03:41, 26 May 2014
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	\| [[range]] \|\| If the matrix <math>A</math> is not positive semidefinite or negative semidefinite, the range is all of <math>\R</math>.<br>If the matrix <math>A</math> is positive definite or (<math>A</math> is positive semidefinite and <math>\vec{b}</math> is in its image), the range is <math>[m,\infty)</math> where <math>m</math> is the minimum value. If the matrix <math>A</math> is negative definite or (<math>A</math> is negative semidefinite and <math>\vec{b}</math> is in its image), the range is <math>(-\infty,m]</math> where <math>m</math> is the maximum value. \|\| The case of "not positive semidefinite or negative semidefinite" does not arise for <math>n = 1</math>. Moreover, all the semidefinite cases must be definite, so we only have to consider the positive definite case and the negative definite case.<br>The positive definite case corresponds to <math>a > 0</math><br>The negative definite case corresponds to <math>a < 0</math>		\| [[range]] \|\| If the matrix <math>A</math> is not positive semidefinite or negative semidefinite, the range is all of <math>\R</math>.<br>If the matrix <math>A</math> is positive definite or (<math>A</math> is positive semidefinite and <math>\vec{b}</math> is in its image), the range is <math>[m,\infty)</math> where <math>m</math> is the minimum value. If the matrix <math>A</math> is negative definite or (<math>A</math> is negative semidefinite and <math>\vec{b}</math> is in its image), the range is <math>(-\infty,m]</math> where <math>m</math> is the maximum value. \|\| The case of "not positive semidefinite or negative semidefinite" does not arise for <math>n = 1</math>. Moreover, all the semidefinite cases must be definite, so we only have to consider the positive definite case and the negative definite case.<br>The positive definite case corresponds to <math>a > 0</math><br>The negative definite case corresponds to <math>a < 0</math>
	\|-		\|-
	\| [[local minimum value]] and points of attainment \|\| If the matrix <math>A</math> is ~~negative~~ definite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is positive semidefinite but not positive definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The positive definite case corresponds to <math>a > 0</math>: Here, the local minimum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The negative definite case corresponds to <math>a < 0</math>, and there is no minimum in this case.		\| [[local minimum value]] and points of attainment \|\| If the matrix <math>A</math> is positive definite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is positive semidefinite but not positive definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The positive definite case corresponds to <math>a > 0</math>: Here, the local minimum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The negative definite case corresponds to <math>a < 0</math>, and there is no minimum in this case.
	\|-		\|-
	\| [[local maximum value]] and points of attainment \|\| If the matrix <math>A</math> is negative ~~semidefinite~~, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is negative semidefinite but not negative definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The negative definite case corresponds to <math>a < 0</math>: Here, the local maximum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The positive definite case corresponds to <math>a > 0</math>, and there is no maximum in this case.		\| [[local maximum value]] and points of attainment \|\| If the matrix <math>A</math> is negative definite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is negative semidefinite but not negative definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The negative definite case corresponds to <math>a < 0</math>: Here, the local maximum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The positive definite case corresponds to <math>a > 0</math>, and there is no maximum in this case.
	\|-		\|-
	\| [[gradient vector]] function (analogous to the [[derivative]]) \|\| <math>\vec{x} \mapsto 2A\vec{x} + \vec{b}</math> \|\| the [[derivative]] is <math>x \mapsto 2ax + b</math> (consistent with the matrix formulation)		\| [[gradient vector]] function (analogous to the [[derivative]]) \|\| <math>\vec{x} \mapsto 2A\vec{x} + \vec{b}</math> \|\| the [[derivative]] is <math>x \mapsto 2ax + b</math> (consistent with the matrix formulation)

Vipul: /* Key data */

2014-05-26T03:41:18Z

Key data

← Older revision		Revision as of 03:41, 26 May 2014
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	\| [[range]] \|\| If the matrix <math>A</math> is not positive semidefinite or negative semidefinite, the range is all of <math>\R</math>.<br>If the matrix <math>A</math> is positive definite or (<math>A</math> is positive semidefinite and <math>\vec{b}</math> is in its image), the range is <math>[m,\infty)</math> where <math>m</math> is the minimum value. If the matrix <math>A</math> is negative definite or (<math>A</math> is negative semidefinite and <math>\vec{b}</math> is in its image), the range is <math>(-\infty,m]</math> where <math>m</math> is the maximum value. \|\| The case of "not positive semidefinite or negative semidefinite" does not arise for <math>n = 1</math>. Moreover, all the semidefinite cases must be definite, so we only have to consider the positive definite case and the negative definite case.<br>The positive definite case corresponds to <math>a > 0</math><br>The negative definite case corresponds to <math>a < 0</math>		\| [[range]] \|\| If the matrix <math>A</math> is not positive semidefinite or negative semidefinite, the range is all of <math>\R</math>.<br>If the matrix <math>A</math> is positive definite or (<math>A</math> is positive semidefinite and <math>\vec{b}</math> is in its image), the range is <math>[m,\infty)</math> where <math>m</math> is the minimum value. If the matrix <math>A</math> is negative definite or (<math>A</math> is negative semidefinite and <math>\vec{b}</math> is in its image), the range is <math>(-\infty,m]</math> where <math>m</math> is the maximum value. \|\| The case of "not positive semidefinite or negative semidefinite" does not arise for <math>n = 1</math>. Moreover, all the semidefinite cases must be definite, so we only have to consider the positive definite case and the negative definite case.<br>The positive definite case corresponds to <math>a > 0</math><br>The negative definite case corresponds to <math>a < 0</math>
	\|-		\|-
	\| [[local minimum value]] and points of attainment \|\| If the matrix <math>A</math> is ~~positive~~ definite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is positive semidefinite but not positive definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The positive definite case corresponds to <math>a > 0</math>: Here, the local minimum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The negative definite case corresponds to <math>a < 0</math>, and there is no minimum in this case.		\| [[local minimum value]] and points of attainment \|\| If the matrix <math>A</math> is negative definite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is positive semidefinite but not positive definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The positive definite case corresponds to <math>a > 0</math>: Here, the local minimum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The negative definite case corresponds to <math>a < 0</math>, and there is no minimum in this case.
	\|-		\|-
	\| [[local maximum value]] and points of attainment \|\| If the matrix <math>A</math> is negative semidefinite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is negative semidefinite but not negative definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The negative definite case corresponds to <math>a < 0</math>: Here, the local maximum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The positive definite case corresponds to <math>a > 0</math>, and there is no maximum in this case.		\| [[local maximum value]] and points of attainment \|\| If the matrix <math>A</math> is negative semidefinite, then <math>c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}</math>, attained at <math>\frac{-1}{2}A^{-1}\vec{b}</math><br>If <math>A</math> is negative semidefinite but not negative definite, it depends on whether <math>\vec{b}</math> is in the image of <math>A</math>. If yes, replace <math>A^{-1}\vec{b}</math> with the solution <math>\vec{v}</math> to <math>A\vec{v} = \vec{b}</math>, so we get a local minimum of <math>c - \frac{1}{4}\vec{b}^T\vec{v}</math> attained at <math>\frac{-1}{2}\vec{v}</math><br>If <math>A</math> is not positive semidefinite or if <math>\vec{b}</math> is not in the image of <math>A</math>, no local minimum value \|\| The negative definite case corresponds to <math>a < 0</math>: Here, the local maximum value of <math>c - \frac{b^2}{4a}</math> is attained at <math>\frac{-b}{2a}</math> (consistent with the matrix formulation)<br>The positive definite case corresponds to <math>a > 0</math>, and there is no maximum in this case.

Vipul: /* Key data */

2014-05-26T03:40:50Z

Key data

← Older revision		Revision as of 03:40, 26 May 2014
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	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Item !! Value !! Consistency with the case <math>n = 1</math>, where <math>f(x) = ax^2 + bx + c</math>, <math>A = (a)</math> (a <math>1 \times 1</math> matrix), \vec{b} = (b)</math> (a 1-dimensional vector)		! Item !! Value !! Consistency with the case <math>n = 1</math>, where <math>f(x) = ax^2 + bx + c</math>, <math>A = (a)</math> (a <math>1 \times 1</math> matrix), <math>\vec{b} = (b)</math> (a 1-dimensional vector)
	\|-		\|-
	\| default [[domain]] \|\| the whole of <math>\R^n</math> \|\| the whole of <math>\R</math>		\| default [[domain]] \|\| the whole of <math>\R^n</math> \|\| the whole of <math>\R</math>