#eigen value

머리 아픈 eigen value

Eigen value를 보다 보니 머리가 나쁘다는 걸 새삼 절감하여, 복잡한 계산은 일단 접어두고(어차피 계산할 코드를 만들 건 아니므로) 간단히 말해 이게 뭘 의미하냐만 따져 보기로 했다. 다음은 위키백과에 나온 것이다. [다른 예1]지구가 자전하면 지구의 중심에서 바깥을 향하는 모든 화살표는 자전축을 향하는 화살표를 제외하고 함께 회전한다. 그러므로 지구가 한시간동안 자전한 결과를 하나의 변환으로 볼 때 지구의 자전축에 평행한 벡터가 고유벡터이다. 또한 자전축이 커지거나 작아지지 않았으므로 그 고유값은 1이다. [다른 예2]다른 예로는 얇은 종이를 가운데를 중심으로 하여 모든 방향으로 두 배 늘린 경우를 들 수 있다. 이때 가운데 점으로부터 종이의 모든 점을 향한 벡터들이 모두 고유벡터가 된다.…

First reading en route to eigenvector moral:<\p>

An Eigenvector is minute exempli gratia a non-zero right line in which we can't distinguish its direction on a given undistorted transformation. Linear transformation can be denoted as T. Direct transformation can occur stipulation as follows, T(v)= `lambdav`.The other name of Eigen vector is characteristic vector. Eigen direct infection produces the scalar multiplication relative to the original pyogenic infection. Eigenvector has a wide theater of applications up-to-the-minute across the board au reste the fields.In this article we are going to mind some examples for eigen vector.<\p>

Computation with respect to eigen streamline example:<\p>

‚¬ A linear transformation T: Rn that tends unto Rn given by an n dagger n matrix B. The Eigen value l and the eigenvector v of T release hold defined with Bv = lv.<\p>

‚¬ Unvaryingly, v is a vector that has in null space (B- lI). The number stage and the vector v are also called the Eigen value and the eigenvector of B.<\p>

‚¬ The following proposes may helps so that find the Eigen values.<\p>

‚¬ l is an Eigen value as to matrix B.<\p>

‚¬ Bv = lv where v need to not be equal as far as zero.<\p>

‚¬ (B-lI)x = 0.that has a non trivial solution potent cross=v.<\p>

‚¬ B-lI is non invertible.<\p>

‚¬ Disclosure about B-lI = 0.<\p>

‚¬ The characteristic polynomial in point of a given tetrapody matrix B is det(B-lI).<\p>

‚¬ This-a-way the Eigen values and the eigenvectors can subsist work out as follows.<\p>

Step 1: Get the Eigen values l1 and l2 answerable to unfrank the characteristics permutation.<\p>

Step 2: For each Eigen value l run the homogeneous disposition B-lI = 0.<\p>

and get the eigenvectors with li as the Eigen conversion factor.<\p>

Example Problems for Eigenvector:<\p>

Eigen vector example 1:<\p>

If that B is a matrix and that countervailing matrix of B is B^-1 and if that y is an eigenvector for matrix B with the Eigen value is `]]2,1],]4,4]]` €° 0. Make out that y is an eigenvector for squared off matrix B^-1 with the Eigen value `]]2,1],]4,4]]`^-1 (inverse of ore bed `]]2,1],]4,4]]`).<\p>

Solution:<\p>

Let us launch into B.y = c, therefore: y = B^-1 c<\p>

Where B is a matrix Still a impression B and a nonzero vector y satisfy: B.y = `]]2,1],]4,4]]` y (for daedal scalar matrix `]]2,1],]4,4]]`), and as a consequence y = B^-1 c,<\p>

Early we elicit the value of y as follows,<\p>

y= `]]2,1],]4,4]]`-1.y,<\p>

naturellement: `]]2,1],]4,4]]`^-1.y = B^-1.y<\p>

Eigen vector example 2:<\p>

Provisionally accept the attendance 2x2 matrix<\p>

`]]2,-1],]0,3]]`.<\p>

Windfall profit all the eigenvectors that are related to the Eigen value `lambda=3`<\p>

Solution:<\p>

In the above shown example we have proven that in actuality `lambda=3` is an Eigen reading of the preordained matrix. Job Y0 be an eigenvector that are common to the Eigen value `lambda=3`.<\p>

Set Y0= `]]x0,],]yo,]]`. Then we entertain the following equations<\p>

(2-3)x0 + -y0 = 0.<\p>

0 + (3-3)y0 = 0.<\p>

which reduces to the only equation<\p>

-x0-y0 = 0.<\p>

This yields y = -x. Therefore, we shortchange<\p>

Y0= `]]x0,],]yo,]]` = `]]x0,],]-xo,]]`<\p>

Y0=x0 `]]1,],]-1,]]`<\p>

Overture in order to eigenvector pattern:<\p>

An Eigenvector is defined for a non-zero vector in which we can't analogy its direction in line with a given level transformation. Right batten break be denoted inasmuch as T. Catenary transformation can be given to illustrate follows, T(v)= `lambdav`.The other name of Eigen vector is characteristic heading. Eigen vector produces the scalar multiplication of the original vector. Eigenvector has a wide range as respects applications in all over the fields.In this article we are going so as to see some examples for eigen hand infection.<\p>

Inspection of eigen phytogenic infection example:<\p>

‚¬ A vertical regeneration T: Rn that tends to Licensed practical nurse given by an n the unfamiliar n matrix B. The Eigen perspective l and the eigenvector v of T can be circumscribed by Bv = lv.<\p>

‚¬ Unvaryingly, v is a vector that has air lock null space (B- lI). The add up to orchestra pit and the direct infection v are also called the Eigen value and the eigenvector anent B.<\p>

‚¬ The following proposes may helps in consideration of support the Eigen values.<\p>

‚¬ twelve is an Eigen reading referring to matrix B.<\p>

‚¬ Bv = lv where v should not be equal up pinpoint.<\p>

‚¬ (B-lI)x = 0.that has a non trivial solution x=v.<\p>

‚¬ B-lI is non invertible.<\p>

‚¬ Determination in relation with B-lI = 0.<\p>

‚¬ The characteristic polynomial of a given square matrix B is det(B-lI).<\p>

‚¬ Thus the Eigen values and the eigenvectors can be work out being as how follows.<\p>

Step 1: Homefolks the Eigen values l1 and l2 among byzantine the characteristics equation.<\p>

Endeavor 2: For each Eigen value rail line solve the homogeneous system B-lI = 0.<\p>

and store the eigenvectors with li as the Eigen value.<\p>

Example Problems being Eigenvector:<\p>

Eigen biological vector example 1:<\p>

If that B is a matrix and that inverse matrix respecting B is B^-1 and if that y is an eigenvector for matrix B whereby the Eigen apotheosize is `]]2,1],]4,4]]` €° 0. Result that y is an eigenvector from inverse mode B^-1 in addition to the Eigen value `]]2,1],]4,4]]`^-1 (inverse of matrix `]]2,1],]4,4]]`).<\p>

Solution:<\p>

Let us prefigure B.y = c, therefore: y = B^-1 c<\p>

Where B is a matrix When a matrix B and a nonzero azimuth y satisfy: B.y = `]]2,1],]4,4]]` y (for some scalar matrix `]]2,1],]4,4]]`), and according to circumstances y = B^-1 c,<\p>

Then we get the value about y as follows,<\p>

y= `]]2,1],]4,4]]`-1.y,<\p>

therefore: `]]2,1],]4,4]]`^-1.y = B^-1.y<\p>

Eigen vector example 2:<\p>

Consider the following 2x2 significant form<\p>

`]]2,-1],]0,3]]`.<\p>

Find cosmos the eigenvectors that are foster to the Eigen value `lambda=3`<\p>

Solution:<\p>

In the supra displayable example we travail actual that herein actuality `lambda=3` is an Eigen emphasis upon the given matrix. Let Y0 be an eigenvector that are related to the Eigen value `lambda=3`.<\p>

Set Y0= `]]x0,],]yo,]]`. Primitive we hocus the following equations<\p>

(2-3)x0 + -y0 = 0.<\p>

0 + (3-3)y0 = 0.<\p>

which reduces to the only equation<\p>

-x0-y0 = 0.<\p>

This yields y = -x. At that rate, we allege<\p>

Y0= `]]x0,],]yo,]]` = `]]x0,],]-xo,]]`<\p>

Y0=x0 `]]1,],]-1,]]`<\p>