## Trending News

# Eigenvalue 是非題

1.If A is a 3x3 matrix with characteristic polynomial p(x)=(x-3)(x-2)^2,

then it is guaranteed that there is a nonzero vector X such that AX=2X

2.If A is a 3x3 matrix with characteristic polynomial p(x)=(x-3)(x-2)^2,

then there is at most one nonzero vector X such that AX=3X

3.If A is a 3x3 matrix with characteristic polynomial p(x)=(x-3)(x-2)^2,

then there are two nonzero,lineary independent vector X1 and vector X2 such that AXi=2Xi

### 1 Answer

- Anonymous1 decade agoFavorite Answer
Answer:

1. True.

2. True.

3. False.

proof:

1.

For the eigenspace (characteristic space) of any eigenvalue is at least of

dimension 1.

2.

For the eigenspace space of any eigenvalue is at least of dimension 1

and the dimension of the eigenspace space is less than or equal to

the algebraic multiplicity 1, the degree of the factor (x-3) of p(x).

3.

Note that take a matrix A which is triangularable but diagonalable,

then it would fit your requirement.

Here is a counterexample.

Let A=

3 0 -1

0 2 0

0 0 2

then the characteristic polynomial of A is p(x) = (x-3)(x-2)^2

Now set X= (x,y,z)^t, 0= (0,0,0)^t

A-2I=

1 0 -1

0 0 0

0 0 0

(A-2I)X = 0 implies x - z = 0 and z = 0.

This leads to x = z = 0 and y is free.

So the eigenspace of 2 can be generated by(0,1,0), it is of dimension 1,

therefore there are no two nonzero, lineary independent vector X1

and vector X2 such that AXi=2Xi.

2008-01-20 15:47:28 補充：

更正打錯 :

Let A=

3 0 -1

0 2 1<-----2列3行為1

0 0 2

A-2I=

1 0 -1

0 0 1<-----2列3行為1

0 0 0

2008-01-20 16:45:45 補充：

remark:

Suppose k be an eigenvalue of A and p(x) = (x-k)^m(k)*…be the

characteristic polynomial of A, then by definitions

2008-01-20 16:46:05 補充：

(a) the eigenspce(characteristic space) of k = ker(A-kI) = {x: Ax=kx}

(b) the geometry multiplicity of k = dim( eigenspce ).

(c) the algebraic multiplicity of k =m(k), the degree of the factor (x-k)

of p(x).

2008-01-20 16:46:29 補充：

the following fact are true:

1.dim( eigenspce )> or =1, for any eigenvalue k.

2. m(k)> or = dim( eigenspce )> or =1, for any eigenvalue k.

3. The algebraic multiplicity m(k) of k may strictly greater than the

geometry multiplicity dim( eigenspce ) of k.

2008-01-20 16:48:03 補充：

How to get your counterexample:

For counterexample, let the main diagonal elements be the correspondent

eigenvalues: 3,2,2 ( note that the algebraic multiplicity of 3 is 1, the

algebraic multiplicity of 2 is 2)

2008-01-20 16:48:31 補充：

and let the entry of A at row 1 column 2 is A(12)=0, put A(13)=-1,

A(23)=1and 0 elsewhere so that

2008-01-20 16:55:23 補充：

"the dimension of the eigenspace of 2 is 1",

" not 2 ",

then it would work.

註:

抱歉得請您回微積分的題目:求 f(x)=x/(2x的絕對值+1)的微分.

看一下我的更正錯誤, 寫在意見(4), 其中x<0 的部分出 錯!