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

Rating
  • Anonymous
    1 decade ago
    Favorite 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!

Still have questions? Get your answers by asking now.