Re: [線代] 矩陣填空使其"不"可對角化

[Sfly (topos)]

※ 引述《cyt147 (大叔)》之銘言：
: Determine all values of a,b,c,d,e,f in R so that
:     1 a b c
:     0 1 d e
: A=( 0 0 2 f ) is NOT diagonalizable.
:     0 0 0 2
: A為upper-triangular，所以它的eigenvalue為主對角線的entry，也就是1跟2，而且
: 它們的algebraic multiplicity皆為2，根據我學到的知識:
: A is not diagonalizable if and only if dim(E_1)=1, or dim(E_2)=1,
: where E_i denotes the eigenspace of A corresponding to i. That is to say,
: A is not diagonalizable if and only if one of its eigenvalues has geometric
: multiplicity unequal to its algebraic multiplicity.
: To attain the above result, I need to find the eigenspaces of A. But ther
: are so many unknowns in the system of linear equations that I can't proceed
: as usual. Can somebody please tell me how to deal with this kind of situation?

Since the eigenvalues of A are 1 and 2,
A is diagonalizable
<=> the minimal polynomial of A is (x-1)(x-2)
i.e. (A-I)(A-2I)=0

<=>
(0 a b c)(-1 a b c) = (0 -a ad ae+bf) =0
 0 0 d e  0 -1 d e    (0  0  0   df )
 0 0 1 f  0  0 0 f    (0  0  0    f )
 0 0 0 1  0  0 0 0    (0  0  0    0 )

<=> a、d、f、(ae+bf)=0.

Hence, A is not diagonalizable <=> a*d*f*(ae+bf)!=0

--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 118.161.37.63
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1514486619.A.808.html
推 cyt147      : 謝謝，請讓我想一下，我看過您第一段用的定理。
→ Desperato   : 欸欸欸欸欸居然qw q
→ Desperato   : 還是覺得怪怪的
→ Desperato   : [[1,1,0,0],[0,1,0,0],[0,0,2,1],[0,0,0,2]]
→ Desperato   : 這傢伙明顯not diagonalizable 吧
→ Desperato   : 啊 應該是 a and f and ad and df and ae+bf all
→ Desperato   : zero 才對吧 這樣就沒錯了
推 cyt147      : 這樣不就等價於a=f=0?之前我的做法還正確嗎?
→ cyt147      : 我是說free variable那個
推 yueayase    : 漂亮~~
※ 編輯: Sfly (114.45.201.160), 12/30/2017 21:14:13