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Urgent: eigenvector problem

  • Thread starter Mathman23
  • Start date
255
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Hi

I have this here matrix

[tex]A = \left[ \begin{array}{ccc} 2 & 1 & 0 \\ 0 & 1 & 0 \\ 3 & 3 & 0 \end{array} \right][/tex]

I calculate the eigenvalues and get (2,1,-1)

Next I calculate the eigenvectors and get (1,0,1) and (-1,1,0) and (0,0,0)

My professor says my third eigenvector is wrong and it should (0,0,1)

My calculation:


[tex]A = \left[ \begin{array}{ccc} (2-(-1) & 1 & 0 \\ 0 & (1-(-1) & 0 \\ 3 & 3 & 1-(-1) \end{array} \right]
= \left[ \begin{array}{ccc} 3 & 1 & 0 \\ 0 & 2 & 0 \\ 3 & 3 & 0 \end{array} \right][/tex]

Then according to the theorem regarding eigenvectors:

[tex]\left[ \begin{array}{ccc} 3 & 1 & 0 \\ 0 & 2 & 0 \\ 3 & 3 & 0 \end{array} \right] \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] [/tex]

then

[tex]3v_1 + v_2 = 0[/tex]

[tex] 2v_2 = 0[/tex]

[tex]3v_1 + 3 v_2 = 0[/tex]

Is my calculations correct ??

sincerley and best regards,

Fred
 
694
0
First of all, the zero vector is never an eigenvector (even though it might behave like one). If x is an eigenvector "belonging" to the eigenvalue -1, then Ax = -x, or equivalently (A + I)x = 0. The right-most entry in the bottom row of this equation:

[tex]\left[ \begin{array}{ccc} 3 & 1 & 0 \\ 0 & 2 & 0 \\ 3 & 3 & 0 \end{array} \right] \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] [/tex]

is wrong, there should be a 1 there instead of a 0.
 
Last edited:
321
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Check your eigenvalues again. I get [tex]\lambda = 0,1,2[/tex]
 
255
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Hi and many thanks for Your answer.

I have a second question I hope You can answer for me.

Finding a matrix P which is invertible and which complies with P^-1 AP ? ?

Isn't P then span of the three eigenvector ???

Sincerley and Best Regards,

Fred

Muzza said:
First of all, the zero vector is never an eigenvector (even though it might behave like one). If x is an eigenvector "belonging" to the eigenvalue -1, then Ax = -x, or equivalently (A + I)x = 0. The right-most entry in the bottom row of this equation:

[tex]\left[ \begin{array}{ccc} 3 & 1 & 0 \\ 0 & 2 & 0 \\ 3 & 3 & 0 \end{array} \right] \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] [/tex]

is wrong, there should be a 1 there instead of a 0.
 
255
0
Corneo said:
Check your eigenvalues again. I get [tex]\lambda = 0,1,2[/tex]

Hi again

thats because I typed my matrix wrong

[tex]A = \left[ \begin{array}{ccc} 2 & 1 & 0 \\ 0 & 1 & 0 \\ 3 & 3 & -1 \end{array} \right][/tex]

Sincerley

Fred
 

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