Eigenvectors of a 3x3 Matrix A: Calculation and Verification

[Mathman23]

Hi

I have this here matrix

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

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:


A = \left[ \begin{array}{ccc} (2-(-1) &amp; 1 &amp; 0 \\ 0 &amp; (1-(-1) &amp; 0 \\ 3 &amp; 3 &amp; 1-(-1) \end{array} \right]<br /> = \left[ \begin{array}{ccc} 3 &amp; 1 &amp; 0 \\ 0 &amp; 2 &amp; 0 \\ 3 &amp; 3 &amp; 0 \end{array} \right]

Then according to the theorem regarding eigenvectors:

\left[ \begin{array}{ccc} 3 &amp; 1 &amp; 0 \\ 0 &amp; 2 &amp; 0 \\ 3 &amp; 3 &amp; 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]

then

3v_1 + v_2 = 0

2v_2 = 0

3v_1 + 3 v_2 = 0

Is my calculations correct ??

sincerley and best regards,

Fred

 

[Muzza]

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:

\left[ \begin{array}{ccc} 3 &amp; 1 &amp; 0 \\ 0 &amp; 2 &amp; 0 \\ 3 &amp; 3 &amp; 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]

is wrong, there should be a 1 there instead of a 0.

 

[Corneo]

Check your eigenvalues again. I get \lambda = 0,1,2

 

[Mathman23]

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

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:

\left[ \begin{array}{ccc} 3 &amp; 1 &amp; 0 \\ 0 &amp; 2 &amp; 0 \\ 3 &amp; 3 &amp; 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]

is wrong, there should be a 1 there instead of a 0.

 

[Mathman23]

Check your eigenvalues again. I get \lambda = 0,1,2

Hi again

thats because I typed my matrix wrong

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

Sincerley

Fred