Urgent: eigenvector problem

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:

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

Then according to the theorem regarding eigenvectors:

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

then

$$3v_1 + v_2 = 0$$

$$2v_2 = 0$$

$$3v_1 + 3 v_2 = 0$$

Is my calculations correct ??

sincerley and best regards,

Fred

Related Introductory Physics Homework Help News on Phys.org

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:

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

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

Last edited:

Corneo

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

Mathman23

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:

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

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

Mathman23

Corneo said:
Check your eigenvalues again. I get $$\lambda = 0,1,2$$

Hi again

thats because I typed my matrix wrong

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

Sincerley

Fred

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving