Find the normalized eigenvectors of this matrix:


My work:

$$\det(A-\lambda I) =(\lambda-1)^2(\lambda-4)$$

$$\lambda_1=1, \lambda_2=1,\lambda_3=4$$

$$\lambda_1\iff x + y +z=0$$ $$\lambda_2\iff x + y +z=0$$ $$\lambda_3\iff -2x + y +z=0$$ $$\lambda_3\iff x + -2y +z=0$$ $$\lambda_3\iff x + y +-2z=0$$

When my lecturer did this the equations were easy to solve... How do I find the normalized eigevectors from these equations?

Please Help :)

  • $\begingroup$ just a hint. There is no unique eigenvector structure for all matrices. This means there might be many solutions to the equations without any problem. $\endgroup$ – Seyhmus Güngören May 11 '13 at 12:40
  • $\begingroup$ I think you wanted the bottom row of your matrix to be 1 1 2 and not 1 2 1. $\endgroup$ – Gerry Myerson May 11 '13 at 12:57
  • $\begingroup$ By the way, is it possible that you are studying eigenvalues and eigenvectors, and you have never learned how to solve a system of three linear equations in three unknowns? $\endgroup$ – Gerry Myerson May 11 '13 at 13:01
  • $\begingroup$ Yeh just corrected it and yes (I don't!), I'm a chemist not a mathmatician I just want to find these eigenvector things and pass my exam. $\endgroup$ – Anon May 11 '13 at 13:05
  • $\begingroup$ Would you expect to pass an exam on organic chemistry, if you didn't know what carbon was? You are not ready for the course you have enrolled in. Run, don't walk, to see your teacher and try to work out a way to get out of this mess. $\endgroup$ – Gerry Myerson May 11 '13 at 13:08

For the matrix:


The CP is given by:

$$|A - \lambda I| = 0$$

For the CP, we get:

$$-\lambda^3+6 \lambda^2-9 \lambda+4 = -(\lambda-4) (\lambda-1)^2 = 0$$

This leads to three eigenvalues as $\lambda_1 = 4, \lambda_2 = 1, \lambda_3 = 1$.

We have a repeated eigenvalue.

To find an eigenvector, for each eigenvalue, we solve:

$$[A - \lambda_i I]v_i = 0$$

When we have repeated eigenvalues, we may need to resort to generalized eigenvectors, which I assumed you learned in class.

Lets find an eigenvector as an example for $\lambda_2 = 1$:

$$[A - \lambda_2 I]v_2 = 0$$

$$[A - \lambda_2 I]v_2 = \begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}v_2 = 0$$

The row-reduced-echelon-form (RREF) is given by:

$$A=\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}v_2 = 0$$

So, we can choose an eigenvector to satisfy the RREF for:

$$x + y + z = 0 \rightarrow (-1, 0 , 1)$$

Since we have a repeated eigenvalue, we can choose a second, linearly independent, eigenvector as:

$$(-1, 1, 0)$$

Notice that both of these satisfy the RREF equation.

Repeat the above process for the other eigenvector so you can practice.

This leads to the eigenvalues and eigenvectors as:

$$\lambda_1 = 4, v_1 = (1, 1, 1)$$

$$\lambda_2 = 1, v_2 = (-1, 0, 1)$$

$$\lambda_3 = 1, v_3 = (-1, 1, 0)$$

However, we are being asked to find the normalized eigenvectors, so we have to do an additional step.

The way we normalize is to divide each eigenvector by it's length.

  • $v_2 = (-1,0, 1)$
  • $|v_2| = \sqrt{-1^2 + 0^2 + 1^2} = \sqrt{2}$
  • Thus, the normalized eigenvector is:

$$\tilde v_2 = \frac{v_2}{|v_2|} = \frac{1}{\sqrt{2}}(-1,0, 1)$$

You should verify that the normalized eigenvector has length one.

Repeat this for the other two eigenvectors.

  • $\begingroup$ It seems to me that you've skipped over one of the problems facing OP, which is that it appears that OP does not know how to get your $v_1$, $v_2$, and $v_3$. $\endgroup$ – Gerry Myerson May 11 '13 at 13:00
  • $\begingroup$ Sorry I did write the matrix wrong its now corrected can you help now :) $\endgroup$ – Anon May 11 '13 at 13:02
  • $\begingroup$ @GerryMyerson: Okay, let me add those details too. Great catch on the incorrect row! Thanks $\endgroup$ – Amzoti May 11 '13 at 13:05
  • $\begingroup$ So how did you get the eigenvectors from the simultaneous equations, you've not really answered my question just told me the answer! :) $\endgroup$ – Anon May 11 '13 at 13:08
  • $\begingroup$ So I just choose solutions to the equations and that is my eigenvector. But surely someone else could choose countless other solutions and there all correct. $\endgroup$ – Anon May 11 '13 at 13:25

To find the eigenvectors of the matrix $A$ you find the null space of $(A-\lambda_1 I)$, $(A-\lambda_2 I)$ and $(A-\lambda_3 I)$. Since $\lambda_1=\lambda_2$, to find the eigenvectors corresponding to those eigenvalues it suffices to find the null space of $(A-\lambda_1 I)x=0$.

Thus $$(A-\lambda_1 I)x=0$$ $$\longrightarrow\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\\\end{bmatrix}x=0$$ $$\longrightarrow x_1=\begin{bmatrix}-1\\0\\1\\\end{bmatrix} \text{ and } x_2=\begin{bmatrix}-1\\1\\0\\\end{bmatrix}$$ Similarly $$(A-\lambda_3 I)x=0$$ $$\longrightarrow\begin{bmatrix}-2&1&1\\1&-2&1\\1&1&-2\\\end{bmatrix}x=0$$ $$\longrightarrow x_3=\begin{bmatrix}1\\-1\\1\\\end{bmatrix}$$ Finally, to answer the question you just normalize them to get $$x_1=\begin{bmatrix}-1/\sqrt{2}\\0\\1/\sqrt{2}\end{bmatrix} \text{, } x_2=\begin{bmatrix}-1/\sqrt{2}\\1/\sqrt{2}\\0\end{bmatrix}\text{, } x_3=\begin{bmatrix}1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}\end{bmatrix}$$


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