I want to find normalized eigen vectors for:
$$ \begin{pmatrix} 1 & -2 & 0 \\ -2 & 5 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix} $$

The eigen values I found are 5.828,2,0.171
When finding the eigen vectors for $\lambda=5.828$ by letting $AX=\lambda X$

I get the following equations:
$$X_2=-2.41 X_1\\ X_1=-0.41 X_2$$
How can I solve these two and find normalized eigen vectors?
I get $X_3=0$

  • 1
    $\begingroup$ The exact eigenvalues are $\,3-2\sqrt2, 2,3+2\sqrt2$. Once you have eigenvectors, divide them by their norms, that's all. $\endgroup$
    – Bernard
    Jun 21, 2015 at 13:40

1 Answer 1


It is clear that the eigenvector for the eiganvalue $2$ is $(0,0,1)$. The remaining two are more interesting.

Notice that the eigenvalues are roots of the polynomial $x^2-6x+1$. (This is what you get from the characteristic polynomial. I assume this is how you get them.)

In particular, any eigenvalue fulfills $\lambda^2=6\lambda-1$.

Now to find an eigenvector for the eignevalue $\lambda$ you solve the linear system (I have omitted the third variable, since we know it must be zero): $$ \begin{pmatrix} 1-\lambda & -2 \\ -2 & 5-\lambda \end{pmatrix}\overset{(1)}\sim \begin{pmatrix} 1-\lambda & -2 \\ 0 & 0 \end{pmatrix} $$

(Since we know that $\lambda$ is an eigenvalue, the above matrix cannot have full rank. That's why we could replace the second row immediately by zeroes in the step (1). But you can also check that the second row is indeed a multiple of the first or. Just use the fact that $(\lambda-1)(\lambda-5)=4$ and multiply the first row by the number $\frac{\lambda-5}2=\frac2{\lambda-1}$.)

Clearly, the vector $\vec x=(2,1-\lambda,0)$ is a solution.

We want a normalized vector, so we calculate the norm of this vector: $|\vec x|^2 = 4+(1-\lambda^2)=\lambda^2-2\lambda+5=6\lambda-1-2\lambda+5=4(\lambda-1)$.

So the normalized eigenvectors have the form $$\frac{\vec x}{|\vec x|}=\left(\frac1{2\sqrt{\lambda-1}},-\frac{\sqrt{\lambda-1}}2,0\right).$$

Remark. I have tried to solve the system in a way in which we get a solution for both eigenvalues at the same time. (To get two eigenvectors for the price of one.) Moreover, in this way I avoided calculations with square roots.

Of course, if you prefer, you can find the solutions for $\lambda_1=3-2\sqrt2$ and for $\lambda_2=3+2\sqrt2$ separately.

In response to the following comment I will add more details:

I don't understand how the last row elements were made $0$? Is it like doing row echelon form? Why is it important to do this? From then how do you say $\vec x =(2,1−\lambda)$ is a solution

Yes, a standard way to solve a system of linear equations is get the matrix to row echelon form. In our case we have a homogeneous system represented by the matrix: $$\begin{pmatrix} 1-\lambda & -2 \\ -2 & 5-\lambda \end{pmatrix}$$ As I have mentioned before, we already know that the matrix does not have full rank. Therefore second row is a multiple of the first row, so by doing one simple row operation we get zeroes in the second row.

However, we might want to go a bit slower and perform the actual operations. (What I wrote above is only true if we calculated the eigenvalues correctly. So in this way we doublecheck that they are indeed eigenvalues.)

So we want add a $c$-multiple of the first row to the second row. We want to choose $c$ in such way that we get zero in the first place of the second row. So we need $c=\frac{2}{1-\lambda}$.

What do we get in the second place of the second row? Well, let's calculate: $$5-\lambda-2c = 5-\lambda - \frac{4}{1-\lambda} = \frac{(5-\lambda)(1-\lambda)-4}{1-\lambda} = \frac{\lambda^2-6\lambda+1}{1-\lambda} = 0$$

So we see that after this row operation we have the new matrix: $$ \begin{pmatrix} 1-\lambda & -2 \\ 0 & 0 \end{pmatrix} $$

This matrix corresponds to the linear system which has only one equation $$(1-\lambda)x_1 - 2x_2=0.$$ How can we find a solution? For any value we choose for $x_2$, the above equation uniquely determines $x_1=\frac{2x_2}{1-\lambda}$. We can choose arbitrary non-zero value for $x_2$, So I chose ${1-\lambda}$, as to avoid fractions.

  • $\begingroup$ I don't understand how this is done. I don't understand how the last row elements were made 0? Is it lke doing row echelon form? Why is it important to do this? From then how do you say $\vec x=(2,1-\lambda)$ is a solution $\endgroup$
    – sam_rox
    Jun 21, 2015 at 16:10
  • $\begingroup$ @sam_rox I have edited my post to add a few more details. (And I have also added the third coordinate, which was missing.) $\endgroup$ Jun 21, 2015 at 16:26
  • $\begingroup$ Thanks now I understand how you have done it. But how I do is if the matrix is A, then by |A-$\lambda$ I|=0 I find the eigen values and from AX=$\lambda X$ I find the eigen vectors.But to solve for eigen vectors you have used a form similar to A-$\lambda$ I=0 . I set AX=$\lambda X$ to obtain $X_2=-2.41 X_1\\ X_1=-0.41 X_2$ . Is there a way to come up with the solution through these equations? $\endgroup$
    – sam_rox
    Jun 21, 2015 at 17:25
  • $\begingroup$ The (systems of) equations $AX=\lambda X$ and $(A-\lambda I)X=0$ are equivalent. $\endgroup$ Jun 21, 2015 at 17:30

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