By definition:

$$Ax = \lambda x$$

for $x \neq 0$

I was using this to calculate the eigenvectors for $A$:

$$A = \begin{bmatrix}2 & 0 & 0\\1 & 3 & 0\\ 2 & 3& 4\end{bmatrix}$$

I already found that the eigenvalues for this matrix are:

$$\lambda = 2,3,4$$

*However*, when I began evaluating the eigenvectors for these eigenvalues, I noticed that for $\lambda = 4$, the resulting eigenvector is:

$$x = \begin{bmatrix}0\\0\\0\end{bmatrix}$$

But how is this possible??

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Please help me find my mistake.

To find the eigenvectors, we must solve the equation:

$$(\lambda I -A)x = 0$$

$$\begin{bmatrix} 4 & 0 & 0\\0 & 4& 0\\ 0 & 0 & 4 \end{bmatrix} - \begin{bmatrix}2 & 0 & 0\\1 & 3 & 0\\ 2 & 3& 4\end{bmatrix} = \begin{bmatrix}2 & 0 & 0\\-1 & 1 & 0\\ -2 & -3 & 0\end{bmatrix}$$

Row reducing, we see that 

$$ \left[
 \begin{array}{ccc|c}
 1&0&0&0\\
 0&-1&0&0\\
 0&3&0&0
 \end{array}
\right] $$

...which tells us that $x\_1=0, x\_2=0, 3x\_2=0$.