# Calculating Eigenvectors of a Diagonal Matrix

I'm feeling dumb even asking this. But there might be a definition for this somewhat like why $$1$$ is not a prime number. Therefor this might be the right place to ask this question anyway.

Given the matrix $$\begin{bmatrix} 4 & 0 \\ 0 & 4 \\ \end{bmatrix}$$

One sees immediately that the eigenvalues are $$4$$ and $$4$$ and the corresponding eigenvectors $$\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$$ and $$\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$$

Assuming one doesn't see that or one tries to program this he would use $$(A-\lambda_i E)v_i=0$$ to calculate the eigenvectors. But using this in this really simple example leads to $$\begin{gather} \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}v=0 \end{gather}$$ So every solution would be correct for $$v$$.

Where is my mistake? I hope it is something obvious. I really hate it when there are special cases and one can not always use one scheme for all related problems.

• In your particular example, every vector is an eigenvector with EV 4 - you're not doing anything wrong. Jun 23, 2019 at 7:58
• Every solution is valid for $v$ but we can easily see that each value within the eigenvectors are independent so we typically use the eigenvectors $[0,1]^T$, $[1,0]^T$ to represent this. Jun 23, 2019 at 7:59
• @PeterForeman that was one of those things which made me suspicious. I could choose two time the same eigenvector since every solution is correct, right? For example v1 and v2 = [1,1]. Wouldn't this mean they are not independent? I assumed the eigenvectors should in this case always be orthogonal to each other to show this independency. Jun 23, 2019 at 8:22
• Yes, if you choose the same one twice, or a multiple of an eigenvector, then they’re not linearly independent. So? That doesn’t prevent you from finding some other pair of eigenvectors that is linearly independent.
– amd
Jun 23, 2019 at 17:22
• If $\lambda$ is an eigenvalue then there is no distinguished eigenvector corresponding to $\lambda$. What you have is the corresponding eigenspace which is the kernel of $A-\lambda I$. This eigenspace is the often represented by a basis, but the choice of the basis is not unique. Jun 23, 2019 at 17:32

There is nothing wrong with your calculations. Note that if $$v_1$$ and $$v_2$$ are eigenvectors corresponding to an eigenvalue $$\lambda$$, so is $$c_1v_1+c_2v_2$$. In your case, note that $$e_1$$ and $$e_2$$ are basis elements for $$\mathbb R^2$$.
This happens for any $$n \times n$$ identity matrix since the eigenvectors are always orthogonal and hence they span the entire $$\mathbb{R}^n$$ space. Thus, any vector in the space is an eigenvector. Therefore, there is no mistake in your solution.