Calculate the eigenvectors of this $2 \times 2$ matrix (problem because all equations result in $0=0$)

Question

What's the eigenvalues and eigenvectors of this matrix $B=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}$?

The characteristic polynomial is $\lambda^{2}-4\lambda+4=0$

The (double-)eigenvalue is $\lambda=2$

Now we want calculate the eigenvector with these, insert $\lambda=2$ here:

$$\begin{pmatrix} 2-\lambda & 0\\ 0 & 2-\lambda \end{pmatrix}= \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}$$

So we have $\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}$

I: $0x+0y=0$

II: $0x+0y=0$

But what would be the eigenvector now?

Can I randomly choose one myself? As example, would this be correct?

Eigenvector is $v= \begin{pmatrix} 1\\ 2 \end{pmatrix}$.

---

An additional question, how would you write the eigenspace?

Answer 1

Yes, for this matrix every nonzero vector is an eigenvector with eigenvalue $2$.

It is a bit of a detour in this case to find that by solving the characteristic polynomial -- but it is certainly a valid way to proceed, until you have the experience to recognize immediately how a matrix of this form behaves.

Generally what you would like to find in this case is a basis for the eigenspace (which in this case is the entire $\mathbb R^2$), so you should choose two linearly independent vectors. The simplest and most boring choice would be $(^1_0)$ and $(^0_1)$, but you can certainly also choose $(^1_2)$ and, for example $(^7_3)$ (or anything else that is not parallel to $(^1_2)$).

(Since question actually asks "what are the eigenvectors", a more strictly correct answer would be "every nonzero vector is an eigenvector" but giving a basis for the eigenspace is conventional and may be the kind of answer that's expected anyway).

Answer 2

In fact, every non-zero vector $(x,y)$ is an eigenvector of $B$ to the eigenvalue $2$. This is because $Bx=2x$ holds for every vector $x$

Answer 3

Simply pick the eigenvectors $\binom{1}{0}$ and $\binom{0}{1}$. These satisfy the given equations canonically: they form a standard basis for the family of eigenvectors in fact.

Yes, you can randomly choose one yourself. Don't forget a second, linearly independent one too.

The eigenspace is the span of the eigenvectors.