Consider the matrix $$B=\begin{pmatrix} 1 & i \\ -i & 0 \end{pmatrix}$$

Compute the eigenvectors and eigenvalues of this matrix viewed as a map from $\mathbb{C^2}$ to $\mathbb{C^2}.$ Determine whether it can be diagonalised.

I understand that in order to find the eigenvalues I must compute det$(B-\lambda I)$ and I have done this, giving me $\lambda=\frac{1+\sqrt{5}}{2} \text{ and } \lambda=\frac{1-\sqrt{5}}{2}.$ I also recognise that there are two distinct eigenvalues and so the matrix is diagonalisable.

How do I find the eigenvectors though?

So far I have done the following;

$$\begin{pmatrix} 1 & i \\ -i & 0 \end{pmatrix}-\begin{pmatrix} \frac{1-\sqrt{5}}{2} & 0 \\ 0 & \frac{1-\sqrt{5}}{2} \end{pmatrix}=\begin{pmatrix} \frac{1+\sqrt{5}}{2} & i \\ -i & \frac{-1+\sqrt{5}}{2} \end{pmatrix}$$

Then set $$\begin{pmatrix} \frac{1+\sqrt{5}}{2} & i \\ -i & \frac{-1+\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}=0$$ which gives $$v_1=v_2\bigg(\frac{-1-\sqrt{5}}{-2i}\bigg).$$ I have come to a bit of a sticking point here though.

  • $\begingroup$ The standard way is to find the nullspace of $B-\lambda I$, for each choice of eigenvalue. $\endgroup$ – vadim123 Mar 2 '16 at 15:59
  • $\begingroup$ Could you show how you would do this please? I have attempted that but I can't seem to find the solution. $\endgroup$ – Si.0788 Mar 2 '16 at 16:11
  • $\begingroup$ There are many examples of this process, on this website and elsewhere. If you share what you tried we can help diagnose where you went wrong. $\endgroup$ – vadim123 Mar 2 '16 at 16:21
  • $\begingroup$ @Si.0788 You are almost there: The relation between $v_1$ and $v_2$ gives the eigenvector associated with eigenvalue $\lambda_1=(1-\sqrt{5})/2$ (Note that if $v\in\mathbb{C}^2$ is an eigenvector of $B$, then so is $cv$, where $c\in\mathbb{C}$). Repeat the same process for $\lambda_2=(1+\sqrt{5})/2$. $\endgroup$ – Katie Imach Mar 2 '16 at 16:38
  • $\begingroup$ A tip: for any square matrix, the sum of the eigenvalues weighted by algebraic multiplicity equals the trace. Your eigenvalues add up to 1 but they should add up to 2, so your eigenvalues couldn't possibly be right. Also, the eigenvalues multiply to give the determinant; in this case you can see that the rows differ only by a factor of $-i$, so the determinant is $0$, which tells you both eigenvalues immediately. $\endgroup$ – Ian Mar 2 '16 at 17:57

For 2x2 matrices, the Cayley-Hamilton theorem is the easiest way to find eigenvectors. For 3x3, it is still very easy, and numerically more stable than solving for the null space.

The Cayley-Hamilton theorem says that if $P(x)$ is the characteristic polynomial of $A$, then $P(A) = 0$. Now if the eigenvalues of $A$ are $\{\lambda_i\}$ (with repetition for multiplicity), then the characteristic polynomial evaluated for $A$ is $$P(A) = (A - \lambda_1I)(A - \lambda_2I) \ldots (A - \lambda_nI) = 0$$ Therefore, the column vectors of $(A - \lambda_2I)(A - \lambda_3I) \ldots (A - \lambda_nI)$ lie in the null space of $A - \lambda_1I$. In particular, the non-zero column vectors are eigenvectors associated with $\lambda_1$.

In your case, letting $\varphi = \frac{1+\sqrt{5}}2$, and noting that the other eigenvalue is $1 - \varphi$, then $B - \varphi I$ contains eigenvectors of $1 - \varphi$ and $B - (1-\varphi)I$ contains eigenvectors of $\varphi$.

  • $\begingroup$ Sorry about that, I made mistake when I wrote the matrix, I've edited it now though. Is my answer so far now correct for the new matrix? $\endgroup$ – Si.0788 Mar 2 '16 at 18:04
  • $\begingroup$ @Si.0788 - No. You only corrected the top $B$. Your calculations have the same error in them. $\endgroup$ – Paul Sinclair Mar 2 '16 at 18:11

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