I am just starting to learn about linear algebra. My lecturer gave me the following exercise, and said 'it's quite an interesting exercise':
$\text{Show that the identity operator $\hat{\mathbb{1}}$ can be written as a sum of outer products of an eigenbasis: $\hat{\mathbb{1}} = \sum_{i = 1}^N \lvert \mathbf{x}^i> <\mathbf{x}^i \rvert$}$
Sorry, I am not sure how to typeset the identity operator or how to do dirac notation without a latex package.
Below is my proof. It seems to simple to be right, so I am hoping to get some feeback.
$\textit{Proof.}$
We know that any linear operator $\hat{\mathbf{A}}$ can be written as a sum of the outer product of its eigenvectors with themselves, weighted by the corresponding eigenvalue: $\hat{\mathbf{A}} = \sum_{i = 1}^N \lambda_i \lvert \mathbf{x}^i \rangle \langle \mathbf{x}^i \rvert$. Using the definition of the identity operator $\hat{\mathbf{}1}\mathbf{x} = \mathbf{x}$, we see that any vector is an eigenvector for the identity operator, and each eigenvalue $\lambda_i = 1$. Thus, we can build a basis from any $N$ linearly independent vectors and they will be eigenvectors for $\hat{\mathbb{1}}$, and we can write the identiy operator like this: $\hat{\mathbb{1}} = \sum_{i = 1}^N 1 \times \lvert \mathbf{x}^i \rangle \langle \mathbf{x}^i \rvert = \sum_{i = 1}^N \lvert \mathbf{x}^i \rangle \langle \mathbf{x}^i \rvert$ as required.
EDIT: The theorem I assumed doesn't actually require the operator to be normal after all, and thus I don't need to show that the identity operator is normal, so I have removed that part of the proof.
QED