Proof of $\mathbb{I}=\sum_iv_iv_i^T$, where $v_i^Tv_j=\delta_{ij}$

Question

In quantum physics I have heard this referred to as "(inserting) a complete set of states," and denoted equivalently by $$\mathbb{I}=\sum_i|v_i\rangle\langle v_i|\textrm{, where }\langle v_i|v_j\rangle=\delta_{ij}.$$

Clearly it holds in the standard basis, but I don't know how to prove this for any arbitrary set of orthonormal vectors.

Answer 1

To show that the matrix $A = \sum_{i} v_{i}v_{i}^{T}$ is the identity matrix, it is enough to show that $Aw = w$ for any vector $w$. Since $\{v_{i}\}$ is an orthonormal basis, we can write $w = \sum a_{i}v_{i}$ for some constant $a_{i}$, then $$ Aw = \sum_{i, j} a_{j} v_{i}v_{i}^{T}v_{j} = \sum_{i, j} a_{j}v_{j} \delta_{i, j} = \sum_{i} a_{i}v_{i} = w. $$

Answer 2

I thought of an alternate method to the one I accepted above. Create an orthogonal matrix $O\in\mathbb{R}^n$ whose columns are the orthonormal vectors $\{v_i\}_{i=1,...,n}$. Then we can rely on the property of orthogonal matrices that $OO^T=\mathbb{I}$.

Also note that $$OO^T\equiv \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix} \begin{bmatrix} v_1^T\\ \vdots \\v_n^T \end{bmatrix} = \sum_{i=1}^n v_iv_i^T. $$

Therefore conclude $\sum v_iv_i^T=\mathbb{I}.$

Answer 3

Here's a third way: for any vector spaces $V$ and $W$, we have a natural isomorphism $V^*\otimes W \cong {\rm Hom}(V,W)$ which takes $f\otimes w$ to $v \mapsto f(v)w$. The inverse can be described using a basis for $V$. Namely, if $(v_1,\ldots,v_n)$ is the basis and $(v_1^*,\ldots, v_n^*)$ is the dual basis, the inverse isomorphism maps $T$ to the tensor $$\sum_{i=1}^n v_i^*\otimes T(v_i). $$If $W = V$ and $T = {\rm Id}_V$, then we have ${\rm Id}_V \cong \sum_{i=1}^n v_i^*\otimes v_i$. Your exercise also sweeps another isomorphism under the rug: vectors become columns and covectors become rows, that's why you have $v_i^\top$ instead of $v_i^*$.