# Nielsen & Chuang Exercise 2.4 - “Matrix representation for identity” [closed]

Reproduced from Exercise 2.4 of Nielsen & Chuang's Quantum Computation and Quantum Information (10th Anniversary Edition):

Show that the identity operator on a vector space $$V$$ has a matrix representation which is one along the diagonal and zero everywhere else, if the matrix representation is taken with respect to the same input and output bases. This matrix is known as the identity matrix.

Note: This question is part of a series attempting to provide worked solutions to the exercises provided in the above book.

## closed as off-topic by Sanchayan DuttaJan 7 at 15:10

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On a vector space $V = \textrm{span}(\{|v_1⟩, \ldots, |v_n⟩\})$, the identity operator $\mathcal{I}$ maps each vector to itself, such that $$\mathcal{I} = \sum_i |v_i⟩⟨v_i|.$$
If $\{|v_1⟩, \ldots, |v_n⟩\}$ is chosen as the basis for the matrix representation, then we set
$$|v_1⟩ = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \quad |v_2⟩ = \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix}, \quad \cdots \quad |v_n⟩ = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix},$$
and $⟨v_1| = (|v_1⟩)^\dagger$ such that $$|v_1⟩⟨v_1| = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 & \cdots & 0 \end{pmatrix} = \textrm{diag}(1,0,\ldots,0) , \quad\textrm{etc}.$$
Summing over all terms therefore produces the desired matrix $$\mathcal{I} = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix}$$