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Let $A$ be a positive semidefinite (P.S.D) matrix with distinct set of eigenvalues. since it is P.S.D its eigendecomposition is as follows for eigenpairs of $(\lambda_i,v_i)$

$$ A= \begin{bmatrix} v_1 & v_2 & \cdots v_n \end{bmatrix} \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 0 & \cdots \\ \vdots & & \ddots \\ 0 & \cdots & 0 & \lambda_n \end{bmatrix} \begin{bmatrix} v_1^T \\ v_2^T \\ \vdots \\ v_n^T \end{bmatrix} $$

where $v_i^Tv_i=1$ and $v_i^Tv_j =0$ for $i \neq j$. $A$ can be written as the summation as the following

$$ A= \begin{bmatrix} \lambda_1v_1 & \lambda_2v_2 & \cdots \lambda_nv_n \end{bmatrix} \begin{bmatrix} v_1^T \\ v_2^T \\ \vdots \\ v_n^T \end{bmatrix} = \sum_{i=1}^{n} \lambda_iv_iv_i^T $$

If $A=I$, is the above holds for any set of orthonormal vectors $\{u_i\}_{i=1}^n$? If so, could you show it?

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Let you have a set of orthonormal vectors $\{u_i\}_{i=1}^n$ building $U$ as follows

$$ U= \begin{bmatrix} u_1 & u_2 & \cdots & u_n \end{bmatrix} $$ Hence, $U$ is a unitary matrix, i.e. $U^TU=I_n$ and $UU^T=I_n$, so by multiplying

$$ UU^TU=UI_n \rightarrow I_nUU^T=UI_nU^T \rightarrow I_n=UI_nU^T=\sum_{i=1}^{n} u_iu_i^T $$

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It follows that $I=\sum_{i=1}^{n}v_iv_i^T$ for any orthonormal basis $v_i$, i.e. any complete orthonormal set. The argument is the same, it makes no difference whether the eigenvalues $\lambda_i=1$ are distinct or not. What matters is that $v_i$ form an orthonormal eigenbasis. For $A=I$ every vector is an eigenvector, so any basis is an eigenbasis. Applying the sum to any vector $x$ we have $$ \big(\sum_{i=1}^{n}v_iv_i^T\big)x=\sum_{i=1}^{n}(v_i,x)v_i=x, $$ because $v_i$ are an orthonormal basis. Which means that the sum acts the same way as $I$.

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