How can we decompose the identity matrix given a set of orthonormal vectors

Question

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?

Answer 1

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

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