[proofplan]
We compute the covariance matrix of the linearly transformed random vector $Y=\Gamma^\top X$ directly from the definition of covariance. Centering $X$ by its mean shows that the centered version of $Y$ is obtained by applying the same [linear map](/page/Linear%20Map) $\Gamma^\top$. The covariance matrix therefore transforms as $\Gamma^\top \Sigma \Gamma$, and substituting the spectral decomposition $\Sigma=\Gamma\Lambda\Gamma^\top$ gives $\Lambda$ because $\Gamma$ is orthogonal.
[/proofplan]
[step:Center $X$ and express the centered principal components]
Let $\mu \in \mathbb{R}^p$ denote the mean vector of $X$, so
\begin{align*}
\mu = \mathbb{E}[X].
\end{align*}
Define the centered random vector $X_0: (\Omega,\mathcal{F}) \to \mathbb{R}^p$ by
\begin{align*}
X_0(\omega) = X(\omega)-\mu.
\end{align*}
Since $Y=\Gamma^\top X$ and $\Gamma^\top$ is a deterministic linear map, the mean vector of $Y$ is
\begin{align*}
\mathbb{E}[Y] = \mathbb{E}[\Gamma^\top X] = \Gamma^\top \mathbb{E}[X] = \Gamma^\top \mu.
\end{align*}
Define the centered principal component vector $Y_0: (\Omega,\mathcal{F}) \to \mathbb{R}^p$ by
\begin{align*}
Y_0(\omega) = Y(\omega)-\mathbb{E}[Y].
\end{align*}
Then, for every $\omega \in \Omega$,
\begin{align*}
Y_0(\omega)
&= \Gamma^\top X(\omega)-\Gamma^\top \mu \\
&= \Gamma^\top \bigl(X(\omega)-\mu\bigr) \\
&= \Gamma^\top X_0(\omega).
\end{align*}
Thus $Y_0=\Gamma^\top X_0$.
[guided]
The covariance matrix is defined using centered variables, so the first task is to identify the centered version of $Y$. Let $\mu \in \mathbb{R}^p$ be the mean vector of $X$:
\begin{align*}
\mu = \mathbb{E}[X].
\end{align*}
Define the centered random vector $X_0: (\Omega,\mathcal{F}) \to \mathbb{R}^p$ by
\begin{align*}
X_0(\omega)=X(\omega)-\mu.
\end{align*}
Because $\Gamma^\top$ is a deterministic matrix, expectation passes through this linear transformation:
\begin{align*}
\mathbb{E}[Y]
&= \mathbb{E}[\Gamma^\top X] \\
&= \Gamma^\top \mathbb{E}[X] \\
&= \Gamma^\top \mu.
\end{align*}
Now define the centered principal component vector $Y_0: (\Omega,\mathcal{F}) \to \mathbb{R}^p$ by
\begin{align*}
Y_0(\omega)=Y(\omega)-\mathbb{E}[Y].
\end{align*}
Substituting $Y=\Gamma^\top X$ and $\mathbb{E}[Y]=\Gamma^\top\mu$ gives
\begin{align*}
Y_0(\omega)
&= \Gamma^\top X(\omega)-\Gamma^\top \mu \\
&= \Gamma^\top\bigl(X(\omega)-\mu\bigr) \\
&= \Gamma^\top X_0(\omega).
\end{align*}
So centering commutes with the deterministic linear transformation: $Y_0=\Gamma^\top X_0$.
[/guided]
[/step]
[step:Compute the covariance matrix after the orthogonal change of coordinates]
By definition of covariance for a random vector with finite second moments,
\begin{align*}
\Sigma = \operatorname{Cov}(X)=\mathbb{E}[X_0X_0^\top].
\end{align*}
Using $Y_0=\Gamma^\top X_0$, we obtain
\begin{align*}
\operatorname{Cov}(Y)
&= \mathbb{E}[Y_0Y_0^\top] \\
&= \mathbb{E}\bigl[(\Gamma^\top X_0)(\Gamma^\top X_0)^\top\bigr] \\
&= \mathbb{E}\bigl[\Gamma^\top X_0X_0^\top \Gamma\bigr] \\
&= \Gamma^\top \mathbb{E}[X_0X_0^\top]\Gamma \\
&= \Gamma^\top \Sigma \Gamma.
\end{align*}
Substituting $\Sigma=\Gamma\Lambda\Gamma^\top$ gives
\begin{align*}
\operatorname{Cov}(Y)
&= \Gamma^\top(\Gamma\Lambda\Gamma^\top)\Gamma \\
&= (\Gamma^\top\Gamma)\Lambda(\Gamma^\top\Gamma).
\end{align*}
Since $\Gamma$ is orthogonal, $\Gamma^\top\Gamma=I_p$, where $I_p$ is the $p \times p$ identity matrix. Hence
\begin{align*}
\operatorname{Cov}(Y)=I_p\Lambda I_p=\Lambda.
\end{align*}
[guided]
The covariance matrix of $Y$ is the expectation of the outer product of its centered version with itself. Since $Y_0=\Gamma^\top X_0$, we compute
\begin{align*}
\operatorname{Cov}(Y)
&= \mathbb{E}[Y_0Y_0^\top] \\
&= \mathbb{E}\bigl[(\Gamma^\top X_0)(\Gamma^\top X_0)^\top\bigr].
\end{align*}
The transpose of the product $\Gamma^\top X_0$ is $X_0^\top\Gamma$, so
\begin{align*}
(\Gamma^\top X_0)(\Gamma^\top X_0)^\top
= \Gamma^\top X_0X_0^\top\Gamma.
\end{align*}
The matrices $\Gamma^\top$ and $\Gamma$ are deterministic, so they factor outside expectation:
\begin{align*}
\operatorname{Cov}(Y)
&= \mathbb{E}\bigl[\Gamma^\top X_0X_0^\top\Gamma\bigr] \\
&= \Gamma^\top\mathbb{E}[X_0X_0^\top]\Gamma.
\end{align*}
By the definition of $\Sigma$ as the covariance matrix of $X$,
\begin{align*}
\mathbb{E}[X_0X_0^\top]=\Sigma.
\end{align*}
Therefore
\begin{align*}
\operatorname{Cov}(Y)=\Gamma^\top\Sigma\Gamma.
\end{align*}
Now substitute the assumed decomposition $\Sigma=\Gamma\Lambda\Gamma^\top$:
\begin{align*}
\operatorname{Cov}(Y)
&= \Gamma^\top(\Gamma\Lambda\Gamma^\top)\Gamma \\
&= (\Gamma^\top\Gamma)\Lambda(\Gamma^\top\Gamma).
\end{align*}
The point of choosing the principal component coordinates is exactly that $\Gamma$ is orthogonal. Orthogonality means
\begin{align*}
\Gamma^\top\Gamma=I_p,
\end{align*}
where $I_p$ is the $p \times p$ identity matrix. Hence
\begin{align*}
\operatorname{Cov}(Y)
= I_p\Lambda I_p
= \Lambda.
\end{align*}
[/guided]
[/step]
[step:Read the variances and covariances from the diagonal matrix]
For every $j,k \in \{1,\dots,p\}$, the $(j,k)$ entry of $\operatorname{Cov}(Y)$ is $\operatorname{Cov}(Y_j,Y_k)$. Since $\operatorname{Cov}(Y)=\Lambda=\operatorname{diag}(\lambda_1,\dots,\lambda_p)$, its diagonal entries satisfy
\begin{align*}
\operatorname{Var}(Y_k)=\operatorname{Cov}(Y_k,Y_k)=\lambda_k
\end{align*}
for every $k \in \{1,\dots,p\}$, and its off-diagonal entries satisfy
\begin{align*}
\operatorname{Cov}(Y_j,Y_k)=0
\end{align*}
whenever $j \ne k$. This proves the claimed variance and uncorrelatedness properties of the population principal components.
[/step]
