Doubt in the range of variance of First Principal Component This is my first problem on this forum. Here is the problem:

The covariance matrix of a four dimensional random vector $\boldsymbol X$ is of the form
$$\begin{bmatrix}1&\rho&\rho&\rho\\
\rho&1&\rho&\rho\\
\rho&\rho&1&\rho\\
\rho&\rho&\rho&1\\
\end{bmatrix}\,,$$ where $\rho<0$. If $v$ is the variance of the first principal component, then

*

*$v$ cannot exceed $5/4$.

*$v$ can exceed $5/4$, but cannot exceed $4/3$.

*$v$ can exceed $4/3$, but cannot exceed $3/2$.

*$v$ can exceed $3/2$.


In this problem, I have been given a covariance matrix of a random variable and I am required to compute the range of variance of first principal component. I am having difficulty dealing with these terms, so I have chosen a specific example where $\rho = -0.5$. So, Now, the resulting matrix is given by:
\begin{pmatrix}
1 & -0.5 & -0.5 &-0.5 \\ 
 -0.5&1  & -0.5 & -0.5\\ 
 -0.5&  -0.5&  1& -0.5\\ 
 -0.5&  -0.5&  -0.5& 1
\end{pmatrix}
If I take $\lambda = 1.5$ then this satisfies the characteristic equation ($|A - \lambda I|$ = 0) of the resulting matrix and hence it will be one of the roots of the characteristic equation. I also know that variance of principal component is nothing but characteristic roots. Now, I only have one root at this point and from there only I can see that $\lambda = \frac{3}{2}$ which does not lead to any of the options given. The correct answer is shown as (b).
 A: Let $\Sigma$ be the covariance matrix of $\boldsymbol X$. Variance of the first principal component is then the largest eigenvalue of $\Sigma$.
Now you can write $\Sigma=(1-\rho)I_4+\rho\mathbf1\mathbf1'$ where $\mathbf1$ is a column vector of all-ones, which means you can use the Matrix determinant lemma to find the characteristic polynomial $\det(\Sigma-\lambda I_4)$ and solve for the eigenvalues of $\Sigma$. Or you might take it as a standard result (see this or this) that there are two distinct eigenvalues of $\Sigma$, namely $1-\rho$ (with multiplicity $3$) and $1+3\rho$. This is perhaps easy to remember if you recall that $\det\Sigma=(1-\rho)^3(1+3\rho)$. Because $\rho$ is negative, the largest eigenvalue is $1-\rho$.
Observe that $$0\le\operatorname{Var}(\mathbf 1'\boldsymbol X)=\mathbf 1'\Sigma\mathbf 1=4(1+3\rho)$$
This yields $$\rho\ge -\frac13 \quad,\text{ i.e. },\quad 1-\rho\le \frac43$$
One can also use $\det\Sigma\ge 0$ to obtain this bound.
Combining this with $\rho<0$, we have $$1<1-\rho\le \frac43$$
There is no other restriction on $1-\rho$, so option (2) is correct.
