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What is the maximum eigenvalue of $mathbf A$ where $mathbf A =mathbf{Q} + mathbf{Q}^{T}$ and $mathbf{Q} = uv^{T}$ ?

$u$ and $v$ are two unit vectors. I know the maximum eigenvalue of $textbf{Q}$ is $u^{T}v$. However, I cannot deduce a relation of maximum eigen value of sum of two outer product matrices.

Is there any relation between Positive semidefinite matrices, max. eigen value and sum?

We know that $A$ has rank at most equal to $2$. Let $lambda_1,lambda_2$ denote the (possibly) non-zero eigenvalues of $A$. These eigenvalues satisfy
$$
begin{align*}
lambda_1 + lambda_2 &= operatorname{tr}(A) = 2(u^Tv)\
lambda_1^2 + lambda_2^2 &= operatorname{tr}(A^2) = operatorname{tr}((u^Tv)(uv^T + vu^T) + (v^Tv)uu^T + (u^Tu)vv^T)\
&= 2[(u^Tv)^2 + (u^Tu)(v^Tv)]
end{align*}
$$
We can solve these equations for $lambda_1$ and $lambda_2$.

In particular, we note that
$$
lambda_1 lambda_2 = frac 12 [(lambda_1 + lambda_2)^2 - (lambda_1^2 + lambda_2^2)] = (u^Tv)^2 - (u^Tu)(v^Tv)
$$
which means that the $lambda_i$ are the two solutions to the quadratic equation
$$
lambda^2 - [2(u^Tv)]lambda + [(u^Tv)^2 - (u^Tu)(v^Tv)] = 0
$$
which means that the larger eigenvalue will be given by
$$
lambda = (u^Tv) + sqrt{(u^Tv)^2 - [(u^Tv)^2 - (u^Tu)(v^Tv)]} =
u^Tv + sqrt{u^Tu}sqrt{v^Tv} = u^Tv + 1
$$

Another approach:

Write $v = au + bv_perp$, where $u_perp$ is a unit vector orthogonal to $u$. Notably, $a = u^Tv$, and $a^2 + b^2 = 1$. We have
$$
A = uv^T + v^Tu =
2auu^T + buu_perp^T + bu_perp u^T
$$
That is, if $U$ is the matrix whose columns are $u,u_perp$, then we have $A = UMU^T$, where
$$
M = pmatrix{2a&b\b&0}
$$
Let $W$ be such that $[U W]$ is an orthogonal matrix. We then have
$$
A = pmatrix{U & W}pmatrix{M & 0\0 & 0} pmatrix{U & W}^T
$$
That is, $A$ is similar to the block-diagonal matrix $operatorname{diag}(M,0)$. Conclude that the eigenvalues of $M$ are the non-zero eigenvalues of $A$.

The characteristic polynomial of $M$ is the equation on $lambda$ given above.

Generally, $Q+Q^T$'s maximum eigenvalue is $langle u,vrangle + lVert urVertlVert vrVert$. We can see this in another (perhaps geometric) way. Note that any symmetric matrix $A$ has maximum eigenvalue equal to
$$
lambda_{max}(A) = max_{lVert xrVert = 1} x^TAx.
$$ In this case, we have $$x^T(Q + Q^T)x = x^T(uv^T+vu^T)x = 2langle x,uranglelangle x,vrangle$$ where $langle a,brangle = a^Tb$ denotes the standard inner product. Suppose $u,v$ are unit vectors with $$cos(theta(u,v)) = langle u,vrangle$$ where $theta(u,v)$ is an angle between $u$ and $v$. Assume $theta(u,x) = phi$. Then we have $theta(v,x)geq theta(u,v)-phi, $ and hence that
$$
2langle x,uranglelangle x,vrangleleq 2cos (phi)cdotcos(theta(u,v)-phi) = cos(theta(u,v)) + cos(theta(u,v)-2phi)leq langle u,vrangle +1,
$$ where equality holds when $phi = frac{theta(u,v)}{2}$. This proves the result. For the general case, let $u'=frac{1}{lVert urVert}u$ and $v'=frac{1}{lVert vrVert}v$ to get $$lVert urVertlVert vrVert(langle u',v'rangle+1)=langle u,vrangle + lVert urVertlVert vrVert.$$