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?
On
Generally, $Q+Q^T$'s maximum eigenvalue is $\langle u,v\rangle + \lVert u\rVert\lVert v\rVert$. We can see this in another (perhaps geometric) way. Note that any symmetric matrix $A$ has maximum eigenvalue equal to $$ \lambda_{\max}(A) = \max_{\lVert x\rVert = 1} x^TAx. $$ In this case, we have $$x^T(Q + Q^T)x = x^T(uv^T+vu^T)x = 2\langle x,u\rangle\langle x,v\rangle$$ where $\langle a,b\rangle = a^Tb$ denotes the standard inner product. Suppose $u,v$ are unit vectors with $$\cos(\theta(u,v)) = \langle u,v\rangle$$ 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 $$ 2\langle x,u\rangle\langle x,v\rangle\leq 2\cos (\phi)\cdot\cos(\theta(u,v)-\phi) = \cos(\theta(u,v)) + \cos(\theta(u,v)-2\phi)\leq \langle u,v\rangle +1, $$ where equality holds when $\phi = \frac{\theta(u,v)}{2}$. This proves the result. For the general case, let $u'=\frac{1}{\lVert u\rVert}u$ and $v'=\frac{1}{\lVert v\rVert}v$ to get $$\lVert u\rVert\lVert v\rVert(\langle u',v'\rangle+1)=\langle u,v\rangle + \lVert u\rVert\lVert v\rVert.$$
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.