# solution involving inverse of a rank-1 matrix

I am looking for $\mathbf{y} \in \mathbb{R}^n$ that minimizes the following objective function that involves a real matrix $\mathbf{V} \in \mathbb{R}^{n\times n}$

$$\tag{*} \begin{array}{c} \text{min} \hspace{4mm} \mathbf{y}^T \mathbf{V}\mathbf{y} \end{array}$$

\begin{align*} \mathbf{y} &= \begin{bmatrix} y_{1} & y_{2} &\cdots & y_{n} \end{bmatrix}^T \end{align*}

I have a single equality constraint i-e $y_1=1$. To accommodate this constraint, I rewrite my objective function as

$$\tag{**} \begin{array}{c} \text{min} \hspace{4mm} \mathbf{y}^T \mathbf{V} \mathbf{y} - 2\lambda(\mathbf{u}^T\mathbf{y}-1) \end{array}$$

where $\mathbf{u}\in \mathbb{R}^n$ and is defined as

\begin{align*} \mathbf{u} &= \begin{bmatrix} 1 & 0 &\cdots & 0 \end{bmatrix}^T \end{align*}

Diffrentiating (**) wrt $\mathbf{y}$ and $\lambda$ gives me the solution

\begin{align*}\tag{***} \mathbf{y} &= \lambda \mathbf{V}^{-1}\mathbf{u} \\ \lambda &= \frac{1}{\mathbf{u}^T\mathbf{V}^{-1}\mathbf{u}} \end{align*}

My problem is that in my setup $\mathbf{V}$ is a $rank 1$ matrix. Therefore $\mathbf{V}^{-1}$ does not make sense. Is there a way to fix this problem?

Of course, since $\mathbf V$ is not invertible one cannot use $\mathbf V^{-1}$. However, already the differentiation seems a problem. I've made some quick computation to get
$$(\mathbf V^T+\mathbf V)\mathbf y=\lambda \mathbf u=[\lambda,0,\dots,0].$$ Then $\lambda$ is the first component of the vect0r in the left: $$\lambda=\sum_i(V_{i1}+V_{1i})y_i.$$ The other equations give the following homogeneous system to obtain $\mathbf y$ $$0=\sum_i(V_{ij}+V_{ji})y_i\quad(1\le j<n).$$ Now if your matrix has rank one, this system is very special. Any rank 1 matrix is of the form $\mathbf V=\mathbf v\mathbf w^T$ for some non-zero vectors $\mathbf v,\mathbf w\in\mathbb R^n$. Then the system simplifies to $$\mathbf w\langle \mathbf v,\mathbf y\rangle+\mathbf v\langle \mathbf w,\mathbf y\rangle=\lambda\mathbf u.$$ To understand this system is better to consider the unknowns are the scalar products $\langle \mathbf v,\mathbf y\rangle$ and $\langle \mathbf w,\mathbf y\rangle$ not just the $\mathbf y$.
For instance, if both scalar products are zero, you're done with $\lambda=0$. And this is something, because in dimension $>2$ the orthogonal complemente of two vectors has dimension $\ge1$.
Another possible consideration: if $\mathbf v,\mathbf w$ are not independent, say $\mathbf w=\rho \mathbf v$, $\rho\ne0$, then you get $$\rho\mathbf v\langle \mathbf v,\mathbf y\rangle+\mathbf v\langle \rho\mathbf v,\mathbf y\rangle=\lambda\mathbf u,$$ hence $$2\rho\mathbf v\langle \mathbf v,\mathbf y\rangle=\lambda\mathbf u.$$ If $\langle \mathbf v,\mathbf y\rangle\ne0$, then looking at the equations $i=2,\dots,n$ we see that $2\rho v_i\langle \mathbf v,\mathbf y\rangle=0$, hence $v_i=0$. Thus $\mathbf v=v_1\mathbf u$ and the equation $i=1$ reads $\lambda=2\rho v_1^2y_1$.
Next suppose $\mathbf v,\mathbf w$ are independent. This can come from the coordinates $i=2,\dots,n$, but then the system would have the solution $\langle \mathbf v,\mathbf y\rangle=\langle \mathbf w,\mathbf y\rangle=0$, already discussed.
The only case remaining is that the vectors are independent by the first coordinate en some other $i=2,\dots,n$, not any else. Thus the system reduces to two independent equations whose unknowns are the two scalar products, and it can be solved for any $\lambda$ you choose.