Rank of a matrix product of two vectors Let $\lambda=(\lambda_1, \lambda_2,...,\lambda_n)\in \Bbb R^n$ a vector no zero. Let $A=(a_{ij})$  the matrix of $n\times n$ so that $a_{ij}=\lambda_i \lambda_j$.  Determine the rank of the matrix $A$.
The matrix $A$ is $$ \left(
      \begin{array}{ccccc}
        \lambda_1^2 & \lambda_1\lambda_2 & \cdots & \lambda_1\lambda_n  \\
        \lambda_2\lambda_1 & \lambda_2^2 &\cdots  & \lambda_2\lambda_n  \\
         \vdots& \vdots & \ddots & \vdots  \\
        \lambda_n\lambda_1 & \lambda_n\lambda_2& \cdots & \lambda_n^2  \\
      \end{array}
    \right)$$
 A: Hint. Every row of $A$ is a scalar multiple of the row vector $\lambda$, and $A\ne0$. So, what is the row rank of $A$? Now $A$ is a square matrix. Therefore its rank is equal to its row rank.
A: Fact: $\text{Rank}(AB) \le \text{min}\{\text{Rank}(A),\text{Rank}(B)\}$
And your matrix is
$$\left[ \begin{matrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{matrix}\right]\left[ \begin{matrix}\lambda_1 & \lambda_2 & \cdots & \lambda_n\end{matrix}\right]$$ 
A: The rank is one because you have in the rows scalar multiples of the original $\lambda$.
A: You may alternatively exploit that the rank of a matrix is preserved under a change of a basis, which is described by similarity transformations.
Choose an orthogonal transformation $S$, thus $S\circ S^T=\mathbb 1$, such that
$$S\big(\underbrace{(1,\dots,0)}_{\quad =\,e_1}\big) \:=\: \frac 1{\|\lambda\|}(\lambda_1, \lambda_2,\dots,\lambda_n)\,.$$
Now consider $A$ and the matrix products
$$A\:=\:\big(\lambda_1,\dots,\lambda_n\big)
\begin{pmatrix}\lambda_1\\ \vdots\\ \lambda_n\end{pmatrix}
 \,=\, \|\lambda\|\,Se_1\,\big(\|\lambda\|\,Se_1\big)^T
 \,=\, \|\lambda\|^2\,S\:e_1e_1^T\,S^T \\[3ex]
 \,=\, \|\lambda\|^2\,S\,
\left(\begin{smallmatrix} 1&0&\dots&0\\ 0&0&&\vdots\\ \vdots&&\ddots\\
0&\ldots&&0\end{smallmatrix}\right)\,S^{-1}\qquad
$$
from which $A$'s rank can be read off.
