# Sum of the outer products of $n$ independent vectors is always full rank? [closed]

If I take $n$ linearly independent vectors ($n \times 1$), $v_1, v_2, \ldots, v_n$ and construct a matrix which is the sum of their outer products,

$$M = \sum_i^n v_iv_i^t$$

Then, can it be proven that $M$ is always full rank?

## closed as off-topic by Namaste, José Carlos Santos, user91500, John B, J. M. is a poor mathematicianAug 28 '17 at 5:15

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• Such a matrix will be positive definite. – achille hui Aug 25 '17 at 22:57
• Can we prove this? – Rohit Pandey Aug 25 '17 at 23:01
• ^Exploit the fact that the $\{v_i\}$ form a basis. – stochasticboy321 Aug 25 '17 at 23:15

If $M = \sum v_i v_{i}^{T}$ then
where $c_i$ is a scalar denoting the inner product of $x$ and $v_i$. Since the $v_i's$ are linearly independent this sum yields the zero vector if and only if $c_i=0$ for all $i$ and hence $x$ is the zero vector. So your matrix clearly has full rank.
First, notice that $M$ being full-rank is equivalent to the induced linear operator being injective, which is equivalent to $\det M \neq 0$.
Now define $T : e_i \mapsto v_i$, where $e_i$ is the canonical basis of $\mathbb R^n$. Then $v_i^T = (T e_i)^T = e_i^T T^T$, hence $$M = \sum_i T e_i e_i^T T^T = T \left( \sum_i e_i e_i^T \right) T^T$$ But the matrix in brackets is just the identity matrix, thus $$M = T T^T \quad \Longrightarrow \quad \det M = ( \det T)^2$$ But $T$ is full rank, thus $\det T \neq 0$ and $\det M \neq 0$.
• But the existence of such a $T$ is contingent on $M$ being invertible, no? – Rohit Pandey Aug 25 '17 at 23:16