Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assuming $F$ is a matrix with full rank but more columns than rows - why is $F\cdot F^T$ invertible?

share|improve this question
    
What just happened?! –  Lord_Farin Oct 27 '12 at 21:23
    
Invertible..."things"? You mean like a dog doing a rolling trick or a capsizing ship? Really... –  DonAntonio Oct 27 '12 at 21:25
    
Counter-example in $\mathbb{Z}_p$ (see Ted's comment): $$\begin{bmatrix}1 & \cdots & 1\end{bmatrix}$$ with $p$ columns. –  wj32 Oct 27 '12 at 21:59
add comment

1 Answer

up vote 5 down vote accepted

It suffices to show the null space of $F F^T$ is $\{0\}$.

Suppose $F F^T x = 0$. Then \begin{align*} & x^T F F^T x = 0 \\ \implies & (F^T x)^T (F^T x) = 0\\ \implies & \|F^T x \|^2 = 0 \\ \implies & F^T x = 0 \\ \implies & x = 0. \end{align*}

(Because $F^T$ is a skinny matrix with full rank, its null space is $\{0\}$.)

share|improve this answer
5  
Assuming the matrix is over the real numbers, of course. (Or other field where sum of nonzero squares cannot be zero.) –  Ted Oct 27 '12 at 0:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.