# Prove that $\det(B^TB) \neq 0$

I have a matrix $$A_{N \times M}$$ such that $$A=U^T_{N \times N} \cdot B_{N\times M} \cdot V_{M \times M},$$ where $$U,V$$ orthogonal and $$B_{ij}$$ may has nonzero values only for $$i\le j \le i+1$$ and $$A$$ is full order matrix. How to prove that $$B^TB$$ also has $$\det(B^TB) \neq 0$$.

I think that it is easy conclusion from the fact that $$A$$ has full order matrix but I don't know how exactly to prove it.

• Have you tried considering $A^TA$?
– zo0x
May 27 '20 at 9:44
• If $N\neq M$, $A$ is a non-square matrix, so it does not have a determinant May 27 '20 at 9:44
• Show that Nullspace of A = Nullspace of $A^TA$. Then use rank nullity theorem.
– Koro
May 27 '20 at 9:46
• Sorry, my mistake, I wanted to write faster that $A$ is a full order, and I forgot that for $A$ the determinant does not exist, I edited the post May 27 '20 at 9:50

I guess that “full order” means that $$A$$ has full column rank, that is, with your notation, $$\operatorname{rk}A=M$$.
This is equivalent to $$\det(A^T\!A)\ne0$$. Indeed, if $$\det(A^T\!A)\ne0$$, you can consider $$L=(A^T\!A)^{-1}A^T$$ and immediately see that $$L$$ is a left inverse of $$A$$. Conversely, it's not difficult to prove that $$A$$ and $$A^T\!A$$ have the same rank.
Multiplying a matrix on the left or on the right by an invertible matrix doesn't change the rank. Since $$B=V^TAU$$ the rank of $$B$$ is the same as the rank of $$A$$.
• Why from $L$ is a left inverse of $A$ we have that $rk A=M$? May 27 '20 at 10:12
• @anatolij3253 Suppose $LA=I_M$; since the rank of a product is at most equal to the rank of the factors, you conclude that $\operatorname{rk} A\ge \operatorname{rk} I_M=M$. But $\operatorname{rk}A\le M$ is true for every matrix with $M$ columns. May 27 '20 at 10:14