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.

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    $\begingroup$ Have you tried considering $ A^TA $? $\endgroup$
    – zo0x
    May 27 '20 at 9:44
  • $\begingroup$ If $N\neq M$, $A$ is a non-square matrix, so it does not have a determinant $\endgroup$
    – Didier
    May 27 '20 at 9:44
  • $\begingroup$ Show that Nullspace of A = Nullspace of $A^TA$. Then use rank nullity theorem. $\endgroup$
    – Koro
    May 27 '20 at 9:46
  • $\begingroup$ 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 $\endgroup$ 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 $A$.

  • $\begingroup$ Why from $L$ is a left inverse of $A$ we have that $rk A=M$? $\endgroup$ May 27 '20 at 10:12
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    $\begingroup$ @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. $\endgroup$
    – egreg
    May 27 '20 at 10:14

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