$z$ is a vector in $R^n$. Given the following functions:

$c_i(z): R^n\to R$

$c(z) = (c_1(z),\dots,c_m(z))^T$

$A(z)^T = [\nabla c_i (z)]_{i\in [m]}$, where $[m] = \{1,2,\dots,m\}$

And $A(z)^T$ has full column rank.

We construct a matrix $Z$, whose columns are a basis of $Nul\ A(z)$.

Please prove $$ \left[\begin{array}{c}{A\left(z\right)} \\\\{Z^{T}}\end{array}\right] $$ is invertible (nonsingular).

Let me know if you have any puzzles about the problem itself.

  • $\begingroup$ Have you ever seen a proof of the rank-nullity theorem? $\endgroup$ Nov 6 '19 at 10:36
  • $\begingroup$ Yes, I have looked up the proof. $\endgroup$
    – Ben
    Nov 6 '19 at 10:38
  • $\begingroup$ Well then :-) I'll let you answer your own question and collect all the upvotes. $\endgroup$ Nov 6 '19 at 10:40
  • $\begingroup$ But what I get is only $rank(A(z)) = m$, $rank(Z) = n-m$ $\endgroup$
    – Ben
    Nov 6 '19 at 10:43
  • $\begingroup$ This is why I asked you to look at the proof, not at the theorem. $\endgroup$ Nov 6 '19 at 10:45

For any subspace $V$ of $\Bbb R^n$, you have a decomposition $$\Bbb R^n=V\oplus V^\perp$$ In the present case, consider $V$ to be the row space of $A(z)$. Then by definition $V^\perp=\operatorname{Ker}A(z)$.

By construction, the rows of $A(z)$ are a basis of $V$ and the rows of $Z^T$ are a basis of $V^\perp$. Since $\Bbb R^n=V\oplus V^\perp$, it follows that the rows of the full matrix form a basis of $\Bbb R^n$, so that the matrix is non-singular.

  • $\begingroup$ What does $𝑉^⊥$ mean? $\endgroup$
    – Ben
    Nov 6 '19 at 11:26
  • $\begingroup$ @Ben It means the orthogonal subspace to $V$, the set of all vectors whose scalar product with any element of $V$ is zero. $\endgroup$ Nov 6 '19 at 11:27
  • $\begingroup$ Why do we have, for any subspace 𝑉 of $ℝ_n$, we have a decomposition? What theorem does this rely on? $\endgroup$
    – Ben
    Nov 6 '19 at 11:45
  • $\begingroup$ @Ben It is well-known, you will find it in any textbook. You first show that by Gaussian elimination the dimension of $V^\perp$ is $n-\dim V$, and then you show separately that $V$ and $V^\perp$ are in direct sum. $\endgroup$ Nov 6 '19 at 11:59
  • $\begingroup$ All right. Thanks. $\endgroup$
    – Ben
    Nov 6 '19 at 12:01

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