Show that the following matrix is nonsingular

$$z$$ 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.

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

• What does $𝑉^⊥$ mean?
– Ben
Nov 6 '19 at 11:26
• @Ben It means the orthogonal subspace to $V$, the set of all vectors whose scalar product with any element of $V$ is zero. Nov 6 '19 at 11:27
• Why do we have, for any subspace 𝑉 of $ℝ_n$, we have a decomposition? What theorem does this rely on?
– Ben
Nov 6 '19 at 11:45
• @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. Nov 6 '19 at 11:59
• All right. Thanks.
– Ben
Nov 6 '19 at 12:01