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Show that if $A \in \mathbb{C}^{m \times n}$ is of full rank, then null$(A^*)$ is the orthogonal complement of range$(A)$.

I saw the above fact listed in this linked MSE proof.

My attempt to prove it:

proof

Assume $m \geq n$.

First, we can show that for any vector in null$(A)$ that that vector is orthogonal to any vector in range$(A)$. So let $y \in \text{null}(A)$. Then

$$y^*Ax = (A^*y)^*x = 0^*x = 0$$

Now here's where I'm a little lost. So the following is not proof but a description of what I think needs to happen.

Since $A \in \mathbb{C}^{m \times n}$ and $A$ has full rank then the range of $A$, $Ax$ can be written

$$x_1A_1 + \cdots+ x_nA_n$$

where $A_i$ denotes a column of $A$. That is, the span$(A) = \mathbb{C}^n$ but range$(A) \subset \mathbb{C}^m$ so I think we need to show that span$(A^*) = m-n$.

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You don't need the hypothesis.

Assume $z \perp \mathrm{range}(A)$. Consider any $x \in \mathbb{C}^n$. Since $\langle z,Ax \rangle_{\mathbb{C}^m} = 0$ you have $\langle A^*z,x \rangle_{\mathbb{C}^n}=0$. Setting $x=e_i,i=1,\dots,n$, you find $A^*z=0$, i.e. $z$ is in the null space of $A^*$. Thus the orthogonal complement of the range of $A$ is contained in the null space of $A^*$.

To finish we do the argument in reverse: assume $A^*z=0$, then for any $x \in \mathbb{C}^n$ you have $\langle A^*z,x \rangle_{\mathbb{C}^n}=0$. Equivalently $\langle z,Ax \rangle_{\mathbb{C}^m}=0$. Since $x$ was arbitrary you conclude the reverse containment you want.

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  • $\begingroup$ Why "iterate through a basis"? If this holds for an arbitrary $x\in\Bbb C^n$, you're done. Better just to start with $x$ rather than with $y$. $\endgroup$ – Ted Shifrin May 29 '18 at 17:44
  • $\begingroup$ @TedShifrin Just depends how you prefer to do the proof. A "mechanical" way to do the proof is to write the system $\langle A^*z,x_i \rangle_{\mathbb{C}^n}=0$ for $x_i$ ranging over a basis of $\mathbb{C}^n$ and use the invertibility of the associated coefficient matrix to solve the system. As for starting with $x$ vs. starting with $y$, it's a matter of style. $\endgroup$ – Ian May 29 '18 at 17:49
  • $\begingroup$ @TedShifrin Since this is a relatively basic result (indeed this is pretty much the first real "result" in Strang's book), I figured that being a bit more explicit than I would usually be would be helpful to the OP. $\endgroup$ – Ian May 29 '18 at 18:00

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