Find unitary matrix given its image.

Question

I'm trying to solve this:

Let $$D_{0}=\begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix}$$

Find a D such that $$D^{*}D=I$$$$\operatorname{Im}D_{0}=\operatorname{Im}D$$ (D* is complex conjugate transpose of D)

I tried as following:

1. Calculate $$\operatorname{Im}D_{0}=\begin{Bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} & \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix} \end{Bmatrix}$$

2. Gram-Schmidt orthogonalization process: $$\begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix}\rightarrow \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix},\qquad\begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix}\rightarrow \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$$

3. Let $$ v_{1}=\begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix},\quad v_{2}=\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix},\quad v_{3}=\begin{bmatrix} a\\ b\\ c \end{bmatrix}$$ We have $$D=\begin{bmatrix} v_{1} &v_{2} &v_{3} \\ \end{bmatrix}$$

Because D*D=I, D is an unitary matrix. So $$\langle v_{1},v_{3}\rangle =0$$ and $$\langle v_{1},v_{3}\rangle =0$$

Solve above inner product equation. I have: $$D=\begin{bmatrix} 1/\sqrt{2} &1/\sqrt{3} & 1/\sqrt{6}\\ 0&1/\sqrt{3}&-2/\sqrt{6}\\ -1/\sqrt{2}&1/\sqrt{3}&1/\sqrt{6}\ \end{bmatrix}$$ Check my result: $D^*D = I$ -> correct

But the problem is: $$\operatorname{Im}D=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\neq \operatorname{Im}D_{0}???$$ Am I wrong?

Answer 1

You cannot have a $3 \times 3$ unitary matrix (which is known to be invertible) with a 2D (dimension 2) image space, because that would mean that the kernel is 1D (by rank-nullity theorem) ; alas, a matrix with a 1D kernel is not invertible. Contradiction...

You must look for a matrix $D$ which is $3 \times 2$ (3 lines and 2 columns).

Answer 2

The equation $\ D^*D=I\ $ only implies that $\ D\ $ is unitary if it is square. As stated in your question without the requirement imposed by the statement in your heading that it be unitary, the problem can be solved if (and only if) you take the matrix $\ D\ $ to be $\ 3\times2\ $. In fact, you've already taken most of the steps to solve it. The columns of $\ D\ $ just need to be an orthonormal basis for the column space of $\ D_0\ $, and if you take those columns to be the first two columns of your attempted construction, that condition will be satisfied. That is, take $$ D=\begin{bmatrix} \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}\\ 0& \frac{1}{\sqrt{3}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}\ \end{bmatrix}\ . $$ You will then have $\ D^*D=I_{2\times2}\ $ and $\ \operatorname{Im}\left(D_0\right)= \operatorname{Im}(D)\ $.