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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?

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2 Answers 2

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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)\ $.

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  • $\begingroup$ Yes, you are right. The equation ${D^{*}D=I}$ does not imply unitary matrix. Thank you $\endgroup$ Commented Dec 31, 2019 at 13:46
  • $\begingroup$ One thing I don't understand is that why D must be 3x2 matrix? Why not 3x3, 3x4 or 3x5? $\endgroup$ Commented Dec 31, 2019 at 16:25
  • $\begingroup$ If $\ D^*D=I_{n\times n}\ $, then $\ D\ $ must have $\ n\ $ columns, and they must be linearly independent (otherwise there would be an $\ n\times 1\ $ non-zero column vector vector $\ x\ $ such that $\ 0_{n\times1}=$$D^*Dx=$$I_{n\times n}x\ $, which is obviously impossible). That is, the dimension of the column space of $\ D\ $ must be $\ n\ $. But if $\ \operatorname{Im}(D)=\operatorname{Im}\left(D_0\right)$, then the column space of $\ D\ $ must have the same dimension as the column space of $\ D_0\ $, namely $2$. So $\ n\ $ must be $2$. $\endgroup$ Commented Dec 31, 2019 at 22:00
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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).

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  • $\begingroup$ Yes, you are right. a 3×3 matrix with 2D image space will have determinant = 0, which means not invertible. Thank you. $\endgroup$ Commented Dec 31, 2019 at 13:42

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