# Looking for $H$ such that $H'DH=\sigma_1 I$ and $H'H=\sigma_2 I$, $D$ is a diagonal matrix.

Let $m \times n$ matrix be $H$. I'm looking for what a matrix $H$ can be to satisfy the following condition.

1) $H'DH=\sigma_1 I_n$ : where $D$ is $m\times m$ diagonal matrix, $I_n$ is an $n\times n$ identity matrix. $\sigma_1$ can be any real number and $H'$ is a transpose of $H$.

2) $H'H= \sigma_2 I_n$ : where $\sigma_2$ can be any real number.

Hence I'm looking for a matrix $H$ to make $H'DH$ and $H'H$ to be proportional to identity, for an arbitrary diagonal matrix $D$, if it exists. Does such a matrix $H$ exist? If not, can you suggest me a good proof that it doesn't exist anyhow?

• What is $H^\prime$? – mathcounterexamples.net Jun 24 '18 at 14:02
• $H'$ is a transpose of $H$. – user568810 Jun 24 '18 at 16:15

## 1 Answer

I believe it is easier to first show when H exists for the second requirement, and then show if/when D exists for the first requirement given that H exists. All this depends upon different cases for the value of ${\sigma _2}$ and the relative sizes of m and n, so I parse the answer along those lines.

If ${\sigma _2} = 0$, then necessarily $H = \underline {\overline {\bf{0}} }$ for any values of m and n; then D exists only if ${\sigma _1} = 0$, and in that case, it can be any diagonal matrix.

If ${\sigma _2} \ne 0$ and $m < n$, then no H which satisfies the second requirement exists. The second requirement effectively requires H to be partial orthonormal, and no partial orthonormal matrix exists when rows have been deleted (i.e., $m < n$)

If ${\sigma _2} > 0$ and $m \ge n$, then H exists; any orthonormal matrix for which the requisite number of columns has been deleted and has then scaled by $\sqrt {{\sigma _2}}$ is a candidate. However, singular value decomposition then insures that the only possible value of D is $\left( {\frac{{{\sigma _1}}}{{{\sigma _2}}}} \right)I$. Any diagonal matrix other than this particular choice cannot result in ${\sigma _2}I$. In general, H will have no rows consisting of all zeros. In special cases, H may have up to $m - n$ rows consisting of all zeros; for any such row, the associated value in D may be arbitrarily chosen.

If ${\sigma _2} < 0$ and $m \ge n$, then H will exist but it will necessarily be complex. I believe only the same scaled D then exists (but I haven’t thought carefully about how all the different possible square roots of $- I$ will play through the first requirement).

Note that in the first and third cases above, the arbitrary portions of D need not be diagonal.

• Thanks. I was surprised at your detailed comment. My interest is focused on the case where $\sigma_2 >0$ and $m \geq n$. I found an example that contradicts with your explanation. For instance, let's consider $m=5, n=3$. $H=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$ and $D=diag(3,3,3,5,1)$. Then $H'DH= \begin{bmatrix}1 & 0 & 0 & 0 &0 \\0 & 1 & 0 &0 &0 \\ 0 &0 &1& 0 &0 \end{bmatrix} D \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} =diag(3,3,3)$. – user568810 Jul 22 '18 at 6:42
• Could you generalize the third case in the way that explains my counter-example too? – user568810 Jul 22 '18 at 6:44
• I realized that I mischaracterized the scope of the ambiguity your counterexample points up, so if you saw my previous post (which I have deleted), please ignore it. I apologize for any confusion. The answer to the question has now been properly corrected. – John Polcari Jul 22 '18 at 17:55
• Thanks. Can I ask you more? Can you give me example where $H$ does not have any rows that have all zeros but satisfy $H'H=\sigma_2 I$ and $H'DH=\sigma_1 I$ if $D=(\sigma_1/\sigma_2)I$? – user568810 Jul 24 '18 at 0:30
• $$H = \left[ {\begin{array}{*{20}{c}} { - 0.4950}&{ - 1.1200}\\ { - 0.7084}&{ - 1.0430}\\ { - 0.9281}&{ + 1.2027}\\ { - 1.0951}&{ - 0.2367}\\ { - 1.0920}&{ + 0.3983} \end{array}} \right]\quad {\sigma _2} = 4\quad \Rightarrow \quad D = 0.5{I_{\left( 5 \right)}}$$ – John Polcari Jul 24 '18 at 1:32