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Given a $3 \times 3$ matrix $A$, we would like to left-multiply or right-multiply unitary matrices to introduce zero elements in specific forms as the following,

$\left( {\begin{array}{*{20}{c}} {\text{x}}&{\text{x}}&0 \\ {\text{x}}&0&{\text{x}} \\ 0&{\text{x}}&{\text{x}} \end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}} {\text{x}}&{\text{x}}&0 \\ {\text{0}}&0&{\text{x}} \\ 0&{\text{0}}&{\text{x}} \end{array}} \right)$

where $\text x$ represents a non-zero element.

My attempt:

For the first form,

First apply a householder reflector for the 1st column of the 2nd and 3rd rows but keeps the 1st row unchanged,then we have $\left( {\begin{array}{*{20}{c}} {\text{x}}&{\text{x}}&{\text{x}} \\ {\text{x}}&{\text{x}}&{\text{x}} \\ {\text{0}}&{\text{x}}&{\text{x}} \end{array}} \right)$, then similarly householder on the last column of the first two rows but keeps the 3rd row unchanged, and we have $\left( {\begin{array}{*{20}{c}} {\text{x}}&{\text{x}}&0 \\ {\text{x}}&{\text{x}}&{\text{x}} \\ {\text{0}}&{\text{x}}&{\text{x}} \end{array}} \right)$. I cannot figure out how to make the center element zero while preserving the zeros introduced earlier.

For the second form,

First apply householder reflectors to make it upper triangular $\left( {\begin{array}{*{20}{c}} {\text{x}}&{\text{x}}&{\text{x}} \\ {\text{0}}&{\text{x}}&{\text{x}} \\ {\text{0}}&{\text{0}}&{\text{x}} \end{array}} \right)$, then apply householder on the first row of the last two columns to have $\left( {\begin{array}{*{20}{c}} {\text{x}}&{\text{x}}&{\text{0}} \\ {\text{0}}&{\text{x}}&{\text{x}} \\ {\text{0}}&{\text{x}}&{\text{x}} \end{array}} \right)$, then householder again to make it $\left( {\begin{array}{*{20}{c}} {\text{x}}&{\text{x}}&{\text{0}} \\ {\text{0}}&{\text{x}}&{\text{x}} \\ {\text{0}}&{\text{0}}&{\text{x}} \end{array}} \right)$. I have the same problem to make the center element zero.

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For the second matrix

Given a matrix $A$ with full rank (invertible). We know that unitary matrices are also of full rank, and multiplication with full rank matrices does not change the rank: $$\text{rank}(BA)=\text{rank}(AB)=\text{rank}(A)$$

Your second matrix scheme does not have full rank. Hence it is impossible to get this scheme.

For the first matrix

This took me a while to figure out. Contrary to my previous thought it is in fact possible to do this transformation.

First perform a Rotation or HH-Transformation or whatever to get rid of the entry at the bottom left, just like you did: $$A⇒Q^1A=\left( {\begin{array}{ccc} {\text{x}}&{\text{x}}&{\text{x}} \\ {\text{x}}&{\text{x}}&{\text{x}} \\ {\text{0}}&{\text{x}}&{\text{x}} \end{array}} \right) = \left( {\begin{array}{c|cc} {\text{x}}&{\text{x}}&{\text{x}} \\ {\text{x}}&{\text{x}}&{\text{x}} \\ \hline {\text{0}}&{\text{x}}&{\text{x}} \end{array}} \right)=: \left( {\begin{array}{c|c} A' & B' \\ \hline 0 & D' \end{array}} \right)$$ Then we write that matrix in block structure: $A'∈ℝ^{2×1}, B'∈ℝ^{2×2}, 0∈ℝ^{1×1}, D'∈ℝ^{1×2}$.

Now we perform a SVD on $B'$, which yields: $$B'=UΣV^\top$$ with $U,V∈ℝ^{2×2}$ unitary and $Σ=\text{diag}(σ_1,σ_2)$. This is equivalent to: $$U^\top B'V=Σ.$$ And Σ has the shape we want to achieve in the top right part. So now we need to make $U^\top$ and $V$ to $3×3$-matrices and we are done:

$$\tilde{U}=\left( {\begin{array}{c|c} U^\top & 0_{2×1} \\ \hline 0_{1×2} & 1_{1×1} \end{array}} \right) \qquad \tilde{V}=\left( {\begin{array}{c|c} 1_{1×1} & 0_{1×2} \\ \hline 0_{2×1} & V \end{array}} \right) $$ (These matrices are unitary as well.)
Using those matrices, it holds $$\tilde{U}Q^1A\tilde{V} = \left( {\begin{array}{*{20}{c}} {\text{x}}&σ_1&0 \\ {\text{x}}&0&σ_2 \\ 0&{\text{x}}&{\text{x}} \end{array}} \right) $$

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