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.


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 $$\tilde{U}Q^1A\tilde{V} = \left( {\begin{array}{*{20}{c}} {\text{x}}&σ_1&0 \\ {\text{x}}&0&σ_2 \\ 0&{\text{x}}&{\text{x}} \end{array}} \right) $$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.