# matrix structures obtainable by multiplication by unitary matrices

Out of "Numerical Linear Algebra" by Trefethen and Bau:

Suppose we have $3\times 3$ matrices and wish to introduce zeroes by left- and/or right-multiplications by unitary matrices $Q_j$ such as Householder reflectors or Givens rotations. Consider the following matrix structures: $$(a)\begin{bmatrix} \times & \times & 0\\ 0 & \times & \times\\ 0 & 0 & \times\\ \end{bmatrix}\ \ \ \ \ \ \ \ (b)\begin{bmatrix} \times & \times & 0\\ \times & 0 & \times\\ 0 & \times & \times\\ \end{bmatrix}\ \ \ \ \ \ \ \ (c)\begin{bmatrix} \times & \times & 0\\ 0 & 0 & \times\\ 0 & 0 & \times\\ \end{bmatrix}$$

(where $\times$ "represents an entry that is not necessarily zero")

For each one, decide which of the following situations olds and justify your claim

(i) Can be obtained by a sequence of left-multiplications by matrices $Q_j$

(ii) Not (i), but can be obtained by a sequence of left- and right-multiplication by matrices $Q_j$

(iii) Cannot be obtained by any sequence of left- and right-multiplications by matrices $Q_j$.

Not sure how to argue, here, and not sure how $Q_j$'s being unitary factors in.

Some help would be hot.

• Does $x$ mean "nonzero-entry here" or does it means "one single special value here!" Can we assume, we start with general full matrices? – Laray Jan 18 '17 at 8:58
• $x$ " represents an entry that is not necessarily zero". I'll edit my post and make $\times$ out of them. I guess we start with a matrix where all entries are $\times$, i.e. "not necessarily zero". – User1291 Jan 18 '17 at 9:02

## 1 Answer

Hints.

(a) Can every rank-1 matrix be reduced to this form?

(b) Handle the first column by Householder reflection. Then try to handle the $2\times2$ submatrix at the top right corner by singular value decomposition. Note that in such a decomposition $USV^\ast$, the unitary matrices $U,V$ may have determinants $-1$. How do you remedy for this?

(c) What is the rank of a matrix in this pattern?

• Hi could you give a more detailed answer please? – JustANoob Oct 4 '17 at 10:25