I am reading a tutorial about the Lanczos method for eigen problem / SVD. It mentioned "Then the tridiagonal matrix $B^∗B$ is unitarily similar to $A^∗A$. " What does it mean?
I can derive this: $A^*A=QB^*P^*PBQ^*=Q(B^*B)Q^*$. Is that an implication of this? What does "unitarily" here mean?
Soure:[2007,Restarted Lanczos bidiagonalization for the SVD in SLEPc.pdf]
2 Answers
$\begingroup$Matrix $Q$ is unitary iff its inverse equals its conjugate transpose $Q^*$. For real matrices it is the same as being orthogonal.
Unitarily similar means similarity by a unitary matrix, e.g. $A = Q^* B Q$ where $Q$ is unitary.
The interest in unitary similarity is two-fold. First, like orthogonal similarity's presevation of matrix symmetry, unitary similarity preserves Hermitianness, that a matrix equals its conjugate transpose (just as for a real matrix, symmetry means a matrix equals its transpose).
Second, unitary transformations (like orthogonal transformations in the real case) preserve the lengths of vectors. A rather immediate benefit of this in numerical methods is that multiplication by a unitary matrix will not increase rounding errors geometrically, since the error component of an intermediate computation would stay exactly the same size if a unitary matrix multiplication were carried out in exact arithmetic. The result is that operations carried out by multiplication by unitary or orthogonal matrices will have better stability and numerical accuracy than with other kinds of matrices (for example, Householder transformations have better error properties than elementary row operations).
$\endgroup$ $\begingroup$For an $n \times n$ matrix $B$ to be unitary it must satisfy \begin{equation} BB^{*} = I \end{equation} Where $I$ is the identity matrix. Similarity between two matrices $A, B$ is the property that \begin{equation} B = P^{-1}AP \end{equation} For some invertible $n \times n$ matrix P. To be unitarily similar we have \begin{equation} A = Q^* B Q \end{equation} where $Q$ is unitary.
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