We know that a complex square matrix V is unitary if \begin{eqnarray} VV^{*} = V^{*} V = I \end{eqnarray}

I want to write matrix V into block matrix form, say $V = [V_1, V_2]$. My question is whether matrix $V_1$ and $V_2$ will also satisfy the same unitary condition mentioned above?

Thank you very much for your help.

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You need to specify what changes you are willing to allow to get the matrix into block shape. Just erasing $V$ and writing an unrelated block matrix in its place for instance will not guarantee much properties for the result. Seriously, any time you replace a matrix by another one you should have some idea of what relates the two matrices. – Marc van Leeuwen Apr 8 '13 at 20:10
$V_1$ and $V_2$ will no longer be square matrices. How do you define unitarity for a non-square matrix? – Lord Soth Apr 8 '13 at 20:16
@MarcvanLeeuwen If I wish to preserve the unitary conditions on matrices $V_1$ and $V_2$, respectively? Is it possible? More specifically I want to diagonalize a particular matrix say $A$ by using unitary matrix $V = [V_1, V_2]$, where I want to preserve unitary structure on $V_1$ and $V_2$ respectively. Is it reasonable? – srijan Apr 8 '13 at 20:19
@LordSoth: I think $V_1$ and $V_2$ are intended to be square blocks on the diagonal, but OP might want to clarify this. – Marc van Leeuwen Apr 8 '13 at 20:37

I am supposing that by "block form" you mean "block diagonal form", with diagonal square blocks $V_1,V_2$, and off-diagonal blocks $0$ that you forgot to write. Otherwise your question does not seem to make sense.

Although you have difficulty explaining what you want to allow doing with the matrix, I think what you want to allow is some form of change of basis. An arbitrary change of basis will possibly destroy the unitary property. However a unitary change of basis (conjugating by a unitary basis, which means changing from the standard basis to another orthonormal basis) will preserve the unitary property. This is easy to see, since unitary matrices form a group, so even an isolated left or right multiplication will preserve the unitary property (such isolated operation are not what you want to allow, since they do not correspond to a change of basis; however conjugation is made up of two such multiplications).

The question arise whether unitary conjugation suffices to bring you unitary matrix into block form. This is the case, and even more: unitary conjugation suffices to completely diagonalise your matrix, which is certainly block-diagonal.

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van Thank you very much SIR. You cleared all of my doubts. Unfortunately, I couldn't explain my question properly. – srijan Apr 8 '13 at 20:43

Certainly not. The size of $V_iV_i^\ast$ is larger than the size of $V_i^\ast V_i$, so the two matrix products cannot compare in the first place. And $V_iV_i^\ast$ is bound to have deficient rank, so it cannot be equal to $I$. Yet you do have $V_i^\ast V_i=I$ because the columns of $V$ are orthonormal.

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1551 I am sorry I couldn't get how the size of $V_i V_i^\ast$ is larger than the size of $V_i^\ast V_i$? thanks for your reply – srijan Apr 8 '13 at 20:33
Let's say $V$ is $n\times n$ and $V_1$ has $m$ columns (hence $n\times m$ with $m<n$). Then $V_1V_1^\ast$ is $n\times n$, but $V_1^\ast V_1$ is $m\times m$. – user1551 Apr 8 '13 at 20:35
Thank you very much for clearing my though and doubts. :) – srijan Apr 8 '13 at 20:43