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Let $V$ be an invertible square matrix, and consider the following two conditions:

  1. $V^\dagger V = I$ (orthonormal columns), or
  2. $V V^\dagger = I$ (orthonormal rows).

My questions are:

  • Must $V$ necessarily be normal (i.e., satisfy $V^\dagger V = VV^\dagger$) under either of these conditions?
  • Does either condition imply that $V$ is unitary (i.e., satisfy both $V^\dagger V = I$ and $V V^\dagger = I$)?
Ryan
  • 525

2 Answers2

3

For square matrices, left invertibility is equivalent to invertibility (same as right invertibility). See If $AB = I$ then $BA = I$ for proof. Meaning that if you show for square matrices $A,B$ that $AB = I$, then both are invertible and $B = A^{-1}$. So even the assumption "Let $V$ be an invertible square matrix" is superficial, since either of the orthonormal columns/rows assumptions leads to $V$ being unitary (and thus normal).

rafilou2003
  • 1,587
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The two conditions you give are equivalent. To see this, first note that we may multiply the equation for condition 1 by $V^{-1}$ on the right to obtain $V^\dagger = V^{-1}$. Multiplying the equation for condition 2 by $V^{-1}$ on the left yields the same result. Because either condition can be rearranged to yield the equation $V^\dagger = V^{-1}$, they must be equivalent.

It then follows that if a matrix satisfies these conditions, it must be normal as $V^\dagger V = I = V V^\dagger$, and it must be unitary because both conditions hold.

riemmanianrascal
  • 138