[Math] rank of a matrix in the real and complex field

linear algebralinear-transformations

Let $\boldsymbol{A}\in \mathbb{R}^{2k\times 2d}$ be a real matrix with $\mathrm{rank}(\boldsymbol{A})=2k$ (on the real field) and $k<d$.
Let $\boldsymbol{T}_{2n}$ be an invertible matrix defined as:
\begin{equation}
{{\boldsymbol{T}}_{2n}} = \left( {\begin{array}{cc}
{{{\boldsymbol{I}}_n}}&{i{{\boldsymbol{I}}_n}} \\
{{{\boldsymbol{I}}_n}}&{ – i{{\boldsymbol{I}}_n}}
\end{array}} \right).
\end{equation}
Let $\boldsymbol{B}\in \mathbb{C}^{2k\times 2d}$ be a complex matrix.

Suppose now that the following equality holds:
\begin{equation}
{{\boldsymbol{A}} = {2^{ – 1}}{\boldsymbol{T}}_{2k}^H{{\boldsymbol{B}}}}{{\boldsymbol{T}}_{2d}}.
\end{equation}

Can we say that, since $\mathrm{rank}(\boldsymbol{A})=2k$ (on the real field), then $\mathrm{rank}(\boldsymbol{B})=2k$ on the complex field?

Best Answer

It should be clear that $A$ and $B$ have the same rank over the complex field, as we just transform one by invertible matrices to get the other. So we have to show that the rank of a matrix doesn't change when making the field bigger. For that, we have the following theorem:

Let $K$ be a field and $A \in K^{n \times m}$ a matrix. Let $L$ be an extension field of $K$ and $A_L \in L^{n \times m}$ the same matrix, just interpreted over the bigger field $L$. Then we have $$rank(A) = rank(A_L).$$

This theorem works for every field (so it also does for the real field and the complex field) and doesn't need anything more, not even quadratic matrices. The reason why it works is the rank normal form:
There are matrices $U \in GL(n,K)$ and $V \in GL(m,K)$, such that $$UAV = \begin{pmatrix}I_k & 0 \\ 0 & 0 \end{pmatrix},$$ where $k = rank(A)$.
Now, by extending the field, we also find $U \in GL(n,L)$ and $V \in GL(m,L)$, so we can transform $A_L$ onto the same rank normal form. As every matrix has a unique such form, we can conclude that both matrices have the same rank.

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