Let $\mathbf{A}$ be $m \times n$ ($m < n$) complex matrix and its SVD be $\mathbf{A}=\mathbf{U}\mathbf{\Sigma}\mathbf{V}^H$.

Then we obtain an idempotent and Hermitian matrix, referred to orthogonal projection to prep of null-space, such as


where $\mathbf{A}^{\dagger}$ is a pseudo inverse matrix and $\mathbf{D}$ is a diagonal matrix containing the first $m$ values equal to $1$ and the others equal to 0 on its diagonal.

What can we discuss about the following rank?


I wish to prove whether or not the following identity is true:

$r \equiv n$ (full-rank)

that is, whether or not the considered square matrix is always invertible.

I consider

$r=\operatorname{rank}(\mathbf{V^H}(\mathbf{VD}-\mathbf{I})) =\operatorname{rank}(\mathbf{D}-\mathbf{V}^H) =\operatorname{rank}((\mathbf{D}-\mathbf{V}^H)^H) =\operatorname{rank}(\mathbf{D}-\mathbf{V})$

since $\mathbf{V}$ is unitary, that is, full-rank and $\mathbf{D}^H=\mathbf{D}$.

Then, if let a $\mathbf{V}$'s eigen deccomposition be $\mathbf{V}=\mathbf{W}\mathbf{\Lambda}\mathbf{W}^H$, we might be able to obtain

$r =\operatorname{rank}(\mathbf{D}-\mathbf{W}\mathbf{\Lambda}\mathbf{W}^H) =\operatorname{rank}(\mathbf{W}(\mathbf{D}-\mathbf{\Lambda})\mathbf{W}^H) =\operatorname{rank}(\mathbf{D}-\mathbf{\Lambda})$ (similarity)

but I haven't succeeded.


Thank you for a counterexample to the above identity is false.

As I see the comment, the case of $r \neq n$ is derived from the very special $\mathbf{A}$.

I suspect that

the necessary and sufficient condition is that the $\mathbf{A}$'s right-singular matrix $\mathbf{V}$ can be chosen as the identity matrix $\mathbf{I}$.

If so, as I replied, since this case leads to decrease the number of valid decision variables in the system of linear equation denoted by $\mathbf{A}$ (What name is the rundown of the number of valid variables or of degrees of freedom?), we can ignore this case by checking whether the number of valid variables are full, or rewriting systems of linear equation without rundown.

Can we prove it?

Any ideas?

The following is an another approach I did.

We can write

$\mathbf{D} = \left[\begin{array}{cc} \mathbf{I}_m' & \mathbf{O}\\ \mathbf{O} & \mathbf{O} \end{array}\right], $

where $\mathbf{I}_m'$ is the $m$-square identity matrix (here prime is written to write the following without prime).

so I express $\mathbf{V}=[\mathbf{V}_{m} ~|~ \mathbf{V}_{n-m}]$ where $\mathbf{V}_m$ is the first $m$ column vectors of $\mathbf{V}$ and $\mathbf{V}_{n-m}=\mathbf{V}$ is the remaining $n-m$ column vectors and $\mathbf{I}=[\mathbf{I}_m ~|~ \mathbf{I}_{n-m}]$ in the same way.

Then I obtain

$ \mathbf{VD}=[\mathbf{V}_m ~|~ \mathbf{O}] $

and consider

$ \mathbf{I}-\mathbf{VD} = [\mathbf{I}_m-\mathbf{V}_m ~|~ \mathbf{I}_{n-m}],$

since we capture canonical basis as criterion and $\operatorname{rank}(\mathbf{VD}-\mathbf{I})=\operatorname{rank}(\mathbf{I}-\mathbf{VD})$.

$[\mathbf{I}_m-\mathbf{V}_m ~|~ \mathbf{I}_{n-m}]$ expresses the space of $\operatorname{Im}(\mathbf{I}_m-\mathbf{V}_m) \cup \operatorname{Im}(\mathbf{I}_{n-m})$, right? (I am not familiar)

If so, I obtain

$r = \operatorname{rank}([\mathbf{I}_m-\mathbf{V}_m ~|~ \mathbf{I}_{n-m}])= \operatorname{dim}( \operatorname{Im}(\mathbf{I}_m-\mathbf{V}_m) \cup \operatorname{Im}(\mathbf{I}_{n-m})), $

and rhs can be written for dimention theorem as

$ r= \operatorname{rank}(\mathbf{I}_m-\mathbf{V}_m) + \operatorname{rank}(\mathbf{I}_{n-m}) - \operatorname{dim}(\operatorname{Im}(\mathbf{I}_m-\mathbf{V}_m) \cap \operatorname{Im}(\mathbf{I}_{n-m})) $

and the known rank $\operatorname{rank}(\mathbf{I}_{n-m})=n-m$ leads to

$ r = \operatorname{rank}(\mathbf{I}_m-\mathbf{V}_m)- \operatorname{dim}(\operatorname{Im}(\mathbf{I}_m-\mathbf{V}_m) \cap \operatorname{Im}(\mathbf{I}_{n-m}))+(n-m), $

implying that $r=n$ (full-rank) is satisfied if

$\operatorname{rank}(\mathbf{I}_m-\mathbf{V}_m)- \operatorname{dim}(\operatorname{Im}(\mathbf{I}_m-\mathbf{V}_m) \cap \operatorname{Im}(\mathbf{I}_{n-m}))=m.$

What space does $\operatorname{Im}(\mathbf{I}_m-\mathbf{V}_m) \cap \operatorname{Im}(\mathbf{I}_{n-m})$ express, null-space of something?

What can we mention of the equality condition more about $\mathbf{V}_m$ or its origin $\mathbf{A}$ ?

These motivations are why we can know the detail that $\mathbf{V}$ and $\mathbf{I}$ consists of the complete orthogonal basis, $\mathbf{V}_m$ consists the right-singular vectors corresponding to non-zero singular values of $\mathbf{A}$, and $\mathbf{V}$ has eigenvalues whose absolute values all equal to $1$ in $\mathbf{\Lambda}$.

I think that $\operatorname{rank}(\mathbf{I}_m-\mathbf{V}_m)=m$, that is, $\mathbf{I}_m-\mathbf{V}_m$ is full-rank, and $\operatorname{dim}(\operatorname{Im}(\mathbf{I}_m-\mathbf{V}_m) \cap \operatorname{Im}(\mathbf{I}_{n-m}))=0$ are usual (for example, $\mathbf{A}$ is generated randomly).

Thank you.

  • $\begingroup$ Shouldn't $D$ have the first $r$ elements equal to 1 where $r$ is the rank of A? $\endgroup$ – Arin Chaudhuri Jul 25 '16 at 15:45
  • $\begingroup$ Sorry I forget to write the assumption of rank(A)=m (row rank), and so D have ther first m non-zero valuese and r is just not the rank of A. $\endgroup$ – Hikaru Kawasaki Jul 25 '16 at 16:41

I will provide a counterexample. Assume $A$ is a real $ m \times n $ of rank $m$ where $ m < n $ and its singular value decomposition is $$ A = \sum_{i=1}^{m} \sigma_i u_i e_i^T, $$ where $e_i$ is the $i^{th}$ canonical basis vector in $R^n$ and $\sigma_1 > \dots > \sigma_m > 0$.

The orthogonal projector to the row space of $A$ (which is the orthogonal complement of null space of A) is clearly the $n \times n$ block diagonal matrix $$ P= \left( \begin{matrix} I_m & 0\\ 0 & 0 \end{matrix} \right). $$ It follows that $$PV - I = P - I = \left( \begin{matrix} 0 & 0 \\ 0 & -I_{n-m} \end{matrix} \right).$$

And the rank of the above is $n - m < n.$

  • $\begingroup$ This is actually a counterexample, and leads to $A=[A' | O_{(n-m) \times m}]$. I suppose A as a coefficient matrix of linear constraints, so I understand that your example is a case to decrease the degree of freedom. Thank you very much! $\endgroup$ – Hikaru Kawasaki Jul 27 '16 at 7:00

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