Sanity check: is this simple formula for pseudoinverse of $[\mathbf{U} \cdots \mathbf{U}]$ correct?

Let $$\mathbf{U}$$ be some matrix, and then consider the "block row vector" $$\underbrace{[\mathbf{U} \cdots \mathbf{U}]}_{N \text{ times}} \,.$$

Claim: The pseudoinverse of this is the "block column vector" $$\frac{1}{N}\begin{bmatrix} \mathbf{U}^\dagger \\ \vdots \\ \mathbf{U}^\dagger \end{bmatrix} = \begin{bmatrix} \frac{1}{N}\mathbf{U}^\dagger \\ \vdots \\ \frac{1}{N}\mathbf{U}^\dagger \end{bmatrix}$$

Proof (?) of claim: I believe I was able to show that this Ansatz satisfies the four properties which uniquely define the pseudoinverse of a matrix by using the following two "lemmas"

$$\begin{bmatrix} \mathbf{F}_1 \cdots \mathbf{F}_N \end{bmatrix} \begin{bmatrix} \mathbf{G}_1 \\ \vdots \\ \mathbf{G}_N \end{bmatrix} = \sum_{n=1}^N \mathbf{F}_n \mathbf{G_n}$$

$$\begin{bmatrix} \mathbf{D}_1 \\ \vdots \\ \mathbf{D}_N \end{bmatrix} \begin{bmatrix} \mathbf{E}_1 \cdots \mathbf{E}_N \end{bmatrix} = \begin{bmatrix} \mathbf{D}_1 \mathbf{E_1} & \mathbf{D}_1 \mathbf{E}_2 & \cdots \\ \vdots & \ddots & \vdots \\ \mathbf{D}_N \mathbf{E}_1 &\cdots &\mathbf{D}_N \mathbf{E}_N \end{bmatrix}$$

Then the proof seems to be simply applying those principles and then using the facts that $$\mathbf{U}^\dagger$$ is the pseudoinverse of $$\mathbf{U}$$ (e.g. $$\mathbf{U}^\dagger \mathbf{U} \mathbf{U}^\dagger = \mathbf{U}^\dagger$$). Is this correct?

• @BaselJ. $U^\dagger$ denotes the pseudoinverse, not the conjugate-transpose. Jul 13, 2020 at 20:27
• @Omnomnomnom, that makes more sense thanks. Jul 13, 2020 at 20:29

The result has a very simple proof if we use the fact that $$(A \otimes B)^+ = A^+ \otimes B^+,$$ where $$\otimes$$ denotes the Kronecker product. Now, let $$\mathbf 1$$ denote the column-vector with a $$1$$ for every entry. It follows that $$\pmatrix{\mathbf U & \cdots & \mathbf U}^+ = (\mathbf 1^T \otimes \mathbf U)^+ = (\mathbf 1^T)^+ \otimes \mathbf U^+ = \left(\frac 1N \mathbf 1\right)\otimes \mathbf U^+ = \frac 1N \pmatrix{\mathbf U^+ \\ \vdots \\ \mathbf U^+}.$$