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?

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

1 Answer 1


I can't be sure that your proof of the properties was correct, but your claim indeed holds.

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 $$ \pmatrix{\mathbf U & \cdots & \mathbf U}^+ = (\mathbf 1^T \otimes \mathbf U)^+ = (\mathbf 1^T)^+ \otimes \mathbf U^+ = \left(\frac 1N \mathbf 1\right)\otimes \mathbf U^+ = \frac 1N \pmatrix{\mathbf U^+ \\ \vdots \\ \mathbf U^+}. $$

  • $\begingroup$ Ohhhh.... This is much slicker, thank you very much! $\endgroup$ Jul 13, 2020 at 20:32

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