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Suvrit
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This is also false. Here is a counterexample.

A = \begin{bmatrix} 1 & 1/\sqrt{2}\\\\ 1/\sqrt{2} & 1 \end{bmatrix}

B = \begin{bmatrix} 0 & -1/\sqrt{2}\\\\ 1/\sqrt{2} & 0 \end{bmatrix}

Then, the said block matrix has eigenvalues $(0,0,2,2)$, while

$\lambda^{1/2}(B^T*B) = (1/\sqrt{2},1/\sqrt{2})$$\lambda^{1/2}(B^TB) = (1/\sqrt{2},1/\sqrt{2})$ and

$\lambda(A) = (.2929,1/\sqrt{2})$$\lambda(A) = (1+1/\sqrt{2},1-1/\sqrt{2}))$

This is also false. Here is a counterexample.

A = \begin{bmatrix} 1 & 1/\sqrt{2}\\\\ 1/\sqrt{2} & 1 \end{bmatrix}

B = \begin{bmatrix} 0 & -1/\sqrt{2}\\\\ 1/\sqrt{2} & 0 \end{bmatrix}

Then, the said block matrix has eigenvalues $(0,0,2,2)$, while

$\lambda^{1/2}(B^T*B) = (1/\sqrt{2},1/\sqrt{2})$ and

$\lambda(A) = (.2929,1/\sqrt{2})$

This is also false. Here is a counterexample.

A = \begin{bmatrix} 1 & 1/\sqrt{2}\\\\ 1/\sqrt{2} & 1 \end{bmatrix}

B = \begin{bmatrix} 0 & -1/\sqrt{2}\\\\ 1/\sqrt{2} & 0 \end{bmatrix}

Then, the said block matrix has eigenvalues $(0,0,2,2)$, while

$\lambda^{1/2}(B^TB) = (1/\sqrt{2},1/\sqrt{2})$ and

$\lambda(A) = (1+1/\sqrt{2},1-1/\sqrt{2}))$

Source Link
Suvrit
  • 28.6k
  • 7
  • 82
  • 150

This is also false. Here is a counterexample.

A = \begin{bmatrix} 1 & 1/\sqrt{2}\\\\ 1/\sqrt{2} & 1 \end{bmatrix}

B = \begin{bmatrix} 0 & -1/\sqrt{2}\\\\ 1/\sqrt{2} & 0 \end{bmatrix}

Then, the said block matrix has eigenvalues $(0,0,2,2)$, while

$\lambda^{1/2}(B^T*B) = (1/\sqrt{2},1/\sqrt{2})$ and

$\lambda(A) = (.2929,1/\sqrt{2})$