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made matrices indefinite
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Dirk
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I suspect that you mean $0 < t < 1$ in your question.

Then the answer in still no under the added condition that both matrices are not of full rank. Consider $$ A_1 = \begin{pmatrix}1 & 0 \\\\ 0 & 0\end{pmatrix}\quad A_2 = \begin{pmatrix}0 & 0 \\\\ 0 & 1\end{pmatrix}. $$$$ A_1 = \begin{pmatrix}1 & 0 & 0\\\\ 0 & 0 & 0\\\\ 0 & 0 & -1\end{pmatrix}\quad A_2 = \begin{pmatrix}0 & 0 & 0\\\\ 0 & 1 & 0\\\\ 0 & 0 & -1\end{pmatrix}. $$

I suspect that you mean $0 < t < 1$ in your question.

Then the answer in still no under the added condition that both matrices are not of full rank. Consider $$ A_1 = \begin{pmatrix}1 & 0 \\\\ 0 & 0\end{pmatrix}\quad A_2 = \begin{pmatrix}0 & 0 \\\\ 0 & 1\end{pmatrix}. $$

I suspect that you mean $0 < t < 1$ in your question.

Then the answer in still no under the added condition that both matrices are not of full rank. Consider $$ A_1 = \begin{pmatrix}1 & 0 & 0\\\\ 0 & 0 & 0\\\\ 0 & 0 & -1\end{pmatrix}\quad A_2 = \begin{pmatrix}0 & 0 & 0\\\\ 0 & 1 & 0\\\\ 0 & 0 & -1\end{pmatrix}. $$

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Dirk
  • 12.7k
  • 6
  • 54
  • 97

I suspect that you mean $0 < t < 1$ in your question.

Then the answer in still no under the added condition that both matrices are not of full rank. Consider $$ A_1 = \begin{pmatrix}1 & 0 \\\\ 0 & 0\end{pmatrix}\quad A_2 = \begin{pmatrix}0 & 0 \\\\ 0 & 1\end{pmatrix}. $$