Solving for Unknown in Matrix

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What number $b$ in \begin{bmatrix} 3 & b \\ 1 & 0 \end{bmatrix} makes $A = Q\Lambda Q^T$ possible?

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HINT Notice that if $A = Q\Lambda Q^T$ then $A^T=A$.

UPDATE

I am assuming $\Lambda$ is real-valued. If you assume $\Lambda$ is diagonal, $A$ must be symmetric so $b=1$. Without any assumptions on $\Lambda$, as Laray indicated, $A$ must be normal...