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Determinant

Let x>0 and  A \in \mathbb{R}^{2 \times 2} such that \det \left(A^2 + x \mathbb{I}_{2 \times 2} \right) = 0. Prove that

    \[\det \left( A^2 + A + x \mathbb{I}_{2 \times 2} \right) = x\]

Solution

Note that

    \[\det \left(A^2 + x \mathbb{I}_{2 \times 2} \right) = \det \left(A + i \sqrt {x} \mathbb{I}_{2 \times 2} \right) \det \left(A - i\sqrt{x}  \mathbb{I}_{2 \times 2} \right)\]

Since A is real, its complex eigenvalues come in conjugate pairs. Thus, in this case we conclude that A has eigenvalues \pm i \sqrt x.

Now, if \lambda is an eigenvalue of A, then \lambda^2 + \lambda + x is an eigenvalue of A^2 + A + x \mathbb{I}_{2 \times 2}. Thus, the matrix A^2 + A + x \mathbb{I}_{2 \times 2} has eigenvalues (i\sqrt {x} )^2 + i\sqrt {x} + x = i\sqrt {x} and (-i\sqrt {x})^2 - i\sqrt {x} + x = -i\sqrt {x}.

Now, \det \left(A^2 + A + x \mathbb{I}_{2 \times 2} \right) is the product of these eigenvalues, which is to say

    \[\det \left(A^2 + A + x\mathbb{I}_{2 \times 2} \right) = (i\sqrt{x})(-i\sqrt{x}) = x\]

as desired.

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