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Let \(A\) be a square real (or complex) matrix. Then \(A\) is invertible if and only if zero is not an eigenvalue. Proof or counterexample.
in Mathematics by Platinum (164,236 points) | 280 views

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The statement is true.

If \(A\) is invertible, then its determinant is non-zero, and the characteristic polynomial of \(A\) is non-zero. This implies that zero is not an eigenvalue of \(A\).

Conversely, if zero is not an eigenvalue of \(A\), then the dimension of the eigenspace associated with zero is zero. This means that the columns of \(A\) are linearly independent, so the matrix \(A\) has a non-zero determinant, and is thus invertible.
ago by Platinum (164,236 points)

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asked Jan 21, 2022 in Mathematics by MathsGee Platinum (164,236 points) | 282 views

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MathsGee Android Q&A

MathsGee Android Q&A