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If $Q$ is an orthogonal matrix, then the matrix has orthonormal columns.

I asked this question to my friend and he says: Let $Q= -\frac{1}{2}I$, then it is orthogonal, and $Q+\frac{1}{2}I$ is zero, so not invertible.

But I think if we set $Q=-\frac{1}{2}I$, then it is not an orthogonal matrix since it doesn't have unit length, right?

Does anyone know whether it is true or false? Let me know the reason or counter example.

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    The eigenvalues of an orthogonal matrix all have absolute value $1$. In particular $-1/2$ is not an eigenvalue.2012-11-18

1 Answers 1

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This is true. We can prove it by contradiction.

Suppose $Q+\frac{1}{2}I$ is not invertible. Then there's a non-zero column $v \in \mathbb{R}^n$ such that $(Q+\frac{1}{2}I)v = 0$, which means that $Qv=-\frac{1}{2}v$. But then $ v^T v = v^T Q^T Q v = \left(Qv\right)^T \left(Qv\right) = \left(-\frac{1}{2}v\right)^T \left(-\frac{1}{2}v\right) = \frac{1}{4} v^T v, $ which is a contradiction because $v^T v \neq 0$.

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    Oh, I misread your answer! Thanks a lot!2012-11-18