If $Q$ is an orthogonal, then is $Q+\frac{1}{2}I$ invertible?

Question

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.

yeah i think you're right: the columns are perpendicular but all length 1/2 — uncookedfalcon, Nov 18 '12 at 07:07
The eigenvalues of an orthogonal matrix all have absolute value $1$. In particular $-1/2$ is not an eigenvalue. — Robert Israel, Nov 18 '12 at 07:19

score 9 · Accepted Answer · answered Nov 18 '12 at 07:15

9

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 $v^T v \neq 0$.

answered Nov 18 '12 at 07:15

Dan Shved

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Can you give a counterexample? I tried to find one but failed.. – pascl Nov 18 '12 at 07:52
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Counterexample to what? I just proved that $Q+\frac{1}{2}I$ is invertible whenever $Q$ is orthogonal, there is no counterexample to that. – Dan Shved Nov 18 '12 at 07:53
Oh, I misread your answer! Thanks a lot! – pascl Nov 18 '12 at 08:00

If $Q$ is an orthogonal, then is $Q+\frac{1}{2}I$ invertible?

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