# Determinant of identity minus product of orthogonal matrix and rank-$1$ matrix

I am interested in calculating, or bounding in some way, the following determinant $$\det\left[\mathcal{I}-Rxx^t\right]$$ Here, $Rxx^t$ is clearly a singular matrix. Im the case that I am particularily interested in, $R$ is an orthogonal matrix with determinant $+1$. I understand that there are ways of doing this using characteristic polynomials and such, but im not educated in those things. Any help greatly appreciated!

NOTE: I do not know $R$, it can be any special orthogonal matrix. I started my derivation with the `matrix determinant lemma' and am not looking to return to this step, but rather to attack the above determinant

## 1 Answer

You can use the theorem $\det\left(I+uv^\mathrm{T}\right) = 1+u^\mathrm{T}v$:

\begin{align} \det(I-Rxx^\mathrm{T}) = 1-(Rx)^\mathrm{T}x = 1-x^\mathrm{T}R^\mathrm{T}x \end{align}

Furthermore $-\left\|x\right\|^2_2\leq x^\mathrm{T}R^\mathrm{T}x\leq \left\|x\right\|^2_2$ if $R$ is orthogonal and no other information is known. Thus, we have for the determinant

\begin{align} 1 - \left\|x\right\|^2_2\leq \det(I-Rxx^\mathrm{T}) \leq 1 + \left\|x\right\|^2_2 \end{align}

where both bounds can be reached for some orthogonal $R$.

• ah yes - infact, im trying to avoid using this as this is the starting point in my derivation. I do not know what $R$ is other than it is special orthogonal. I am trying to show, eventually, that i can make the above determinant negative by choosing my $u$ and $v$. – jdizzle Jul 19 '14 at 16:33
• I adapted my answer. If you do not have any further information on $R$, the bounds in the answer are least restrictive. I hope I am understanding your question correct. – Dimitar Ho Jul 19 '14 at 16:45