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Although I cannot prove it rigorously no matter which unit vector I try, I get: $$\det(Q)=-1$$ $$\det(uu^T) = 0$$ I found $Q$ to be orthogonal but how to prove or disprove these?

üzeyir saçıkay
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  • $\det(uu^T) = 0$ is easily proven using rank (assuming $n>1$). – Arthur Nov 24 '17 at 21:53
  • See https://math.stackexchange.com/questions/1514408/matrix-determinant-lemma-with-adjugate-matrix?rq=1 . – darij grinberg Nov 24 '17 at 21:54
  • Every column of $uu^T$ as a multiple of $u$. What do you know about the determinant of a matrix that has linearly dependent columns? – amd Nov 24 '17 at 23:42

3 Answers3

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You have for any pair of n-dimensional column vectors $u,v$ and the $n \times n$ identity matrix $I_n$ that $\det(I_n + uv^T) = 1 + v^Tu. \label{e:1} \tag{*}$

So if $\|u\|^2 = 1$ then $\det(I_n + (-2u)u^T) = 1 + u^T(-2u)=-1.$

To see why $\eqref{e:1}$ is true note that $\begin{pmatrix} I_n & 0 \\ -v^T & 1 \end{pmatrix} \begin{pmatrix} I_n & -u \\ v^T & 1 \end{pmatrix}$ and $\begin{pmatrix} I_n & u \\ 0 & 1 \end{pmatrix} \begin{pmatrix} I_n & -u \\ v^T & 1 \end{pmatrix}$ have the same determinant.

Arin Chaudhuri
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We have $Q^T=Q$ with an easy computation, so \begin{align} Q^TQ &=(I-2uu^T)^T(I-2uu^T)\\ &=(I-2uu^T)(I-2uu^T)\\ &=I-2uu^T-2uu^T+4uu^Tuu^T\\ &=I-4uu^T+4\|u\|^2uu^T \end{align} so $Q$ is orthogonal precisely when $\|u\|=\sqrt{u^Tu}=1$. Under this hypothesis we have $$ (2uu^T)u=2\|u\|^2u=2u $$ which means that $u$ is an eigenvalue of $2uu^T$, relative to the eigenvalue $2$. This implies that the characteristic polynomial of $2uu^T$ is $$ \det(\lambda I-2uu^T)=(\lambda-0)^{n-1}(\lambda-2) $$ because $2uu^T$ has rank $1$ and so $0$ is an eigenvalue of geometric multiplicity $n-1$. The geometric multiplicity is at most equal to the algebraic multiplicity and the degree of the characteristic polynomial is the sum of the algebraic multiplicities of the eigenvalues. This forces the algebraic multiplicity of the eigenvalue $0$ to be $n-1$ and of the eigenvalue $2$ to be $1$.

Now evaluate for $\lambda=1$.

See also Householder transformation on Wikipedia.

egreg
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Just take an orthonormal basis of $\mathbb R^n$ such that $u=e_n$ is one element of such basis. Then we have $uu^Te_i=0, \forall i=1...n-1$ and $uu^Tu=u\|u\|^2=u$. So we have that the eigenvalues of $A=uu^T$ : $\sigma(A)=\{ 0,...,0,1\}$, so $det(A)=0$. Now you just have to use the fact that $\sigma(P(A))=P(\sigma(A))$, which means that if you take a polynomial of the matrix $A$ its spectrum will be formed by the polynomial of each element of the spectrum, so we have: $\sigma(Q)=\sigma(I-2A)=1-2\sigma(A)=\{1,...,1,-1\}$ so we have that $det(Q)=1 \cdot ...\cdot 1 \cdot -1=-1.$

chak
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