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Here is something I read and I didn't understand :

Let $n \in \mathbb{N}^{\ast}$ and $\Gamma$ be a $n \times n$ positive definite matrix. Let $X \in \mathbb{R}^{n}$, $X \neq 0$ and $\big(X, \varepsilon_{2}, \ldots, \varepsilon_{n}\big)$ an orthogonal basis of $\mathbb{R}^{n}$ such that, in this basis, $\Gamma$ writes :

$$\begin{pmatrix} x_{0} & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & S & \\ 0 & & & & \end{pmatrix}$$

Then, the matrix $\Gamma$ is :

$$ \Gamma = x_{0} \frac{X X^\top}{\Vert X \Vert^{2}} + \Big(\mathrm{I}_{n} - \frac{X X^\top}{\Vert X \Vert^{2}} \Big) \begin{pmatrix} \ast & \ast & \ldots & \ast \\ \ast & & & \\ \vdots & & S & \\ \ast & & & & \end{pmatrix} \Big( \mathrm{I}_{n} - \frac{X X^\top}{\Vert X \Vert^{2}} \Big)^\top \tag{$\clubsuit$}$$ where $\ast$ represents some real coefficients.

I don't understand why the expression $(\clubsuit)$ is true. Here is what I tried :

Let $B_{\mathrm{can}}$ be the canonical basis of $\mathbb{R}^{n}$ and $B=\big(X,\varepsilon_{2},\ldots,\varepsilon_{n}\big)$. Let $P$ be the matrix of change of basis from $B_{\mathrm{can}}$ to $B$. Then :

$$ \Gamma = P \begin{pmatrix} x_{0} & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & S & \\ 0 & & & & \end{pmatrix} P^{-1} $$

and :

$$ \frac{X X^\top}{\Vert X \Vert^{2}} = P \begin{pmatrix} 1 & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & \mathbf{0} & \\ 0 & & & & \end{pmatrix} P^{-1} \quad \mathrm{and} \quad \Big(\mathrm{I}_{n} - \frac{X X^\top}{\Vert X \Vert^{2}} \Big) = P \begin{pmatrix} 0 & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & \mathrm{I}_{n-1} & \\ 0 & & & & \end{pmatrix} P^{-1} $$

Therefore, $\Big(\mathrm{I}_{n} - \frac{X X^\top}{\Vert X \Vert^{2}} \Big) \begin{pmatrix} \ast & \ast & \ldots & \ast \\ \ast & & & \\ \vdots & & S & \\ \ast & & & & \end{pmatrix} \Big( \mathrm{I}_{n} - \frac{X X^\top}{\Vert X \Vert^{2}} \Big)^\top$ is equal to :

$$ \begin{eqnarray*} & & P \begin{pmatrix} 0 & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & \mathrm{I}_{n-1} & \\ 0 & & & & \end{pmatrix} P^{-1} \begin{pmatrix} \ast & \ast & \ldots & \ast \\ \ast & & & \\ \vdots & & S & \\ \ast & & & & \end{pmatrix} P \begin{pmatrix} 0 & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & \mathrm{I}_{n-1} & \\ 0 & & & & \end{pmatrix} P^{-1} \\[3mm] & = & P \begin{pmatrix} 0 & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & \mathrm{I}_{n-1} & \\ 0 & & & & \end{pmatrix} \begin{pmatrix} \diamond & \diamond & \ldots & \diamond \\ \diamond & & & \\ \vdots & & \tilde{S} & \\ \diamond & & & & \end{pmatrix} \begin{pmatrix} 0 & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & \mathrm{I}_{n-1} & \\ 0 & & & & \end{pmatrix} P^{-1} \\[3mm] & = & P \begin{pmatrix} 0 & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & \tilde{S} & \\ 0 & & & & \end{pmatrix} P^{-1} \end{eqnarray*} $$

So, in the end I get :

$$ \Gamma = P \begin{pmatrix} x_{0} & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & \tilde{S} & \\ 0 & & & & \end{pmatrix} P^{-1} $$

which is not the expected expression. Where am I wrong ?

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1 Answer 1

up vote 2 down vote accepted
+50

Considering that $(X,\epsilon_2,\dots,\epsilon_n)$ forms an orthogonal basis, we have direct sum decomposition of $V=\mathbb R^n$. We write $$V=\mathrm{span}\{X\}\oplus\mathrm{span}\{\epsilon_2,\dots,\epsilon_n\}=\mathrm{span}\{X\}\oplus\mathrm{span}\{X\}^\perp$$ It's not hard to write down the projection operator $\mathcal P_X=\frac{XX^T}{||X||^2}$ on subspace $\mathrm{span}\{X\}$ and the rejection operator $\mathcal R_X=I-\mathcal P_X=I-\frac{XX^T}{||X||^2}$. Now the decomposition becomes $$V=\mathrm{Im}\mathcal P_X\oplus\mathrm{Im}\mathcal R_X$$ Next your question is readily followed by direct computations. $$\begin{align}&x_0\frac{XX^T}{||X||^2}=x_0\mathcal P_XI_n\\ =&x_0\begin{pmatrix} 1 & 0 \ldots & 0 \\ 0 & & \\ \vdots & \mathbf O_{n-1} \\ 0 & \end{pmatrix}=\begin{pmatrix} x_{0} & 0 & \ldots & 0 \\ 0 & & & \\ \vdots & & \mathbf O_{n-1} \\ 0 & & & \end{pmatrix}\end{align}\\$$

$$\begin{align}&\Big(I_n-\frac{X X^T}{\Vert X \Vert^{2}} \Big) \begin{pmatrix} \ast & \ast & \ldots & \ast \\ \ast & & \\ \vdots & & S \\ \ast & & & \end{pmatrix} \Big(I_n-\frac{X X^\top}{\Vert X \Vert^{2}} \Big)^T\\ =&\mathcal R_X\begin{pmatrix} \ast & \ast & \ldots & \ast \\ \ast & & \\ \vdots & & S \\ \ast & & & \end{pmatrix}\mathcal R_X^T =\begin{pmatrix} 0 & 0 & \ldots & 0 \\ \ast & & \\ \vdots & & S \\ \ast & & & \end{pmatrix}\mathcal R_X^T\\ =&\begin{pmatrix} 0 & 0 & \ldots & 0 \\ 0 & & \\ \vdots & & S \\ 0 & & & \end{pmatrix}\end{align}$$ Hence $$\Gamma=x_0\mathcal P_XI_n+\mathcal R_X\tilde S\mathcal R_X^T$$

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Thanks a lot ! I can't figure out where I was wrong. Was it my expression for $\mathrm{I}_{n} - \frac{X X^\top}{\Vert X \Vert^{2}}$ ? –  jibounet Nov 22 '13 at 15:26
    
@jibounet I guess you forget the transpose of that. –  Shuchang Nov 22 '13 at 15:35
    
Yes you're right ! I think I made a mistake there with the transpose. Thanks for your explanations. –  jibounet Nov 22 '13 at 15:39

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