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Verify that if $P$ is an orthogonal matrix and $x = Py$ then $y^Ty = x^Tx$. Let $A$ be a real symmetric $n × n$ matrix. Then we know that there exists a real orthogonal matrix $P$ such that $P^TAP$ is diagonal. By using the transformation $x = Py$, or otherwise, prove that for every $x\in {\mathbb R}^{n}$,

$$mx^Tx ≤ x^TAx ≤ Mx^Tx$$ where $m$ and $M$ are the smallest and greatest eigenvalues of $A$ respectively. For which $x$ is it true that $x^TAx = Mx^Tx$?

$\begin{bmatrix}5 & 1&\sqrt2\\1 & 5&\sqrt2\\ \sqrt2 &\sqrt2 &6\end{bmatrix}$

Find the maximum and minimum values of $x^Tx$ for those $x$ for which $x^TAx = 1$. Giving no heed to orientation, sketch the surface $S$ with equation $x^TAx = 1$, and indicate on it those vectors $x$ at which $x^Tx$ attains its maximum and minimum values on $S$.

$x^TAx=y^TDy$ and $\lambda_{min}y^Ty\leq y^TDy\leq \lambda_{max}y^Ty$ but how to link it to $1$ and the surface?

stedmoaoa
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1 Answers1

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Your eigenvalues are $(4,4,8)$

So you your biggest $x$ will be something that loads exculsively on 4 eigenvalues

and its magnitude will be $\frac 12$

$(\frac 1{2\sqrt 2}, -\frac 1{2\sqrt 2}, 0)$ will work.

So you your smallest $x$ will parallel to the eigenvector associated with the 8 eigenvalue.

and its magnitude will be $\frac {1}{2\sqrt 2}$

$(\frac {1}{4\sqrt 2},\frac {1}{4\sqrt 2}, \frac 14)$

As for the surface, it is an ellipsoid. It has 2 equal axes and one short axis. The big eigenvector is the minor axis.

Doug M
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  • Can you explain why the biggest s will load on 4 eigenvalues and the magnitude? – stedmoaoa May 18 '17 at 19:50
  • Consider from high school algebra $a x^2 + b y^2 = 1$ (with $a,b>0$) is an equation for an ellipse. Put it in standard form. $\frac {x^2}{\frac {1}{\sqrt a}} + \frac {y^2}{\frac {1}{\sqrt b}} = 1$ We can see how these coefficients affect then axes. – Doug M May 18 '17 at 20:18
  • Thank you! And another question: for which x is it true that $x^TAx=Mx^Tx$? Should it be all the eigenvalues are the same? – stedmoaoa May 19 '17 at 06:59