More fun in 2015, Part 28

Algebra Level 5

$5x^2+6y^2+7z^2=4xy+4yz+2015$

Let $M$ and $m$ be the maximum and the minimum of $x^2+y^2+z^2$ subject to the constraint above, where $x,y$ and $z$ are real numbers. Find $\frac{M}{m}$ .

The answer is 3.

1 solution

Otto Bretscher
Nov 9, 2015

Make a substitution $x=(u-2v+2w)/3, y=(-2u+v+2w)/3, z=(2u+2v+w)/3$ . Now we are asked to find the extrema of $u^2+v^2+w^2$ subject to the constraint $u^2+2v^2+3w^2=2015/3$ . We have $m=2015/9=(u^2+2v^2+3w^2)/3\leq u^2+v^2+w^2\leq u^2+2v^2+3w^2=2015/3=M$ , so that $M/m=\boxed{3}$

You're missing a lot of crucial parts. I'm already lost in your very first line.

Pi Han Goh - 5 years, 7 months ago

As I told you earlier, my work is based on eigenvalues, but I wrote my solution without explicit reference to them, which makes it a bit mysterious. The solution is very systematic and straightforward, however.

The matrix of the quadratic form $5x^2+6y^2+7z^2-4xy-4yz$ is $\begin{bmatrix} 5&-2&0\\-2&6&-2\\0&-2&7\end{bmatrix}$ We "split" the mixed terms $-4xy$ and $-4yz$ off the diagonal to make the matrix symmetric. The eigenvalues are 3, 6, and 9, and the answer to our problem is the ratio of the extreme eigenvalues, 9 and 3.

Here is another problem that you could solve quickly this way, without resorting to these weird inequalities that seem so forced and arbitrary.

Otto Bretscher - 5 years, 7 months ago

Thanks. You should put that in your solution.

Question 1 : What is the reasoning behind

$\dfrac{\text{max}}{ \text{min} } =\text{ ratio of extremes of eigenvectors} ?$

Question 2 : How do you form a matrix such that their eigenvectors are "nice numbers"? Or you must (and is the only way) go through $\det(A - \lambda I_n) = 0$ ?

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – You ask interesting questions, as usual.

I have another busy day of work ahead of me, so, let me (try to) answer Question 2 for now. As a linear algebra teacher, I'm often faced with the problem of constructing a nice matrix $A$ (with integer entries, say) with given (integer) eigenvalues. It's pretty easy: Just make $A=S^{-1}DS$ , where $D$ is diagonal with the desired eigenvalues on the diagonal, and $S$ is an integer matrix with determinant 1 or -1 (that guarantees that the inverse has integer entries). One way to construct such matrices $S$ is to start with a triangular matrix with 1's and -1's on the diagonal, and then add multiples of rows (or columns) to other rows (or columns) to "mix things up".

I hope to get to Question 1 later. In the meantime, I posted another problem addressing this issue.

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher –

Help please. I'm not getting a symmetric matrix!

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – You asked "How do you form a matrix such that their eigenvectors are "nice numbers"? "... You did not ask for a symmetric matrix and I did not give you one, Comrade. ;) I have some special tricks for those that I will gladly share when I have a little free time.

To get a more interesting (non-symmetric) matrix with eigenvalues 2,6,7 you could do $\begin{bmatrix}1&1&1\\2&3&4\\3&6&10\end{bmatrix}^{-1}\begin{bmatrix}2&4&5\\0&6&6\\0&0&7\end{bmatrix}\begin{bmatrix}1&1&1\\2&3&4\\3&6&10\end{bmatrix}=\begin{bmatrix}51&90&142\\-32&-58&-96\\6&12&22\end{bmatrix}$

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – .

I have some special tricks for those that I will gladly share when I have a little free time.

Please share your wisdom! Comrade Otto =D

Pi Han Goh - 5 years, 7 months ago

@Otto Bretscher – Can you write up a full solution for any of the practice problems here so I can replicate the result?

Pi Han Goh - 5 years, 7 months ago

@Otto Bretscher – But how do you make it symmetric? (So I can convert it to a quadratic form)

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – I love to share some of my techniques with a Comrade ;)

One simple technique is to use $A=S^{-1}DS$ where $D$ is diagonal and $S$ is orthogonal with rational entries... you can then scale $A$ to make the entries integers.

Matrices that I often use for $S$ are variants of $\begin{bmatrix}0.6&-0.8\\0.8&0.6\end{bmatrix},\frac{1}{3}\begin{bmatrix}2&2&1\\-2&1&2\\-1&2&-2\end{bmatrix},\frac{1}{2}\begin{bmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&-1&1\\1&-1&1&-1\end{bmatrix}$

For example, if you take $D=\begin{bmatrix}4&0\\0&9\end{bmatrix}$ and the $S$ above you get $A=S^{-1}DS=\begin{bmatrix}7.2&2.4\\2.4&5.8\end{bmatrix}$ ... multiply by 5 if you want an integer example. That's how I constructed this problem ...

I'm giving away my "secrets" here (for Comrades only!) ;) many of my recent problems are based on this construction.

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – Can you give an explicit example? I'm new to this. Thank you Comrade Otto =D

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – Here are a few remarks to Question 1, brief as usual... that is all you need.

The question boils down to finding the maximum and minimum of $q(\vec{x})=\vec{x}^TA\vec{x}$ , where $A$ is a symmetric matrix and $\vec{x}$ is a unit vector. By the Spectral Theorem, there exist perpendicular unit eigenvectors $\vec{u},\vec{v},\vec{w}$ . with corresponding eigenvalues $a\leq b\leq c$ . Now $q(\vec{w})=c$ is the maximum, etc.

Otto Bretscher - 5 years, 7 months ago

Can u post a simple solution that a rookie can understand. Sorry for the pain 😆

neelesh vij - 5 years, 7 months ago

Do you like Lagrange multipliers? All that is required are partial derivatives...

Otto Bretscher - 5 years, 7 months ago

I have never heard of it. Please can u explain or attack a link for the following. Thanks

neelesh vij - 5 years, 7 months ago

@Neelesh Vij – You may benefit from reading Brian's excellent solution here

Otto Bretscher - 5 years, 7 months ago

So how do we know what type of surface this is without finding its eigenvalues (by completing the squares alone)? I have some feeble attempts but all boils down to randomly factoring terms after this step:

$5x^2 + 6y^2+ 7z^2 -4xy -4yz = 2015 \; .$

Note : I'm asking for a systematic approach, not some random "Oh I factorized it nicely by accident!" approach.

And yes, I know there are 3 positive eigenvalues, which implies that it is an ellipsoid.

Pi Han Goh - 5 years, 3 months ago

Yes, completing the squares is a very systematic process, although it can get a bit messy if done by hand. You start with the terms involving $x$ , in our case $5x^2-4xy$ $=5(x^2-\frac{4}{5}xy)$ $=5(x-\frac{2}{5}y)^2-\frac{4}{5}y^2$ . Now you take care of the terms involving $y$ , and you are done.

Otto Bretscher - 5 years, 3 months ago

ohhhhhhhhhhhhhhhh this is simpler than I thought!!! I was working on completing all 3 squares at the same time!

Here's the answer:

$\begin{aligned} 5x^2 + 6y^2+ 7z^2 -4xy -4yz &= &5 \left( x^2- \dfrac45 xy\right) + 6y^2+ 7z^2 -4yz \\ &= &5 \left ( \left( x- \dfrac25y\right)^2 - \dfrac4{25}y^2 \right) + 6y^2+ 7z^2 -4yz \\ &= &5 \left( x- \dfrac25y\right)^2 + y^2 \left( 6 - \dfrac45 \right) + 7z^2 -4yz \\ &= &5 \left( x- \dfrac25y\right)^2 + 5.2y^2 - 4yz + 7z^2 \\ &= &5 \left( x- \dfrac25y\right)^2 + 5.2 \left( y^2 - \dfrac{4}{5.2} yz\right) + 7z^2 \\ &= &5 \left( x- \dfrac25y\right)^2 + 5.2 \left( \left( y -\dfrac2{5.2} z \right)^2 - \dfrac4{5.2^2} z^2 \right) + 7z^2 \\ &= &5 \left( x- \dfrac25y\right)^2 + 5.2 \left( y -\dfrac2{5.2} z \right)^2 + z^2 \left( 7 - \dfrac4{5.2} \right) \\ \end{aligned}$

Since $5,5.2,7 - \dfrac4{5.2}$ are all positive numbers, then the surface is an ellipsoid.

Pi Han Goh - 5 years, 3 months ago

@Pi Han Goh – Yes exactly! You can type much faster than me ;)

Otto Bretscher - 5 years, 3 months ago

More fun in 2015, Part 28

The answer is 3.

1 solution

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