let-qmathbfx7-x_12x_227-x_32-8-x_1-x_2-4-x_1-x_3-8-x_2-x_3-find-a-unit-vector-mathbfx-in-mathbbr3-at

Question

Let $Q(\mathbf{x})=7 x_{1}^{2}+x_{2}^{2}+7 x_{3}^{2}-8 x_{1} x_{2}-4 x_{1} x_{3}-8 x_{2} x_{3}$
Find a unit vector $\mathbf{x}$ in $\mathbb{R}^{3}$ at which $Q(\mathbf{x})$ is maximized,
subject to $\mathbf{x}^{T} \mathbf{x}=1 .[	ext { Hint: The eigenvalues of the matrix of }$ the quadratic form $Q$ are 9 and $-3 . ]$

Runpeng Li · Accepted Answer

Okay, so four problem eight again. Well, given the code red warm Excuse me. We&#x27;re given a corseted form and we need to find a unit vector that account that can mix, move, maximize Q X subject to x transpose time Maxie was doing was one. So, uh, from our routines were already given the argon values off. The of course on the Matrix is knife and negative three. So we can just directly use thes second values so personally, still need to write down our matrix to find out the again vectors. So our matrix will be 717 on the diagonal and 12 the coefficient for ex likes to will be active eight. So we have four here next, four here and also have connected to here connected to here and next four here. Next four year. Okay, now so we can take our I gave up arguing Victor&#x27;s corresponding to I recommend it. I&#x27;m the one which is again just you are to be in order to find the Agon Victor. We just considered system a Times B equals langa. Times be so our first lycan victor will be 0.91. Active 0.37 negative 0.18. Now we just plugging over V one and we will get loved a one, which is the maximum you can get.

Jack Chen · Answer

Okay, So for this program of the symmetric metric disassociated with this quadratic form, you&#x27;ll be minus 2 to 0 two for minus 12 and 0 to 0. Since we want to make the minds this quadratic form are subject toe a constant. First, we need to figure out the Agon value, and I get back. Turns off this matrix a cyclical characteristic equation will be the determinant off a miner&#x27;s lamp. Die. It was a zero. Now eso that is minus two minus lambda minus one minus lambda and the miners Lambda and the miners for minus NMDA my nose, Um, four times minus two minus lambda equals zero. So this characteristic question has the race solutions than the one in Costa to them that we close to minus one in the Lambda three coastal minus four. So the maximum value off Q X, um subject two x transpose times X equals to one. You will be the maximum again barely, which is to so the corresponding vector. We will be the Agon vector associated to this ag Emilio. So a minus two I equals two R minus 4 to 0 to minus 3 to 0 to minus two sell a corresponding idea. Better will be U equals two Cross minus 1/3 to third and the 2/3. So when our conclusion is one X equals to this unit. Vector you on a pathetic A phone Q X, a chief say, to make Mama Leo, too.

Jack Chen · Answer

she&#x27;s so, uh, given this quadratic form the next video off Q x ah subject two X transpose times X equals to one is, um the maximal I am value off the metrics A. So, in this case, a close to seven minus one minus 13 So the agon value off a is the solution to this characteristic equation that would be seven minus lambda times three minus lambda minus one equals zero. So Lamda one equals 25 plus Ruutel five and the Lambda Chewy, close to five minus root. If I so sees Lambda Wise Greater Lambda to sell a maximum value equals 25 plus Ruto five.

Runpeng Li · Answer

All right. So in this problem, work even for skipping a, uh, a quadratic a form that is ny uh, I x one squared us seven x two squared us 11 x three squared, minus eight. Ex one next to and plus eight. Excellent. Next three. So the pristine we need to do is to write out the matrix bye, which is a three by three matrix. Uh, so So the way we need to, uh, right out, right how the Matrix A is to compare the coefficient of the quadratic born to our general coded form for three by three matrix, which we have already done in the previous work. So I would not go through the approach here, so I&#x27;ll just write down the matrix a year. So it&#x27;s nine naked four or and next, four, seven, zero, and or zero 11. Okay, So now the next step is to find out the wagon vectors corresponding to the given Aiken value. So? So to do that, the approach that I would recommend to do it&#x27;s to consider linear system A. My list loved I because they&#x27;re so by that way, um, we just do the reduced to this system and find out a solution. So I won&#x27;t. I will write down the pros. The process here and because it&#x27;s just a rule, reduced reduction won&#x27;t be too hard and will be a mine. It&#x27;s not that I was zero. And turns out we have three again vectors for a second doctor. ISS effective to give to one with again value love. No one wishes three and second of Victor is like to 1/2 and one and one and which, which correspond to the vector. Sorry. Uh, Lambda too. That is night also. Half we&#x27;d be three. We should call respond to, uh, one negative 1/2 of one and correspondent, which corresponds to London. Three. That is 15. So Okay, so right now we recall the&#x27;re, um uh, This The room is given our given by our textbook, so I&#x27;ll just write down here we have. We have a symmetric and each idea values are distinct. Distinct. Then the Eigen vectors. Our fourth Avenue or something. Okay, So that means Thea Aiken vectors. We get these three back wagon victors are all definitely view. Uh, you want you can just check whether their authority this theory tells us that, um it must be or, you know, so we can just write down the matrix p directly by these. These three factors Just negative too. Give to one. And next. One, three. Sorry. 91 two, two. Oh, sorry. Uh, so remember to just a right down the matrix on the ortho one matrix in, uh, normal lines to form. So I&#x27;ll just directly right down it all my life form. By my calculation, that would be, uh, 1/3 which is still scaler. We have nothing negative to negative too one and elected. One, two, two. And the last column is too active. One. I do. So this is our or a token of matrix speak. So the last thing we need to d&#x27;oh is too. Find the change of variable X equals P y transforms X t x transpose X transpose a axe into a quadratic form with no course product term. So this by we can just write down. There&#x27;s write down the correct equaled uniform directly by our theory. Um, in the textbook, it&#x27;s just Ah, the Eiken values off, eh? Times. Sorry. Uh, unbearable should be. Why, Yeah, I can values number one why one squared so that, uh, lambda and times why ends where where off Ramdas are I can values. I can&#x27;t bel us off a and the work even the Yangon medics off a to B three Say it&#x27;s, uh, 39 15. So we will just write down directly. That will be three. Why one&#x27;s weird? I my two squared and 15 Why three squared. So where?

Amrita Bhasin · Answer

Okay. We know that if we&#x27;re looking at this graft, we have a ship that looks like this, and this is on the X axis is You can see so four coordinate points we find the vergis is zero car 02 times zero plus zero equals 01 comma 02 times one plus zero equals two to calm itude two times two plus two equals sex. Zero coma for two times zero plus four equals four. Now we know the minima ax. Come. Why is when X equals zero y equals zero and has a value of zero? No, The minimum for X equals two y equals two has a value of sex. So the answer is the men is zero, and the max is sex for this question.

Jack Chen · Answer

eso in this program. We know that, um, for Patty we know that. And less recourse to t lesser because to capture him, eso if we step track this lower, Um, case in for off these three terms. So we have zero there. So he calls to t minus AM Yes. So he caused to capture and miners him? No, it divided by ah, this term for this three. Since this term is known active, so we don&#x27;t have to change the direction off BSE equality. So we have the oral Ezrin t minus and over capital and minus him. This link was toe. What? So if we just let our for you close to this number, we have t closed toe one minors off our times lower case and plus off our times kept William, which proofs the first statement in the for a second time. And if we let X equals to root off one minors off our times, you Zaban plus route off off our times to use that one. Um so x chance post times X equals two. Um, since the chance poses linear. So we have what route off one minus. Oftentimes use transposed time plus route off fraud times You one chance post times himself. And don&#x27;t forget that we have one condition Here you&#x27;re seven. So we were right out here since you seven is perpendicular to your supple one. That is you, Saban transpose times Use of one equals 20 eso This product equals two one minus often times Use your stubborn on chest post empty self plus our front times Use a bed. I use a bunch Is post times years of one This is both vectors are unit vectors So we have one minors off our process. For us, it&#x27;s one. Um, if this product is one, them is x defined in this way is also a unit vector. Now X transpose a X equals two Root off What mine is Lambda U N transpose plus route of land. You one chance force times a times myself since, um, you won again value I give actors with Begum Elio kept her in the lower case in. So we have or minus Landrieu and Chase Post, plus our route off Alfa you one chance goes times route off one mine is lambda So eight times User Bernie, close to um lower case in times the self and, uh, eight times you want enclosed the opera upper case. In times you won again. We used effect of that. You end chance post times you want because zeros away have one minus Lunda and plus off a one minus afra him plus Alfa kept him. And the by our definition before this equals to keep which proofs are last statement.

Problem

Find the maximum value of $Q(\mathbf{x})=7 x_{1}^…

Problem 8 Easy Difficulty

Answer

$\mathbf{u}_{1}=\mathbf{v}_{1} / \sqrt{5}$
$\mathbf{u}_{2}=\mathbf{v}_{2} / \sqrt{2}$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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$\mathbf{u}_{1}=\mathbf{v}_{1} / \sqrt{5}$$\mathbf{u}_{2}=\mathbf{v}_{2} / \sqrt{2}$

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Caleb E.

$\mathbf{u}_{1}=\mathbf{v}_{1} / \sqrt{5}$
$\mathbf{u}_{2}=\mathbf{v}_{2} / \sqrt{2}$

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications