use-the-inner-product-axioms-and-other-results-of-this-section-to-verify-the-statements-in-exercis-3

Question

Use the inner product axioms and other results of this section to verify the statements in Exercises $15-18$ .
$\langle\mathbf{u}, \mathbf{v}\rangle=\frac{1}{4}\|\mathbf{u}+\mathbf{v}\|^{2}-\frac{1}{4}\|\mathbf{u}-\mathbf{v}\|^{2}$

Anthony Ramos · Accepted Answer

Hello there. So for this exercise, we need to show these expression here. So let&#x27;s focus on the right hand side so we can regret this expression in a different way. So let&#x27;s remember that the norm off any vector you is defined at the square root off the inner product off you with itself. Okay, so in this case, we&#x27;re considering the squares off the norms. So that means we&#x27;re just considering the inner product off the factor with itself. Okay, so this so, yeah, let us regret this expression here in terms off the inner products. Okay, so let&#x27;s take each component page component. So first, we&#x27;re going to just take into account this expression here, so this will be equal to the inner product off you plus v u plus V, divided four. But this is just you squared, plus two times you be Plus, So here is you with you on the here is a V to be divided four. Okay, on the other expression here. So this is the first expression. Then let&#x27;s consider the second one. So in the second one is do minus V, divided four. So these he&#x27;s just taking in their product off U minus V with U minus v. Divided four on This is equals to you with you minus two times the product off you with V on plus the in their product off B with be divided four. Okay, so we got these two expressions on. We need to subtract them. Okay, so at the end, we got the following we got you with you. Invited four plus one half off the inner product with off you with V plus the be divided four on from the other expression we got minus you, you divided four. Plus, I&#x27;m going to put this on the next line plus one half you movie and minus dinner product off. We with itself divided for So here is clear that this expression is going to canceled with this expression here on this one. With this one on, these two expressions here are going to sum up on this will become just you, V. That is exactly what we want to show. And that&#x27;s it.

Angelo Rendina · Answer

let&#x27;s combine in just one computation, both the plus and minus case. So the length of you plus or minus V squared. So this is you plus or a minus V against you, plus or minus B. But with these notation, we mean that we either always take the blasts. All we always take the miners now is the linearity in the first argument that&#x27;s you against you. Plus or minus V plus or minus V against you, plus or minus B. Now we&#x27;re still in a charity. In the second argument, we got you against you plus or minus you against V Blaster miners. And now we put in brackets the whole second in their product. So we have V against you blossom minus V against me. Now observe that you against you is just the length of your squared. And then we have plus reminders twice you ve where the two comes from. The fact that both you and we and we and you are the same because of the symmetry. And we&#x27;re something them with the same sign. So it&#x27;s other two pluses or two minuses. So it&#x27;s plus or minus two you ve and then the last term is a plus in green because he re multiplying pastor minors by plus or minus so either that both plus or the boat minus. When we multiply to turn together, you always get plus, So the remaining term is plus V against Visa that the length of B squared an hour done. We just write the sum. And so the plus reminder simplifies out zero, and you have twice the length of you squared, plus twice the length of the square as we wanted.

Angelo Rendina · Answer

you and we form a more formal basis for the vectors pays every considering we want to study the lamps off humanity. So instead of considering the length of your ministry, that&#x27;s considered square. And we know that disease the inner product off humanist be against itself against humanity. Now use the linearity in the first argument so that you against humanity, then minus V against humanity. And now again, we used dealing. Art is in the second argument. So listen, old Wheat in right, the first in the product and we green the second in a product. So for the first in a product and right, we have you against you, minus you against the then we have minus now in green for the second product V against you. And then we have minus minus. That becomes a plus V against V. Now you against you is one. Because remember you Visa Ortho. Normal basis. So you&#x27;re against you is one that the normal part you against V zero. That&#x27;s the orthogonal part Be against you. Similarly, zero. And we against V is one. It&#x27;s all in all this is too. And now we&#x27;re done because the land of you minus V is just the root of these sorrow too

Donald Albin · Answer

I&#x27;m going to prove that the inner product of V and the zero vector is zero. Well, first of all, the zero vector is the same as zero times any vector. And so I can write that one vector times. Another vector is the same as this vector times. This vector that is using, um, property, too. We can reverse the order. Um, then using property three using property three. This is zero times you V, which is zero. And this is V zero. And so if you write this, you might want to start here and proceed in this direction. So what I see here is just using a definition. Then using this equals that that&#x27;s property to this equals that this that&#x27;s gonna be property three and then zero times anything is zero.

Donald Albin · Answer

all right now we&#x27;re going to figure out what the inner product of you plus V and W plus X would be well, first of all, according to Property four, that&#x27;s the same as you w plus X plus V w plus X And then according to what we did in Problem 30 that&#x27;s the same as you w plus you, the inner product of you and X plus the inner product of V and W plus the inner product of the and X.

Donald Albin · Answer

So I need to prove that you the inner product of you with V plus double you is the inner product of U and V plus the inner product of you and W. Okay, so this is very much like, um, Property three, which says that, um So I&#x27;m gonna write property three. I am going to write V plus W comma. You equals you w plus V w. Okay, well, this is the same as you, V plus W because of property, too. So Property three is here, okay? And one property, too is here. And then we can use just, um this is supposed to say you up here. Oh, my goodness is supposed to say that&#x27;s re do this. I think that is very wrong. Um, you double you. All right? I wrote this completely wrong. It&#x27;s supposed to save you and w you all right. So let&#x27;s let&#x27;s keep going. Um, Then we can use property to to just switch these two around. So that&#x27;s going to give us you the plus. You w Okay. So, again, you could write the&#x27;s just like last time in a different order, probably starting here and then writing that and then writing that and then writing that

Problem

Use the inner product axioms and other results of…

Problem 17 Easy Difficulty

Use the inner product axioms and other results of this section to verify the statements in Exercises $15-18$ .
$\langle\mathbf{u}, \mathbf{v}\rangle=\frac{1}{4}\|\mathbf{u}+\mathbf{v}\|^{2}-\frac{1}{4}\|\mathbf{u}-\mathbf{v}\|^{2}$

Answer

$\frac{1}{4}\|\mathbf{u}+\mathbf{v}\|^{2}-\frac{1}{4}\|\mathbf{u}-\mathbf{v}\|^{2}=\langle\mathbf{u}, \mathbf{v}\rangle$

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

Linear Algebra and Its Applications

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Caleb E.

Kristen K.

Samuel H.

Vectors Intro

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$\frac{1}{4}\|\mathbf{u}+\mathbf{v}\|^{2}-\frac{1}{4}\|\mathbf{u}-\mathbf{v}\|^{2}=\langle\mathbf{u}, \mathbf{v}\rangle$

Linear Algebra and Its Applications

Lily A.

Caleb E.

Kristen K.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Caleb E.

Kristen K.

Samuel H.

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

Linear Algebra and Its Applications