suppose-t-and-u-are-linear-transformations-from-mathbbrn-to-mathbbrn-such-that-tu-mathbfxmathbfx-for

Name: suppose t and u are linear transformations from mathbbrn to mathbbrn such that tu mathbfxmathbfx for
Uploaded: 2020-12-11
Duration: 5 min 25 s
Channel: Numerade
Description: Suppose $T$ and $U$ are linear transformations from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ such that $T(U \mathbf{x})=\mathbf{x}$ for all $\mathbf{x}$ in $\mathbb{R}^{n} .$ Is it true that $U(T \mathbf{x})=\mathbf{x}$ for all $\mathbf{x}$ in $\mathbb{R}^{n}$ ? Why or why not?

Question

Suppose $T$ and $U$ are linear transformations from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ such that $T(U \mathbf{x})=\mathbf{x}$ for all $\mathbf{x}$ in $\mathbb{R}^{n} .$ Is it true that $U(T \mathbf{x})=\mathbf{x}$ for all $\mathbf{x}$ in $\mathbb{R}^{n}$ ? Why or why not?

Ahmad Reda · Accepted Answer

in problem 37. Given that that he transformation off that transfer, that you transformation for the victor X equals the Victory X for all X in our in and supporting that he and you our linear transformations from our end to our end. We want to see that if that the U transformation off the tee transformation of the victor Victor X equals X. This is the question we can conclude that from this fall we can rewrite it in the algebraic for any transformation off a function. For example, why equals the transformation off any victor? Why, for example, equals the standard metrics for that transformation multiplied by the victory, Then tee off any function see off any victor equals the standard metrics off the transformation t while deployed by the victim, New X equals X and using the same concept. That transformation off any vector Z, for example, equals the standard metrics for the U transformation. You can assume it. It&#x27;s to be want to blow by the victories that then we can trump make it in algebraic full metrics a multiplied by the metrics be multiply by the victory X equals X. We can see that a B multiplied by the Victory X equals the Victory X, which means that a B equals the identity metrics. And now we can see that the metrics A has an inverse, which is B because when we multiply each other, we get the identity matrix. This is the definition off the embers, which makes which means that A and B are convertible but receives well and be or in veritable. But this is then A and B are a vertebral mattresses, and they are in veritable mattresses because they both are square mattresses square but receives because the transformation is from our end to our end, which means that the standard metrics is end by any and here be is MBA in. Then we have the identity matrix in This is the conclusion from the given. Let&#x27;s see what&#x27;s required toe be proven. We want to to see if this is true. Let&#x27;s take the left hand side. The left hand side means that we have the standard metrics be multiplied by the victor off the transformation TX equals X, and from this assumption from this form, we will transform toe a job for which makes be multiplied by A which is the standard metrics for the transformation function. T the driver Victor X equals X not it&#x27;s not equal X because we want toe. See that if it will give us X, then we have b equals p Multiply by t equals B multiply by a X and we know that e b r in vertebral mattresses and they are the inverse of each other. B is the inverse of a and A is in front of me which makes b equals oy equals I and multiplied by X equals X and it equals the right hand side, which means that this statement is true and this is the final answer of our property.

Runpeng Li · Answer

problem. We have no transformation from our end to our end. Now we&#x27;re given the assumption that, um, tur t u is equal to TV for some distinctive actors view. And you wanna be so or son, you nice to be We have t view you cross TV. So that means this lean your transformation is not 121 by the commission off went to one transformation and further. Um, since we know linear transformation, T can be expressed as matrix a times the vector X and that way, uh, by our convertible matrix to your room. The negation Because, uh, Lena transformation is not one to why he&#x27;s actually the negation off. Art has off our in veritable Mitchard&#x27;s Kira. So it just applied it r f but its indication. So this means since we have, um, indication enough off after convertible Mitri here. Um, so this means a is not convertible. Okay? And that implies, um, telling your transformation X I&#x27;ve been to a axe. Is not I&#x27;m too. This this is the vacation off heart. Um, it&#x27;s a part. I bye aren&#x27;t high. Uh, I am t So that means that Lena transformation t cannot be. Cannot be on thio. Um are into our in. So that&#x27;s our final conclusion

Gideon Idumah · Answer

here one is should&#x27;ve sits s equaled Sue X In our end, Given that f off ex isn&#x27;t see, he&#x27;s on fine subsides off our end. So let&#x27;s p com a Q B in s to show that this is an offense off subsets we need So sure that&#x27;s one minus tse more to abide by p lost See more to ply My q isn&#x27;t s four and you really number now to show us something is an s by different by definition of s we need to show sure dots f off eggs because if anything is an s, then I miss f off one minus c p plus c Q Most be in C. So now let&#x27;s show that this isn&#x27;t see so off f off one minus c be plastic You will be equal to because we know that f is a linear transformation. So this we can break this down to f off one minus c p clothes F off c Q. This can also be broken down. Sue f off I can ride is us p minus C p less. I can pull out it see because it&#x27;s Lena transmissions after c f off cue this. I can write us F p minus a comm plotted C F P close C F off Cuba and then we can ride these oz one minus c Pull off the F off P lost C f off Cue knows things become a Q Isn&#x27;t s dune by definition of s it&#x27;s implies that&#x27;s f off p comma f off. Few isn&#x27;t see because that is our s is defined. No scenes, See is our fine subsets off our M? This would imply by definition, off Alfonso sits. This implies that they are fine. Combination off f p on f off cue use who seem on this is an offering combination off f o p N f off cubes. If you combine these, this would imply dots. Bizweek implied dots one minus c fo being lost C f Q using C, which is what we wanted to show. If you go back to the force Page 12 should this which we have shown. So this implies that&#x27;s this is, um I&#x27;m fine. Some sense

Runpeng Li · Answer

We need to show that, as is a linear transformation and were given the condition that, um, t and s satisfy the credibility creations and tease and enough inspiration. Okay. How? Just write down our goal here. You near transformation. Okay. All right. He ordered to do that. We take arbitrary, take any you and B from our in. So we pluck. Are you gonna be to transferred to the ingenious information as we will have X? And why accident. Why, Okay, now we consider t off acts. Now, remember, acts can be represented as ass off you. So plugging as self you now, since ass is it, it&#x27;s the number. As an S and T said it&#x27;s by the embers in credibility questions. So this is this keeps us directly back to you. So similarly, we have, um, t apply is factor VII. Okay, so now we consider ask you plus. Yeah. So since you plus V can be represented as a transformation of x and y under T, So that will be t ext plus t y. Now remember the inner transformation by our assumption, the transformation transformation T is a leaner transformation so we can apply the property. Help me near says Formacion here at his t x s t y t X plus y okay. And by our convertibility equations, these keeps us express What? And recall that X is s you And why is SP so that ask you Class sp and our first condition It said his buy. So the next condition we need to check it&#x27;s whether s off the escape, their off vector under the interview and then you&#x27;re under the transpiration Information off ass will be the same as the skater times times the transformation off s so that is directly to show that c o s u because s self see you as I see you, that&#x27;s it. And to do that first consider considered TLC X because he&#x27;s a linear transformation. So it&#x27;s just a tee off a sea of t x c times t x and T acts by our previous by our previous reasoning here, T X will be you. It&#x27;s C times. You okay? So s off. See, Time&#x27;s view can be written as as soft tee time. See x right, which is a substitute c times you by. I don&#x27;t trust for Michel t so That&#x27;s, uh, tee times tfc times eight times X. Now we can we can apply the convertibility equations, so this gives us see X. Now, remember, ax by over by this equation here, acts can be represented as ask you so at a same as C times Ask you so as I see you is C times as here. So we&#x27;re done. That means ass, he said.

Ibrahima Barry · Answer

you function d the linear transformation We have to prove first property of a linear transformation and second property of aware 1st 1 you Victor me, I think quite women. And we&#x27;re gonna take the dirt of each recording of defector the body. When you see this is there were you of the first component of you plus first input into the up to the spirit of the first complaining of the loss, complaining a year plus the dirt off the lost input into the this right here. Then you see that this is just de times the derivative of you, plus de times the this proves the first property of a living transformation. We need a same thing for probably two. We have C signs vector you so we don&#x27;t like each element of the vector you by sea and we take the turret of each element In fact, up he because we take the road of seeking doctor out constant the S e c times the spirit of each element inside selector A second property of a Libyan transformation showing properly wanting to you. So he&#x27;s clearly a linear transformation

Rakvi . · Answer

So, first of all observer that he told the one is a transformation from M in our two women are there is it takes. And anyway, in Matrix and our ports and in very in matrix and you&#x27;re given to you to anyone individually So we just have to compute. What happens for each of them interests is so t two t one off is just even off a and then you applied took even if it so he won off is just a minus a transpose. So this is summon bay in metrics and they do off a matrix is a matrix place It&#x27;s trance fools. So you take this and you add X transpose No transport off some of two matrices. It&#x27;s some off transports off individual metrics so you can open up records like this. A transport a V minus eight transports is just gay transpose minus transport off transports, which is just e. Okay, now, as you can see, this is zero. So no matter what metrics, what matrix You input into t 2 50 when he will get zero. So therefore Kato off even is the zero transformation

Problem

Suppose a linear transformation $T : \mathbb{R}^{…

Problem 37 Easy Difficulty

Suppose $T$ and $U$ are linear transformations from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ such that $T(U \mathbf{x})=\mathbf{x}$ for all $\mathbf{x}$ in $\mathbb{R}^{n} .$ Is it true that $U(T \mathbf{x})=\mathbf{x}$ for all $\mathbf{x}$ in $\mathbb{R}^{n}$ ? Why or why not?

Answer

true

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Linear Algebra and Its Applications

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Absolute Value - Example 1

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Linear Algebra and Its Applications

Grace H.

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Absolute Value - Example 1

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

Grace H.

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

Linear Algebra and Its Applications