suppose-a-linear-transformation-t-mathbbrn-rightarrow-mathbbrn-has-the-property-that-tmathbfutmathbf

Name: suppose a linear transformation t mathbbrn rightarrow mathbbrn has the property that tmathbfutmathbf
Uploaded: 2019-08-06
Duration: 2 min 53 s
Channel: Numerade
Description: Suppose a linear transformation $T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ has the property that $T(\mathbf{u})=T(\mathbf{v})$ for some pair of distinct vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^{n} .$ Can $T$ map $\mathbb{R}^{n}$ onto $\mathbb{R}^{n}$ ? Why or why not?

Question

Suppose a linear transformation $T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ has the property that $T(\mathbf{u})=T(\mathbf{v})$ for some pair of distinct vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^{n} .$ Can $T$ map $\mathbb{R}^{n}$ onto $\mathbb{R}^{n}$ ? Why or why not?

Runpeng Li · Accepted 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

Ashley Boni · Answer

So we&#x27;re given that the vectors v one v p. Span are and tease alone your transformation on We want to show that teest zero transformation. So, um, if X is any vector in on our end t of X is equal to zero. So if X is been any vector on our end and all the vector of these span our end, we can write our vector X as a linear combination of the V&#x27;s. So, in other words, um, just for some Constance C, we can write X equals C one V one plus C to be to all these reserve vectors plus dot, dot dot Ah, C p VP. So now let&#x27;s take t of both sides. T of X is equal to t of C one V one plus up to C P V P Tease Lanier s so we can pull Constance out. So we have C one t of V one close that dot dot c P t of v p, and were given that each individual t of the eye is equal to zero for I goes from one to pee. So from star, which is ah this given Here we have C one time zero plus c two times zero plus all the way up to C P Times zero, which is just equal to zero. So therefore t of any vector X is equal.

Henry Carnick · Answer

so to solve this problem, I implicit out the properties of the enterprise that we&#x27;re going to be referencing in our solution. So the first thing you do is one you need to show. T of a post TV is equal to t of a plus B where A and B are factors. So by definition of our transformation t of a us to you being is equal to the inner product you want common Be okay, comma the enterprise you two comma a plus the inner product you want comma, be comma the inner product. You too. Com A B So the thing to know is ah in the first component of each term. If we look at their inner product, first term of their inner product is both you want, which means my prop free. We can add these two terms in the second and in their second component while fixing them in the first component. And then it&#x27;s also true for the second component of each of these terms. So what this means is that keep property three. We get, uh, enterprise. You want comma A plus B palla in approx you to come on a posting which is of course, equal to tee off a plus B. And so the second thing we need to show is that C times t of A is equal to t of C times A where a is a vector and C is a constant. So what this means, by definition, definition of transformation is that c times t of a is equal to see Time&#x27;s the in a direct Monica you want common a comma, the inner product of you to come at now by P two proper to We know that, for instance, you want common A is equal to the enterprise A comma you want and the same is true for you to come in a. So this is actually equal to see times the inner product Yeah, a common you one comma see times, you know, product a Colin. You, too. And then by P to we could distribute the p one we could distribute. And so we end up with see, Time&#x27;s a common you one and comma the proxy times a college you, too. And then by P to we could flip these in a product that 19 equality and we end up with you in a product you want Colisee times a come on in a product, you too gonna see times A which is, of course, equal to tee off, See times at.

Michael Jacobsen · Answer

in this exercise were given a set of linear dependent vectors in R N Let&#x27;s start out by writing just what it means for this set of vectors to be linearly dependent, it means is that what this means is there exists scale. Er&#x27;s x one. Let&#x27;s go with C one rather see to and C three, which are not all zero such that the following linear combination is satisfied. Where we take C one times V one plus C two times V two plus C three times V three and this will be equal to the zero Vector where we said not all of these vectors air Zero. And that means that this vector equation has a nontrivial solution. Let&#x27;s inject a new element into this situation. Let&#x27;s define T, which will map from the vector space are in into RM. So that t is linear here. The most important statement about T is that this transformation is linear. So now we wonder about that are set of vectors that could be the images of tea that would be specifically TV one TV too, and TV three we know that the set of vectors v one v two v three is already linearly dependent. So the question becomes, What about this set? Should it be linearly, independent, linearly dependent? Or do we have not enough information to tell? Well, one way to determine the outcome here, using our transformation T is to start back up at this equation here. Let&#x27;s apply the transformation t on the left hand side. So that&#x27;s tea of C one times V one plus C two times V two plus C three times V three. Then on the right hand side will use the fact that since T is linear t of the zero Vector is still the zero vector so we can write in the zero back there here. Well, this equation so far, it&#x27;s giving us closer to determining if this set is linear, independent or dependent. But we haven&#x27;t gone further enough. Let&#x27;s begin using more properties of linearity. Linearity allows us to write this left hand side here in the form of C one times T a. V one plus C two times TV, too, plus C three times t at V three and this is set equal to zero vector. Sylvie in your itty allows us to break apart the vector addition as well as the scaler multiplication so that we get to this new equation. Now let&#x27;s recall what we said about C one, C two and C three. We said that they are not all zero, and here they are in this equation. So that means that this vector equation has a non trivial solution. And it&#x27;s also a homogeneous vector equation because the right hand side is equal to the zero vector. Now see what it means for a set to be linearly dependent. This set is linearly dependent now because we have a nontrivial solution to this equation, and so we can now state our conclusion TV one TV, too, and TV three as a set is linearly dependent. So the punchline to this example is that linearly dependent sets are mapped into linearly. Independent sets provided our transformation t is linear now, sadly, we might want to extend this. What if the set was linearly independent? Well, tragically linearly. Independent sets could still be mapped to a linearly dependent sets. So the statement here that this set is linearly dependent on Lee works for dependence

Angelo Rendina · Answer

So we have these linear transformation tea from a vector space v T o r n, and we have, ah interpret well where the finding. I mean a product on V given by the brother of you against V is by definition to you dot tv. We remember that the dot there is in a product that standard in the product inside our end because the U. N TV are factors in our end and so we just need to check that he&#x27;s in a product against your V defined in the vector space. Big B is actually the product, so we need to check the four actions. So what are we, the 1st 1? Which is the symmetry? So we want to check that V against you is by definition tv dot to you and these by the symmetry off the dot product in rn. This is TV at you. Sorry, not TV and all these by definition you against me. So dysfunction on V is symmetric they only to try that it&#x27;s leaner. So x plus y against a vector V We want to check the disease X against being plus wagons being so we use the definition of the inner product. Well, these in a product t x plus y dot d y. And now again by linearity off T and by the charity off the dot product. In our end, this becomes tx dot tv bluff the y dot TV And now again, by definition, did his ex against v blast. Why against me Now for the third property we need to check it is or more genius So CX against V. It is, by definition, tee off, see axe dot tv. But now both t and dot product are linear. Well, that I&#x27;m a genius. So this becomes C t X dot tv and again, these by definition is c times X against me. And finally, we need to check that the dot product, while the inner product is positive definite off. We need to change X against ex. He&#x27;s got their equal to zero so exciting. Sex is by definition, tx dot tx. And now because dot is another product on our end, we know that this is definitely great unequalled in zero. And also this can only be zero if t x zero because the season in the product So did your brother have something against itself can be zero only director itself. Zero. But now remember that tea is want one. So the only victory gets within the vector in V that gets maps to the zero vector. Our end zero is zero factor in we and that means that X must be zero. And therefore we have checked for actions showing that isn&#x27;t a problem is indeed

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.

Problem

Let $T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n…

Problem 38 Hard Difficulty

Suppose a linear transformation $T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ has the property that $T(\mathbf{u})=T(\mathbf{v})$ for some pair of distinct vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^{n} .$ Can $T$ map $\mathbb{R}^{n}$ onto $\mathbb{R}^{n}$ ? Why or why not?

Answer

cannot

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

Linear Algebra and Its Applications

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Kayleah T.

Caleb E.

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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.

Kayleah T.

Caleb E.

Maria G.

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

Linear Algebra and Its Applications