suppose-a-is-an-m-times-n-matrix-with-the-property-that-for-all-mathbfb-in-mathbbrm-the-equation-a-m

Question

Suppose $A$ is an $m 	imes n$ matrix with the property that for all $\mathbf{b}$ in $\mathbb{R}^{m}$ the equation $A \mathbf{x}=\mathbf{b}$ has at most one solution. Use the definition of linear independence to explain why the columns of $A$ must be linearly independent.

Michael Jacobsen · Accepted Answer

in this video, we&#x27;re starting off by assuming that the Matrix equation a X equals B will have at most one solution for all vectors, be that we might choose. Let&#x27;s take some logical steps to see where we can get from this statement. Well, if this works where we have at most one solution for all be, let&#x27;s let be be equal to the zero vector. It should work as well. For the zero vector by the statement, then a X equals zero vector has at most or at most is probably the most important key word here. One solution. So we know X equals zero vector is going to be consistent and we&#x27;re only going to get to one solution. But we already know that the Vector X equals zero vector is the trivial solution to this equation. So because we have a homogeneous equation with those euro vector here, placing as euro vector in this position gives us a solution which we call the trivial solution. But that means we have found one solution and there is only one solution because of the word at most. So let&#x27;s say what that means Next were right to therefore a X equals zero vector has on Lee the trivial solution thanks to the at most statement here again, probably the most important statement in this entire problem. No, when we say X equals of zero Vector has on Lee the trivial solution that gives us another conclusion immediately about this matrix A. We have now that the columns of a are linearly independent. So we&#x27;re able to say a very strong statement about the columns of a really, just from this information, which is quite surprising.

Jimmy Yao · Answer

don&#x27;t just the problem here we want to argue are a Is that many independent? So So there The corner of a linear independents is equivalent to say a s incredible. So it is actually a is inevitable if, in the only if agents pose a it is also you, Bertel. So they said speakers If we see this terrorist in so the so so here for this demolition If a the colon is they&#x27;re Carlin&#x27;s of a it&#x27;s linear independence. Then we know a is you&#x27;re irritable. So if a is the convertible and so we also know a transpose is inevitable So this is because the rink of a it&#x27;s ricotta rink of transposed So yes, so they have the humors so safe. But he knows the U S of a transport be ancient spell beavers and the humorous of a will be a uber&#x27;s. So we say a transpose Huber&#x27;s and a universe we&#x27;ll be the humors of a transpose eight. This is because it&#x27;s their time. Together, it&#x27;s echo to a uber&#x27;s in a transpose universe, a transposed in a and become to a transpose i A. If you go to a inverse, a issue could you buy. So this means it is the inverse matrix of a chance was eight. So agents pose a is convertible, so for two this direction so we know a chance. Close A. It&#x27;s the Invincible due to their rink knowledge t serum. We know that the rink of agents pose a trust that not a team of aid you This chance was a It&#x27;s Nico to in the hair agents will agency in veritable So we know this rink. It&#x27;s because you in so the knowledge he should be sure that you come to narrow So we know if agents pose a eggs. If you co 20 you have to have X You cut your hero, right? They were going to consider the now space of eighth. So for the now space of a so it looks like it&#x27;s satisfies a X equal to zero. So we choose any experience now space okay thing than now of age. So this aches had a property. A ex coaches there. So also, if we transposed, there&#x27;s we&#x27;re gets shows X transferal agents bill should also be there. Oh, if we time that to them together we will get X transport a gentle a X if you cut through there. Mm, No, I don&#x27;t want to do that into this is enough. So if X it&#x27;s lifting this now space a X is equal to zero and we times a transpose to two this vector. So we know agents will A X is equal to zero since we already know if there now space for a transpose a is contents in the their oh, so we can derive that extra butte there. So if exit zero then we means this means this now space for any the X in the snow space We can deduce that X is zero. So this means there&#x27;s now space only contend content zero So according to their well so well, I don&#x27;t want to do that. So for any So we know there no space here, so only condemned zero. So the knowledge t for a it&#x27;s their so called into the rank nal ity. Mr Room. We know the rank of a it&#x27;s in. So here we know the the colon, the colon that&#x27;s a his denia independence

Runpeng Li · Answer

problem. Um, we want to show that if a X equals zero has only the trivial solution, she will solution. Then way. Want to show Dad? Hey, X, Because be must I have this ocean for every Be okay. Now, The way to do it without the convertible matrix serum is to observe the structure off A. So we consider let&#x27;s consider the reduced row ish in form off, eh? Say a prime. Okay, so scuse me. So that means, um, a prime times X equals zero. This system has has no pre variable because this system has has only has only intriguing solution that it means the system has no free bearable. Because we we know from our previous material we know that, uh, the system has nontrivial solution. It has at least a work repairable. So he&#x27;s just a negation of the statement. So has I&#x27;ll just write down here just for your information, A prime axe times. They&#x27;re a prime axe equal zero has has on Lee Trivial solution. It&#x27;s equivalent to say that this system has no free bearable. No, for you bearable. Okay, So for the Moore Demi&#x27;s matrix A has, um, has an people in positions. Of course. We&#x27;re assuming Matrix a had Matrix A&#x27;s On By and Square matrix. Yeah, this is our assumption. Um, people positions. Excuse me, people. Positions. And this is our conclusion here. Now, from here, we can say, um, since there are n people, people in positions and the matrix structure off, eh, Um, or we consider a prime and B and be seen as some some some values on the diagonal and zero for the lower triangular. An appetite for dinner. We have some numbers. I don&#x27;t. I just, uh, using darts to represent them. So, hey, in terms off a a prime time zags, we can see that of all the terrible here we are using a, uh, using this form of a prime then our nest our last variable se X end must be determined by the value we&#x27;re taking for B. So that means So we&#x27;re saying be one. So let&#x27;s suppose this is our vector B until fiend is equal to a prime times axe. And this is some sort of early on the diagonal times the vector X. So that means we were supposed defector ads his ex Juan x two txn Okay, So that means the last the last term Other day that I&#x27;ll be there diagonal If we multiply, multiply this This value by the nest. Last entry off. This factor acts. We have some. Aye, aye aye. Which is the entry for the last in the center of the diagonal of ape. Right? So? So there&#x27;s a II times the last the last available off the vector acts, which is X end is exactly the end. So we can solve for accent here. So that is bien divided by a II and so on so forth. So on so forth we can solve for everything Go bearable x one Until we were done, we have done for X one because, um, there&#x27;s no free bearable. So that&#x27;s why we can do this. So that means xn. That means acts the vector must exist. So what that means for the more damages the system a axe? It was B has a solution or every single b we&#x27;re taking the knob should be from are in

Angelo Rendina · Answer

we begin with the observation that the columns of a are literally independent, even a leave the rose off they transposed ordinary, independent. That&#x27;s simply because the rows of a transport are the columns of a because that&#x27;s how transports works. Now let&#x27;s find an elementary metrics, eh? That reduces a transpose so hee, a transposed, is reduced by Rose. And we know that the elementary mattresses are in veritable. So let&#x27;s keep that in mind. Okay, Now we show that the columns of a and therefore, well, the rose or transposed our little independent by contradictions. So we assume they&#x27;re they&#x27;re not. So if the roles of a transfers are not literally independent, there means that when we reduce a transport by Rose and we look at the very last row So let&#x27;s say the end raw off e a transposed, we get a zero. So these are absurd hypotheses, and we show that if we assume that this is the case, then we have a contradiction and therefore the rolls off they transposed must be literally independent. Now with these absurd assumption, we look at the last So the end raw off the square metrics e a transpose A. But we know that the role of a product is the last throw off E a transpose That motive lies, the medics say, but now by Apophis is the last drop off E a transpose is just zero the multiplies they therefore and so we get zero. So the last throw off the m by n matrix e a transpose a zero. But that means that e a transpose a is not in vegetable because it has an entire row of zeros. But I remember that he was in elementary metrics. So Hee itself is in vertebral and that means that the problem lies in a transpose a so a transpose, a square matrix which is not in vertebral. But this is against our assumptions because we assumed to begin with that a transpose a is in vegetable. And so the only problem is are absurd. I bought it Is that the rose of a transpose are not linearly independent

Gideon Idumah · Answer

nine. This question we have that we given that they warn AP our Linda Lee Independence. So we&#x27;ve given that. So we have these also recall. That&#x27;s cool moms off A It is these columns off, eh? Ponds? The column space off a ready off. That&#x27;s so therefore by definition of a busies probate busies. Because we have that&#x27;s it is literally, literally independence. And we have that. It&#x27;s also spawns the column space off a. So this would imply that Hey, Juan damping he&#x27;s a busies four column space off.

Runpeng Li · Answer

Okay, so in this problem, work, even new tricks, eh? Which column? Which columns? Already there. You know the independent. Oh. Which columns are you? Nearly independent. So what does this mean? Well, I over on our convertible Matrix Spear. Um, So what this means hey, is convertible. And there again by our incredible matrix fear. Um, there must exist. Matric. Uh, I&#x27;m back. And make sure you speak such that a times B his identity? No. Now we can see there a squared. So that is Hey, time&#x27;s a So we consider it&#x27;s cedar a times, eh? Times be times be And we consider this when we remember the matrix Multiplication is associate social social TV so we can calculate the time speedy inside first so that it&#x27;s a times he times v will be I didn&#x27;t hear So that&#x27;s ight nd times be and I didn&#x27;t he times a is just a can a time speeds identity. All right, so that means netting plies a squared that implies he squared ties b squared. He&#x27;s I don&#x27;t know. Furthermore, be square is in verse of a squared, so that means a square things is inevitable. And by our incredible matrix fear. Um, if we have a murder mystery stuff than a column of the squared must span our end. So a square conclusion is a square, you convertible and columns all those spin are in, so

Problem

Suppose an $m \times n$ matrix $A$ has $n$ pivot …

Problem 39 Hard Difficulty

Suppose $A$ is an $m \times n$ matrix with the property that for all $\mathbf{b}$ in $\mathbb{R}^{m}$ the equation $A \mathbf{x}=\mathbf{b}$ has at most one solution. Use the definition of linear independence to explain why the columns of $A$ must be linearly independent.

Answer

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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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.

Catherine R.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications