let-a-be-an-m-times-n-matrix-such-that-at-a-is-invertible-show-that-the-columns-of-a-are-linearly-in

Question

Let $A$ be an $m 	imes n$ matrix such that $A^{T} A$ is invertible. Show that the columns of $A$ are linearly independent. [Careful: You may not assume that $A$ is invertible; it may not even be square.

Angelo Rendina · Accepted 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

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

Griffin Johnston · Answer

suppose I had a square matrix. A So a is an in by in matric it has the same number of rows as column. And suppose also that the road space of a is equal to the null space of a no space of a And just as a quick reminder than no space of a is all the vectors x the set of all the vectors x such that a times the vector X is equal to zero. It&#x27;s all solutions to the homogeneous equation with this matrix. No, we want to prove in this video that a prove a the A cannot be convertible now, like I talked about in previous videos Ah, good first initial investigation for proof problems especially like this one. We&#x27;re trying to show that a or some some some mathematical object in this case AIDS and matrix. We&#x27;re trying to show that the A cannot have a certain property. In this case, we&#x27;re trying to show that they cannot be in vertebral, usually a good first initial approach to kind of get your hands dirty, and sometimes it ends up being the proof. It ends up turning into a foolproof, but regardless a good initial approaches to attempt to prove this by contradiction. So what is our goal? We try to prove something by contradiction. Well, we first suppose the contrary to suppose suppose a can be convertible can be convertible, then our goal, our goal when we suppose this sister derive a chain off implications. So if a driver change implications, what I mean by that is supposed taken, be in verbal without implies something. And that that&#x27;s something implies another something, and so on and so forth and to eventually we this the end of this chain of implications. We arrived at the end of his chain of implications. We arrive at some contradiction and maybe not the end, but one of these statements in here in this chain of contradiction, sometimes it&#x27;s we we stop when we arrive at a contradiction, so it ends up being the end of our chain. But it&#x27;s not necessarily the end of the Cheney. This infiltrated implications could go on forever, but regardless we eventually arrive at some contradiction. And when we arrive at a contradiction, that means that if we were supposed, if we were to suppose a can be in vertebral, then that would imply a contradiction. It would imply a statement that is absurd. That contradicts our original assumptions Are original assumptions that were given to us to be true for this particular problem. And thus this statement right here Supposing a can be in vertebral that can&#x27;t possibly drew a cannot be in vertebral if you were to arrive at a contradiction. Therefore, A Like I says, I said a cannot be convertible and we&#x27;ve proven a statement. So think for a minute what would happen if a is in verbal and make sure we want to leverage the fact that the road space of a in the north face of a are equal. So what would happen if a is and vertebral and we also have that the roast base of a is equal to the no space of a So I&#x27;m assuming you&#x27;ve had a go at it. So when a is in vertebral by the in vertebral matrix here, Um, all right, the in vertebral matrix here, um, there&#x27;s two things that we know. One of the things that we know if that if an matrix is in verbal, then the Onley solution x to the homogeneous equation A time of the vector X is equal to the zero vector. The only solution is the trivial solution. So X equals zero Vector is the Onley solution to this equation the Onley solution to this equation the Onley solution to this equation. But if X is equal to this year vector that that&#x27;s the only solution to this equation. Remember the set of all acts such that access solution to this equation is the knoll space of a So that means that the north face of a the north face of a is equal to the set containing zero vector and that&#x27;s it. That&#x27;s the only vector in the north place of a now something else that thean vertebral matrix theorem says is that if I&#x27;m matrixes in vertebral, which we are supposing a to b in vertical, remember by our by our our attempted a proof of contradiction proof by contradiction, we&#x27;re supposing that a is in vertebral. So if they were to be in vertebral then this will be true and also by the in vertical Matrix syndrome A is row equivalent to the end by in identity matrix And check out what we know we know that if two matrices are row equivalent, then that means they&#x27;re row spaces are equal. So the road space of a is equal, since a is row equivalent to the identity matrix because a convertible by the in vertebral matrix serum again, we&#x27;re supposing it to be in vertebral here, just just to see what happens to see if we arrive at a contradiction. So since we&#x27;re supposing it to be in vertebral, we know by the invariable major theme it&#x27;s row equivalent to the identity matrix. And we know there&#x27;s a theory that says that if two matrices aero equivalent, then that means that there rose spaces are the same. So we can say that the Roadway survey is the same as the rose base of the identity matrix, but the road space of the identity matrix, the in by an identity matrix. I&#x27;ll just write a kind of a scratch example of the identity matrix, so each vector looks a little bit like this, and finally we get to the last vector. In this I didn t matrix the identity matrix, the road space of the identity makers. It&#x27;s the span of all the row vectors now the span of all these row vectors is all of our in and namely this family&#x27;s row vectors for sure has the vector 1000 and so on, all the way to in or in minus one number of zeros. So the span of all these row vectors is definitely not equal to the set containing Onley, the viewer vector. Moreover, the span of all these row vectors actually equals all of our in So So let me let me emphasize this year. So So this span of all these reflectors is all in your combinations of these row vectors. So, for instance, one vector that would be in the row space off a would be the vector 100 all the way up two in minus one zeros bluff 01 all the way up two in minus one zeros, including this first year right here. And this vector is equal to 11 and then zeros for the rest of the entries. And this vector right here the fact that this factor is in the road space of a because any linear combination of the row vectors off the identity matrix, any linear combination of the row vectors. Identity matrix will lie in the span of the row vectors, a k a in the rose base off the identity matrix, which is the same as the space of a So therefore, the fact that this vector the linear combination of the other of the row vectors in this matrix, these two vectors plus zero times all the other vectors this vector lies in the road space of a. But remember, we suppose that the road space of a is equal to the no space of a. That means if the roads base of a sequel to the North based pay we were given this to be true. That means that every vector on the road space of Asia is in the north face of a and every vector in the north face of a is also in the rose face of a. But check this out. I found ah, vector in the road space of a That is definitely not in the north face of a. So the fact that a is in vertebral implies a contradiction, namely, the fact that we found a vector in the road space of a that is not in the north face of a implying that the road space of A is not equal to the North Bay survey which contradicts what we were given to be true. So if we were to suppose that a can be in vertebral, we would arrive at a contradiction at a contradiction. So therefore a cannot be in vertebral and we&#x27;re done.

Michael Jacobsen · 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.

Angelo Rendina · Answer

we want to show that the power off the inverse is the inverse off the power. So let&#x27;s proceed by induction on end, so I find is equal to one. Well, that&#x27;s done. There&#x27;s nothing to show because a inverse is equal to a mers. Now let&#x27;s assume that the proposition is true for some value of n Gayatri golden. One we want to show is that a mers to the power and plus one will want to show the teacher is equal to a to the power when pass one in verse. So let&#x27;s start by writing these as well, a members to the power of end times a inverse just once. Now let&#x27;s use the inductive step right chambers and as eight again in verse, that multiplies a inverse. Now these is just equal to eight times a n in verse where we use the property that the product of the inverse ease off two matrices is the embers off the products in the opposite order. All right, And now, inside the brackets, we can write these as a and plus one, everything humors and a profits done

Pam Russell · Answer

so to be in vertebral, the inverse of a matrix Times itself has to equal the identity matrix. And so we want to prove that the identity matrix minus a is in vertical. So to do that, we&#x27;re gonna take the inverse of this matrix, multiply it by itself, and we&#x27;re hoping that will get the identity matrix. Now we were given that the inverse of this matrix waas the inverse plus a plus matrix, a squared plus matrix a cute we&#x27;re going to multiply that by our matrix. So, using our multiplication, we will take the identity times the identity that&#x27;s just going to give us an identity, the identity times matrix A. But notice that it is theon position of a so we&#x27;ll get negative a the identity times a And then we&#x27;ll have a times the opposite of itself. And that&#x27;s going to give us a square a square times. The identity will be a square, a squared times matrix a but its opposite that will give us a opposite. A cube, a cube times the identity will give us a cube in a cube times the opposite of a will give us A to the fourth Now we can go ahead and reduce some of this and simplify it down, and we get the identity minus a to the fourth. But we were given in the problem statement that aid to the fourth equaled zero. And so if we substitute that in, we get that is equal to the identity. And we have just proven that the identity matrix minus a is in veritable.

Problem

Let $A$ be an $m \times n$ matrix whose columns a…

Problem 20 Hard Difficulty

Let $A$ be an $m \times n$ matrix such that $A^{T} A$ is invertible. Show that the columns of $A$ are linearly independent. [Careful: You may not assume that $A$ is invertible; it may not even be square.

Answer

the columns of $A$ are linearly independent.

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Linear Algebra and Its Applications

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the columns of $A$ are linearly independent.

Linear Algebra and Its Applications

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

Lily A.

Caleb E.

Michael J.

Joseph L.

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

Linear Algebra and Its Applications