suppose-columns-135-and-6-of-a-matrix-a-are-linearly-independent-but-are-not-necessarily-pivot-colum

Name: suppose columns 135 and 6 of a matrix a are linearly independent but are not necessarily pivot colum
Uploaded: 2019-08-09
Duration: 49 s
Channel: Numerade
Description: Suppose columns $1,3,5,$ and 6 of a matrix $A$ are linearly independent (but are not necessarily pivot columns) and the rank of $A$ is $4 .$ Explain why the four columns mentioned must be a basis for the column space of $A .$

Question

Suppose columns $1,3,5,$ and 6 of a matrix $A$ are linearly independent (but are not necessarily pivot columns) and the rank of $A$ is $4 .$ Explain why the four columns mentioned must be a basis for the column space of $A .$

James Chok · Accepted Answer

supposed columns 135 and six of the matrix, a linearly independent and the rank of ace for Explain why these four columns must make a basis for the common space off, eh? Well, we know that the rank off a is equal before. So what this means is that the column space on a is equal to you are full. Now, what the basis do tells us is that any linearly independent sets off exactly p elements in hey H automatically is a basis for her age. So in this case, we have four literally independent sets which are in the which are in the column space off A and so therefore they must automatically make a basis for college face off.

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.

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.

Mike Verwer · Answer

okay here were asked to show that if the columns of b r linearly dependent, then so are the columns of a B. And so this is ah, approved question. Have to prove this statement. Um, but this proof is really just unpacking a lot of unpacking and a lot of knowing the definitions. So we&#x27;ll start with the idea of the columns would be our linearly dependent. So that means that, for example, this column here, this first column can be represented as a linear combination of all of the other columns. It&#x27;s in particular. It means that this single element here is a linear combination of these elements and sing with this element in these elements and so on. And so what that means is, if you write it out, for example, be 11 is equal to some constant C two times be one too, plus some constant C three times be +13 plus and all the way up to see um and be one end. And this will be the same for B. Thio. Ah, one will be equal to again the same. It&#x27;ll be the same constant, actually, because this entire column Ah, this entire first column is only a combination of all of the other entire columns. So this constant seats, who will be the same in each in each expression of of these elements be Jay. I urge a one. So be to one we&#x27;d be equal to see two times be to to plus C three times be 23 and the same all the way up to C and B to end. And this will continue in this manner. And so we go all the way to the bottom two b one b m. One week will to see to e m too, plus C three b m three and all the way up to C and B m n. Okay, um, I&#x27;m gonna go to the next page here, but we&#x27;re gonna remember these equations, Okay, because now we&#x27;re going to use the fact that the columns of a times B are linear combinations of the columns of a using the corresponding element or ah, using this course wanted. Call him a B as the weights in the linear combination. So, for example, the vector a B one will be equal to, um, be one one times a one plus be to one times eight to the column. Vector eight to and so on. All the way up Until be Mm M one a. M. Okay, but now we&#x27;re going to sub in what we found. Where the equations of these e entries. This calling these entries in the in the first column of B. So this is gonna be ah, very long. Some. So I&#x27;m gonna break up in the lines. Eso 1st 1 will be multiplied by the vector. A one. And it will be, um, see too. Be one too. Plus C three, I mean 13 right? All the way up until C n be one end. And this will be plus the column vector a two time see too be to to plus C three be 23 see and be to end Plus and then all the way until we get to a mm time. See, too e m too. Plus C three since b m three all the way up and so see and b o. And this is yet and it read e mn. Okay, So this is all one big sum you&#x27;ve just subbed in our equations from before and Now what I&#x27;m gonna do is I&#x27;m going to multiply in the column vectors. I&#x27;m going to distribute the call in vectors into this product here. I&#x27;ll do that on the next page again. Pardon? Because this gets very long. So again, we&#x27;re still looking at the first column of The Matrix a Times B and we&#x27;ll distribute in this thes Colin vectors, so that will be equal to C two. Be one too a one plus c three b +13 a one plus that dot CNN Be one n times a one plus right. That&#x27;s the first line, Uh, plus C two. Me, too, too. A two this time. Let&#x27;s see. Three. Beat 23 still Times Column Vector eight to all the way up to see n to end a to again. And this will continue until we get to the last one, which is seats you b m 2 a.m. Let&#x27;s see. Three be mm. Three a three. All the way to see. And that&#x27;s B M and a M. Okay, so now what I&#x27;m gonna do is I&#x27;m gonna factor out these. See, I cz As you can see, I kind of drew them in a nice call him here intentionally. So when a factor those out and we&#x27;ll see what we get left with. So again, I&#x27;ll use yet another page to do this and the result of that well, look like this. So we see too times all of be one end a one plus B hoops or be one to not be one end. You write that. Sorry. Be one too of times a one plus B 22 times a two plus and so on until B and ah, yes, or B M too. So I was AM and the next line will be C three times all of B 13 times a one plus B 23 times a two plus all we have to be m three a. M. And this will continue. So we get all the way down to CNN Times, all of B one n a one plus B two en tons, eight to all the way up until be. Mm. And since a m, I don&#x27;t know why I drew parentheses there. They&#x27;re not necessary. Okay, But now if we take a look at this year, this piece in the brackets is exactly a B two and this in the rackets here is exactly maybe three. And so what we found is at the first column of a B one, right, Because this whole analysis has been on the first column of Baby One is equal to C two times, maybe two plus C three times a B three all the way up until C n. So I&#x27;m just a B N. And this is exactly the formula. Or this in this show is exactly that the first column of A is a linear combination of the first column of a B one or a baby is a linear combination of the other columns of a B. And so we&#x27;ve shown what we intended to show and we&#x27;re done.

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

Donald Albin · Answer

All right. So prove from definition 4.5 point four that if v one v two So this is a set containing V one v two dot dot dot v an is linearly independent. Um, if they are linearly independent, then and I didn&#x27;t finish reading it, but anyway, I&#x27;m gonna work on this linearly independent thing for now. If those air linearly independent, then um d one v one plus de two v two plus dot, dot dot plus de and the and does not equal zero for ah d one d two that ah de and that are rial numbers. Okay, Now, um, and if a is an in vertebral n by n matrix than the set a V one a V two that thought all the way up Teoh a V an is linearly independent. So we&#x27;re gonna prove this by contradiction. Let that be then nearly dependent. Okay, so we&#x27;re trying to prove that that is linear Lee independent that that set. So we&#x27;re going to start with the contradiction with the opposite statement that the contradiction with the opposite statement, which is that, um, that is the nearly dependent, If that is linearly dependent then some see one times a times V one plus some see two times a times V two all the way up to some. See an times a times V an must equal the zero factor and noticed that Ah, C one is just a constant so we can switch this around and just write a time. See one V one plus a time See two v two. Plus that that that plus a CE envy and must equal the zero vector noticed that a is in every term. So a see one V one plus c two v Teoh that dot c N v n must equal zero of actor. That&#x27;s not very neat. If we take the in verse of a times both sides than the left side is still going to be the zero vector. But the inverse of eight times a is the identity matrix and the identity matrix times Another matrix is just itself for some c one c two see, and that are rial numbers. But that contradicts something that we know to be true here and here are contradictory and therefore a V one a V two God, I don&#x27;t a v n are linear Lee, independent

Problem

Suppose vectors $\mathbf{b}_{1}, \ldots, \mathbf{…

Problem 26 Easy Difficulty

Suppose columns $1,3,5,$ and 6 of a matrix $A$ are linearly independent (but are not necessarily pivot columns) and the rank of $A$ is $4 .$ Explain why the four columns mentioned must be a basis for the column space of $A .$

Answer

By the Basis Theorem, this set of four vectors is a basis for the column space.

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

Linear Algebra and Its Applications

Alayna H.

Caleb E.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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By the Basis Theorem, this set of four vectors is a basis for the column space.

Linear Algebra and Its Applications

Alayna H.

Caleb E.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Alayna H.

Caleb E.

Maria G.

Michael J.

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

Linear Algebra and Its Applications