suppose-the-columns-of-a-matrix-aleftbeginarrayllla_1-cdots-a_pendarrayright-are-linearly-independen

Question

Suppose the columns of a matrix $A=\left[\begin{array}{lll}{a_{1}} & {\cdots} & {a_{p}}\end{array}\right]$ are linearly independent. Explain why $\left\{\mathbf{a}_{1}, \ldots, \mathbf{a}_{p}\right\}$ is a basis for
Col $A .$

Gideon Idumah · Accepted 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.

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

James Chok · 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.

Doruk Isik · Answer

in this problem, we&#x27;re given a set of vectors. It was represent those this B and we&#x27;re set up. These factors are the news independence and be half in number of factors and barely needle about independent in end dimensional spice. So it be work you former system out of these wreckers, Uh, you would have a number of factors intimations by sword on the cooling system will be an n by n square metres so that we would have given element for vitro since the number of columns is equal number rolls. And since b vectors are the nearly abundant and since you know that number of columns is think of a number of roles so that this given, ah, system given set up back to roll spend space. We can combine this and say that this given system is actually a faces or, uh, the n dimensional space

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

Problem

In Exercises 31–36, respond as comprehensively as…

Problem 30 Medium Difficulty

Suppose the columns of a matrix $A=\left[\begin{array}{lll}{a_{1}} & {\cdots} & {a_{p}}\end{array}\right]$ are linearly independent. Explain why $\left\{\mathbf{a}_{1}, \ldots, \mathbf{a}_{p}\right\}$ is a basis for
Col $A .$

Answer

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Anna Marie V.

Alayna H.

Caleb E.

Kristen K.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 2

Video Transcript

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

Linear Algebra and Its Applications

Anna Marie V.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Linear Algebra and Its Applications

Anna Marie V.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Anna Marie V.

Alayna H.

Caleb E.

Kristen K.

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

Linear Algebra and Its Applications