suppose-a-is-a-3-times-n-matrix-whose-columns-span-mathbbr3-explain-how-to-construct-an-n-times-3-ma

Question

Suppose $A$ is a $3 	imes n$ matrix whose columns span $\mathbb{R}^{3}$ . Explain how to construct an $n 	imes 3$ matrix $D$ such that $A D=I_{3}$ .

Michael Jacobsen · Accepted Answer

in this video, we have a three by N Matrix. So that&#x27;s three rows and columns, and we&#x27;re going to assume that the columns of a are going to span all of the space are three. Next, let&#x27;s addict something extra to this picture. Let&#x27;s say assume de is a three or N by three matrix, for which eight times D equals I three. So what we want to determine here is what does this matrix de equal in particular? Well, let&#x27;s start with I three. We know that I three is the three by three identity matrix, so it is formed by columns E one E two e three, where e one is 100 E two is 010 anti three is 001 so altogether were then looking for a matrix D, such that when we multiply a times d, we obtain this three by three identity matrix we see displayed here. First, let&#x27;s also note that since the columns of a span or three each time we write a matrix equation eight times X equals B, where be is in are three. This matrix equation is always consistent. This is because knowing that the columns of a spin or three is logically equivalent to the statements that a X equals B is always a consistent system of equations. Let&#x27;s use that fact next to construct our matrix de here, we can say that the systems eight times X equals e one eight times X equals e to eight times X equals e three were e one e two e three are the columns of our identity matrix here. Each one of these systems are consistent, since we could have placed anything in these positions on the right hand sides and they&#x27;re still consistent. Let&#x27;s suppose then that let&#x27;s call this d one de to d three our solutions. So in other words, let&#x27;s say what we mean exactly were saying that since eight times X equals e one e two or three here is always consistent. It follows that a Times D one must be equal to the one. Since this is our first solution. Eight times D, too, is E to and likewise, eight times D three is E three. Now let&#x27;s then define the matrix D By columns D one do to d three, where these came from. The solutions that we had constructed. Then when we take a times D, this is what will produce by the definition of a matrix times another matrix. We take a Times first D one, which came from here. Then the second column of the product is a times D to and lastly, we have eight times d three for the third column of this product. So this is what eight times D looks like. But we also know that eight times d produces e one e two, any three so we could make a substitution here, here and here to get the following. We have the one e two and e three, but this is now the three by three identity matrix. So we&#x27;ve just shown how can to construct the particular matrix de such that eight times d equals I three.

Michael Jacobsen · Answer

In this example, we have a three by three matrix which has been constructed. This particular matrix is not initial inform, and that&#x27;s strictly due to this negative six quantity, which is represented here, highlighted in yellow. Let&#x27;s analyze this matrix further by putting it in echelon form. Once it&#x27;s a natural inform, we will be able to analyze its pivots. So for my first row operation, we&#x27;re going to eliminate this negative six. So let me copy Row One and grow to next multiply row two by three so that we can add it to Row three, the result of B zero zero and one. Now that the Matrix is in its echelon form, we have a pivot here that we can analyze here and here. In other words, this matrix. If we decide to call it a allows us toe show that A has a pivot in every row having a pivot and every row is now very special for this matrix. Because we have a pivot in every row, that&#x27;s logically equivalent. It&#x27;s all right to the bi directional implication, simple to the statement that the columns of a spans are three where the three comes from specifically the fact that we have three rows. So for this particular matrix, we have an example where there&#x27;s a pivot in every row and therefore the columns of this matrix will spend its space are three.

Runpeng Li · Answer

All right. So your problem 30. We need to construct a three by three matrix, not in addition form, but whose columns do not spend our three. So that is actually a beauty, eh? Um, a matrix. That column vector. They&#x27;re the column vectors are they know any dependent? So, too there are many off them, so I&#x27;ll just try. I just try one. Say I have the column, vector y 12 connective one. And we, let&#x27;s say, uh, the second column out due to four and negative, too. And Third column. I&#x27;ll do 36 and negative three. So you can say you can see from from this Matrix, that of the second column is twice off the first column, Third column, third, the third column. He said, um, three times on the first call. So face, make sure excuses three by three matrix, which is, um, clear here. And it is not in the issue of warm, because I have not reduced it yet. And this but this mission is definitely the column. Vectors of columns definitely do not spend our three because we actually we only have one. We only have one vector, and we only have one. Meanwhile, we don&#x27;t have one people position or people call him So So this this column actors doesn&#x27;t spend our three at all.

Runpeng Li · Answer

Okay, so for problem, tell me too, when you to construct three by three matrix A with none. Zero entries and a veteran be in our three, such that he&#x27;s not set Spend by columns of a So he&#x27;s sexually just to construct a matrix A and Victor bees such that the system is inconsistent. So there are a lot of matrices and vectors like this. I&#x27;ll just give give my own matrix and you can try your own parcel So many trees high gave it so that will be one neck of three and two and or one fight and 271 And the vector B, I consider is next. If I 11 elective seven Okay, just to just to show why this, uh, why&#x27;s this system is inconsistent is you just need to consider the argumentative Matrix. So his case, it will be one left of three and two and five. Sorry, negative. Five and four or one high. And 11 271 active seven. So if we if you do the you do the couching elimination and to find out a Rory to station form, then I&#x27;ll just write it down here. So the real redo station form will be 10 17 3rd and zero second. Rory is 01 like 23 13 and zero. So that&#x27;s rule 000 And the one so clearly this term on the third row should not be one. Because we have all zero back already. There is on the left inside. And if we are given any vector, the nef s I will always be zero. But the right hand side is accounts in one, so the system is not consistent in this case.

Runpeng Li · Answer

All right, so in problem 34 suppose A&#x27;s of three by three matrix and A&#x27;s B&#x27;s a vector in our three with the property. That Exodus p D has a unique solution. And we want to explain why the column Self Bay must spend our three. So Okay, so in this case, since we know the leaner system they X equals B is has a unique solution. So that means if we consider the or commended matrix and furthermore, if we considered the rule reduced form off this matrix, it must has has to ruin form like this 81 1 a two to something on the diagonal, diagnose and have some stuff there some stuff here. But it doesn&#x27;t matter. What are they? And we have zeros here, and I&#x27;m on the right hand side. We have B one b two B three as a constant. Okay, So this is the only impossible for this system to have a unique solution because only in this way we can solve a unique by the By the last row, we can solve the unique, unique x three, and furthermore that from the second road that X two will be determined by X three and x one would be determined by X two and x three. So this the only impossible way that we can have a unique solution for this system. So that means by observing this matrix, we have three people, two columns and three people. Columns means, uh, means axes. Um, yeah. So three people columns actually gives me, gives us the information that all the columns because A said three by three matrix And since every every single column is, is that people column So these columns are linearly independent, very independent, and furthermore, they must spend our three because in this case, we actually have three vectors and our three estimation three and those three Rector&#x27;s Arlene Early independent so thes three directors must spend our free so

Patrick Vaughan · Answer

Hello. So today let&#x27;s do a more theoretical type problem. So let&#x27;s show that a arbitrary three by seven matrix of a, uh, has banality of four and it must have a column space. Let&#x27;s show that it must have a column space, um, where it&#x27;s in the, ah, third dimension and is the road space of a then also in the third dimension. So let&#x27;s review our rules. Rank ability zero. So our rank milady here, um, says that e rank of some arbitrary nature. A plus, the reality, the dimension of in all spaces day equals the number of columns. So in this three by seven matrix, we have three rounds, seven columns that we know the number of columns of seven. We know that the utility we&#x27;re assigning that to four and the rank we don&#x27;t know. Yeah, but using this rule, this tells us that the bank of a equals those seven columns minus the formality that were assigned, so that gives us a record three. So let&#x27;s build this arbitrary matrix. And let&#x27;s just assume that it&#x27;s already in produced garage one form just because that makes everything a little be easier. So we have one to you. Three rows. We have our leading one. Our next one. When I went after that, then we have seven columns. So 12 three for five, 67 And we&#x27;ll just fill battle up zeros to make this nice and easy. So this is our three by sudden Matrix. So and reduce little up. Throw up, Sean form The, uh we have the rank is three. So that&#x27;s the number of tipping point 12 three. And then that told us that the column space is the, uh in the reduced Raj lawn form is the Matrix Farm. But we&#x27;re just assuming about this matrix and the original Given. So not only is it released Rachael on form, it equals our original matrix. So our colons stays is actually going to be these three columns because that&#x27;s where a pivot points are. So the column space is the one. Oops, let&#x27;s undo that. See, one equals one 00 see, to equals euro 10 and C three equals euro 01 So we have three doctors in the column space, so this makes it such that the column space all day equals three dimensions because you could just arbitrarily say that this is you&#x27;re x dimension, or why dimension and your son dimension. So yeah, it must have a concept of that. So let&#x27;s done. Look at the road. Space doesn&#x27;t have to be three dimensions. Or is it three dimension? So the road space of our original matrix, But we just arbitrarily chose Do you already be in this reduced row? Actual on form 1234567 Okay, you&#x27;re 10 There&#x27;s one. So the roast space is the non zero rose. So we look at it. We have one hears looking on zero a non zero and a non zero. So the roast space in land such that. And let&#x27;s do this in blue or green. No one equals one 0 to 3 for 567 wrote to equals zero one 03456 seven And then our road three. I&#x27;m sure you can see where this is going. No. Is zero euro one, 4567 So, again, we can look at this and say that this is our excess. Why space? And once again we see that Yes, indeed, for this three by seven matrix that we arbitrarily chose. That the row space zooming back in. Now we don&#x27;t need to see that the row space, okay does indeed show up in three dimensions.

Problem

In Exercises 27 and $28,$ view vectors in $\mathb…

Problem 26 Hard Difficulty

Suppose $A$ is a $3 \times n$ matrix whose columns span $\mathbb{R}^{3}$ . Explain how to construct an $n \times 3$ matrix $D$ such that $A D=I_{3}$ .

Answer

The matrix $D$ can be obtained from the matrix $A$ of order $3 \times n$ which spans $\mathbb{R}^{3}$ .

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

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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The matrix $D$ can be obtained from the matrix $A$ of order $3 \times n$ which spans $\mathbb{R}^{3}$ .

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Anna Marie V.

Heather Z.

Caleb E.

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

Linear Algebra and Its Applications