with-a-as-in-exercise-3-find-a-nonzero-vector-in-nul-a-and-a-nonzero-vector-in-operatornamecol-a

Name: with a as in exercise 3 find a nonzero vector in nul a and a nonzero vector in operatornamecol a
Uploaded: 2019-08-11
Duration: 2 min 17 s
Channel: Numerade
Description: With $A$ as in Exercise $3,$ find a nonzero vector in Nul $A$ and a nonzero vector in $\operatorname{Col} A .$

Question

With $A$ as in Exercise $3,$ find a nonzero vector in Nul $A$ and a nonzero vector in $\operatorname{Col} A .$

Doruk Isik · Accepted Answer

in this problem. We&#x27;re asked to use men Take saying from exercise tree and they were asked to find knows your victory and no, they don&#x27;t. They&#x27;re vectoring problem. Hey, so many column is actually a nuncio vector in call eight. So is an answer to properly It could be any Cole. No, partly a time. Thanks. I should say zero. So then the system is your enroll. Echo for will be 1001 negative seven or 6200 The way I got this is I said near Rome or will be row one minus three row too. So now we have a system off recreations here. We can say that Explorer is equal to seven x three Minus execs or and x two, The sequel to Negative four X Tree. It&#x27;s a negative two plus two eggs for now. Thanks. What? The next two depends on x three MX four but X three units for our our free breakables. So plus, assume a three is one and x four is one that from you questions second equation you find next to to be 94 plus do like a dream. And from the 1st 1 We plan X one to be 7.6. That is one. So what? And we find the zonal director to be one naked too. 11

Michael Jacobsen · Answer

in this example, we&#x27;re going to be working with the four by two matrix A we see displayed here. The first thing we want to determine is the following. Let&#x27;s find X in are to such that the Vector X is in the no space of a well. First, if we say we want extra B in the null space of a that implies this matrix, a Times X must equal zero vector. Now what we could do is take this matrix equation here, augmented with a zero vector and begin row reduction to find all vectors X that satisfy the equation. If we did that, we&#x27;d be effectively computing the null space. That is a fine approach, but sometimes there&#x27;s a quicker way to do things if you look before you leave notice. If we multiply this by three added to this century, we get zero. If you do the same thing here, here and here, it looks like that&#x27;s always the case, so that gives us a hint. Let&#x27;s do the following end. Determine what X should be so that&#x27;s certainly in the null space. So we triple this column and we add to it. This column so I&#x27;ll put a one here. Now you can check a Times X is going to be the following. So to negative one negative for three negative 63 12 negative nine is multiplying X, which is three and one. So for every entry will go row by row. We tripled the first entry and add to it the second entry here that would give a zero. Then on the second row will have three times negative one plus three. So it zero again on to the third row we have four times three, which is negative 12 plus 12 times one. So we have zero again. And, like Wise, take three times free plus one times negative nine and we get a zero. This gives us the zero vector. So we have shown that if X is equal to 31 this implies immediately X is in the null space of a. If these entries here were harder to work with, then I probably would have gone gone with the row reduction method. Notice also, if we just want to find a vector X that&#x27;s in the null space of a You could have gone in with the route of the zero vector, but that&#x27;s typically boring because the zero vector is always in the null space. We found a non zero vector X also in the null space. Next, let&#x27;s consider how we might find a vector in the column space. First, let&#x27;s write down what the calm spaces sold me. Put up border here, and we know that the calm space of A is equal to the span of the columns of that matrix. A. That&#x27;s important because the span of thes two columns is a set of all linear combinations of these columns, and one particularly in your combination could be found by taking one times this column, plus zero times this column that would just give us the very first column vector. So we can say X equals to negative one. Negative 43 is in the calm space of a, as is the second column, and if we add that comes together, that&#x27;s also in the calm space. In fact, every lean your combination. So this is how to find a vector in the null space that&#x27;s non zero, as well as a non zero vector in the calm space

Gideon Idumah · Answer

this problem Got to find in on zero Vitto in the north. Off, eh? On in non zero of that saw in the column off a So let&#x27;s do for the known off a first not to do this. We solved for eggs team a X quarts zero. And how do we do this? We do these by solving the documented many trees. E come on. Zero. So in this case, we off. 14 minus five soup. So five minus 17 370 11 The odd 0000 So we try and turn this into a reduced matrix. So what do we do? We force a vote on everything yet? Zero. Because this is a pipe one stone decency in private school. Um, so we have That&#x27;s world Sue equal to road Sue minus for room Juan. Rule three equal to row three plus five for one under one roof full. The quarter for minus minus Our soo. Rule one. When you do this, you&#x27;re gonna off want 000 Sue minus 393 three minus five 15 of five. I&#x27;m gonna have 0000 No, the next one of these one ton dysentery one and thereafter, everything before bm above and below it zero to make This are also a private&#x27;s. So all we do is we foster for sit at the road soup equal to rude sue divided by minus three. We do that. We&#x27;re going off. 1230 015 over. 30 09 15 0 0350 You know, lastly want It&#x27;s on everything above this pie votes to zero and zero. So I would do is we see for a war on it Calls or warn Minus tsu road Sue Earl three equals over three minus nine or two on then Earl four equals or four minus Terry root roll three. So minus three road sue. So when you do this, we&#x27;re gonna have umm one zero minus one over 30015 over. 300000 0000 Now, with these, we can solve this and this implies That&#x27;s recorder. This is X one. This is extremely so that&#x27;s the implies. X one minus one over a three x three equals zero. This will be ex too lost five over three ex Terry in court zero. And then this is zero quote zero. Now, when you saw for this, you&#x27;re gonna have that. The general solution. The general solution will be x one equal to one of our three x three ext. Suit equal to minus five over a very extremely. Now we don&#x27;t have an ex, Terry, so you can take extra to anything. So we choose would choose any value for X three. See extra Because the one say ex Terry equal to one. So this implies that the value in the note and non zero vet so would be so off. One of the three extremely said I&#x27;ll be one of the three minus five over three. And then I chose this to be one. So this is wants a D&#x27;s is in the north off because you can actually write this as ext. Three quartz Eck story. I can ride these ours one of the three minus five of a theory and then one more supplied by ex Terry. So this Viktor is the victor in the north space off, eh? Now, lastly for the column space off A. We have that so far, the column space off eh? We have by definition of column space that any column any cologne in a is in non zero of Epsom in column space off, eh? So you can choose any column so I can&#x27;t speak. For example, I can pick the first column, which is 14 minus five suit.

Carson Merrill · Answer

So one way we could have a three by three matrix and a vector B Such that B is not in the calm space of a would be as if we constructed, um, a Matrix that in this case has only one pivot column. So an example of this would be 100 and then we&#x27;ll just have a bunch of zeros. So this is a simple example. Um, but here we see that there is only one pivot column. Um, and we can call this a so that way we know that the column space of a or at least the basis of the column space of a is 100 which means that it could be 200300 But there&#x27;s always going to be a number here, and then zeros are on the bottom. Um, then let&#x27;s just pick a B vector to be 010 We know that this is not possible because there&#x27;s no way that we can make that second value in the second row. There&#x27;s no way we can make that value one. So this would be a very simple example, but an effective one

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.

Michael Jacobsen · Answer

in this example, we&#x27;re considering the homogeneous equation. X equals zero, and we&#x27;re told that we have a nontrivial solution right out of the gate. It&#x27;s X equals 111 Let&#x27;s suppose also that is, of size three by three. And the question is four. What matrix A is X equals 111 A solution to our homogeneous equation. Well, in order to determine such a matrix, say, Let&#x27;s first right out a X equals zero in Amore expanded form. Let&#x27;s put in a matrix. Say, I&#x27;m going to leave it blank For the moment, X is going to be 111 and this has to be said equal to the zero Vector, which is 000 Now we could make a completely the zero matrix, but that feels as though we&#x27;re cheating. Really, we wonder, Can we do this and get 111 a solution when a is not the zero Matrix? What one thing we can do is just because it doesn&#x27;t have to be. The Zero Matrix doesn&#x27;t prevent us from zeroing out a row or two, but I wouldn&#x27;t get too out of hand. I want to make the matrix a as simple as possible. So I&#x27;m going to make this 111 in this column. Now, if we think about what would happen if we multiplied out this much, it would look like this so far, we would take the vector entry here, which is a one, and multiply it with the first column of a So will be a column, 111 going here. We still have to decide what should go here. But we know that whatever we put will be plus one times those entries where I left question marks and for the last vector will be putting plus zero time. Excuse me. Plus one times that zeroed out column and all of this has to be equal to zero. So this is the format so far. And if we decide what can go here and then here, ultimately the vector equation to produce all the zeros will be all set will notice. That assumes we chose ah one here. If we put a negative one here, then add up, We&#x27;ll you will definitely get to zero, especially since there&#x27;s no contribution here. Some going to put negative one here and here and here in the corresponding vector equation. We then have a solution, and that tells us that this matrix A allows 111 to be a solution so effectively. If we make one column, the combination of another and zero out a column, that would be one way to determine such a matrix a.

Problem

Let $A=\left[\begin{array}{rrr}{-8} & {-2} & {-9}…

Problem 22 Hard Difficulty

With $A$ as in Exercise $3,$ find a nonzero vector in Nul $A$ and a nonzero vector in $\operatorname{Col} A .$

Answer

$\left[ \begin{array} { l } { 1 } \\ { 0 } \end{array} \right] \cdot \left[ \begin{array} { l } { 3 } \\ { 1 } \end{array} \right]$

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$\left[ \begin{array} { l } { 1 } \\ { 0 } \end{array} \right] \cdot \left[ \begin{array} { l } { 3 } \\ { 1 } \end{array} \right]$

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Anna Marie V.

Kristen K.

Joseph L.

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

Linear Algebra and Its Applications