each-statement-in-exercises-3338-is-either-true-in-all-cases-or-false-for-at-least-one-example-if--5

Question

Each statement in Exercises 33–38 is either true (in all cases) or false (for at least one example). If false, construct a specific example to show that the statement is not always true. Such an example is called a counterexample to the statement. If a statement is true, give a justification. (One specific example cannot explain why a statement is always true. You will have to do more work here than in Exercises 21 and 22.)
If $\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}$ are in $\mathbb{R}^{4}$ and $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\}$ is linearly dependent, then $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}, \mathbf{v}_{4}\right\}$ is also linearly dependent.

Michael Jacobsen · Accepted Answer

In this example, we have a set of four different vectors, each coming from the space are for Let&#x27;s make a specific assumption about this set of vectors. Let&#x27;s assume that the set v one, v two, v three. So just the 1st 3 vectors is a linearly dependant set of vectors. Let&#x27;s see what this would mean specifically if, for example, we let a be a matrix formed from the columns V one V to V three, which we said already is linearly dependent as well as before. So this introduces a little bit of extra drama in this linear algebra problem. If these vectors are linearly dependent and we threw in 1/4 1 can we save this and make the set linearly independent? Well, to determine that, let&#x27;s first consider the Matrix equation eight times X equals zero vector. Well, we know that V one v two V three is linearly dependent hoops thought that was the high. Later there we go. And since b one b two B three is linearly dependent, it follows that this equation has nontrivial solutions by the definition of linear dependence. But let&#x27;s say what that would mean, since there are nontrivial solutions that would imply that A does not have a pivot in every column where the word every is probably the most important word here, since A does not have a pivot in every column that next implies immediately by the definition of linear dependence that the columns of A are linearly dependent. So just because we knew that these are dependent, throwing in 1/4 vector does not change the dependence of the set. If we start with a linearly dependent set, extending the set still is going to be linearly dependent, it&#x27;s our conclusion is that the set of vectors V one, the two V three and before is linearly dependent. The set of vectors could have been really something like 100 vectors of anything we wanted. But as soon as we know that the 1st 3 a linearly dependent, the rest are linearly dependent as well

Michael Jacobsen · Answer

In this example, we have a set of four different vectors, each coming from our four for the set of vectors. Let&#x27;s make an extra assumption here. Let&#x27;s assume that V three The Third Vector is a linear combination of the 1st 2 vectors. Specifically, let&#x27;s say, for example, V three is something like two times V one plus V two. Let&#x27;s say what this would mean altogether. We can write down a Matrix A, which is formed from the given vectors as a columns. So will be V. One V two, V three and before for this matrix A. Then consider the Matrix equation eight times X equals zero vector. Now, for my next step, I&#x27;m going to copy just the Matrix A. But put in a particular vector X set equal to the zero vector as well. And the vector X I&#x27;m going to choose is motivated by this information here. If I put two here than when we take a Times X, we produce to V one. If I put a one here, we also get plus a V to the second column and now I want to put a negative one here. They&#x27;ll give me negative e three zero here would give no contribution down here in this equation. But then to V one plus V two is given to be V three. So we have the three minus fee three and this is equal to the zero vector. So what we&#x27;ve just had found is rather pretty interesting. It shows that this matrix a or rather this homogeneous equation has nontrivial solutions. In fact, this Vector X is one example of a nontrivial solution to the Matrix. But we also know since this is a homogeneous equation, do the zero vector that the columns of a must be linearly dependent. So the punchline to this problem is, if we have a set of four vectors and we find any linear relationship with one vector and the others, then the conclusion is always dependence for that set of vectors.

Viswesh Uppalapati · Answer

in this video, I&#x27;m gonna be explaining how to solve a problem. Number 34 of section 1.7, which is based on win your independence. So this question says that, uh, it gives us a statement and tells us tow prove it either. True for all cases are proof of falls by giving one single counter example. Um so for question over 34 the given statement is, uh, if you have four vectors view one through before they&#x27;re calling vectors and are for and we know that V three equals zero. Then the set that I wrote out here, the one the 233 and the four, um, is linearly dependent. So this set is linearly dependent. Um, Then victory is a zero column vector or a zero vector for short. Um, so this statement is always true. Bye. Here. Um, nine. What states that in the book, which states that if a set of vectors contains zero vector among it, then, uh, the entire set is linearly dependent because zero vector causes the, um, the variable correspondent to that column. So let&#x27;s say if you put this in my augmented matrix in the form of X equals zero Ah, to solve the matrix to see if it to check for linear independency. Ah, then then I envy Assign A B C B as the variables that were selling for in this system, then sea will be a free variable, as there would be no as a V three would not, could not possibly Do you have a column as it zero vector and even through element here.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be solving question number 36 from section 1.7, which is based on the new Independence. So the book gives us a statement and asked us the prove it true for all possible cases O r prove it falls by using one counter example. And the same for Question 36 is, uh, given for Colin vectors and our four. And if v three is not a linear combination of V one v two and before than the set V one b two B three before it is literally independent. This is what the statements claiming. But, um, this is false as, um just cause just cause of the three is linearly independent doesn&#x27;t guarantee uh, the set v one, V two, and before to be linearly independent either. Ah, for a simple counter example, we can choose. Ah, let&#x27;s say V three equals 100 uh, for victory and be one, uh, we can choose 011 You too can d&#x27;oh! 055 And before you can be like is your 11 11 here. It is clear that, um, even though the three is linearly independent, as it cannot be formed with by any combination of V one, V two and before combined together. We know that V two is a linear combination of one as it is just five times one R 500 B one and before is just 11 times be one. So V two equals five b one. So sorry about the fives and the four goals 11 be born. So here, even though the one or be three is literally independent, the other three vectors air not linearly, independent of each other. So that makes the entire set. Do you want me to be there and before be linearly dependent, not linearly independent. So this statement is false.

Michael Jacobsen · Answer

in this example, we have two vectors provided, which came from our for for these two vectors, let&#x27;s make an assumption about V to assume V two is not a multiple of V one. This is a pretty dangerous assumption to make. Let&#x27;s explain why I&#x27;ll put a border here and consider a different set statement where we might assume that V one and V two are not multiples of each other. These statements feel nearly identical, but the difference with the statement in green as we said that they&#x27;re not multiples of specifically each other. In this case, the statement immediately implies that V one V two are linearly independent, since a set of two vectors, not multiples of each other, are always forming a linearly independent set. Now let&#x27;s see what happens when we say simply V two is not a multiple of the one, rather than both vectors not being a multiple of each other, we could have a case where, for example, V one is strictly the zero Vector and V two is not the zero vector then, for this particular example, V one and V two or not multiples of each other. Since there is no vector. I could multiply V one in orderto Patane v two, but the set V one V two is linearly dependent. Now that warrants an exclamation mark because I think personally, that&#x27;s pretty surprising. But the reason the set is linearly dependent is because in the case of V one, if we have a zero vector in any set of vectors whatsoever that&#x27;s said is always linearly dependent. So the punchline to this problem is watch out for a statement that says just this much information. It&#x27;s not the same as this statement at all, which is thes stronger statement giving linear independence.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be solving problem number 38 from section 1.7, which is being some linear independence. So in this problem were given Ah, a statement which states that which gives us four backers everyone through before. Uh, sorry, uh gives us four vectors be won through before. And, um, it states that if the set of the one of the two of you three and before is linearly independent down the set of we won&#x27;t be to envy three with a weight of zero on before, but because it&#x27;s x excluded is also literally independent. So it asked us to consider this equation. This is basic. Just the next expanded form of matrix equation X equals zero. Um, we&#x27;re obviously the right side is equal to zero. And the left side a here would just be, uh, a matrix of B one, b two, b three, and before combined. It&#x27;s an augmented matrix. And X is a solution set which can only be 0000 if, um, the entire set of you one through before is linearly independent. Because since therefore, column backers, we&#x27;re gonna solve it before variables causing this have four entries. So if you multiply these two together ah, and some chemicals zero, we would get zero of you won plus zero v two plus zero B three plus zero before equals zero. So even though this, um, problem tells us that consider this hint where this form of the equation where where you consider it zero to be the weight of before the only possible Bates for our waits for B one B two b three are zeros. Because the set is already linearly independent, any other value for these 1st 3 waits would contradict the statement, and it would be false, as the entire set together is already video. The only independent so different parts of the set, uh, better taken from the original set in this case, if you want me to be three, should also be literally

Problem

Each statement in Exercises 33–38 is either true …

Problem 37 Hard Difficulty

Answer

True

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

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

Linear Algebra and Its Applications