5

-4 Find the dimension of the subspace H of R2 spanned by "H -5 20 dim H=...

Question

-4 Find the dimension of the subspace H of R2 spanned by "H -5 20 dim H=

-4 Find the dimension of the subspace H of R2 spanned by "H -5 20 dim H=



Answers

Find the dimension of the subspace $H$ of $\mathbb{R}^{2}$ spanned by $\left[\begin{array}{r}{2} \\ {-5}\end{array}\right],\left[\begin{array}{c}{-4} \\ {10}\end{array}\right],\left[\begin{array}{r}{-3} \\ {6}\end{array}\right]$

Be solving question number 12 in section 1.7, which is based on a linear independence where it gives us three column. Victor's and Asa's values of H right here would make this set of vectors literally independent. So for her to be literally independent, the Matrix equation X equals zero or the concatenation of these three vectors with zero must have only the trivial solution, which in this case is X equals 000 Because, um, here three doctors, three lawmakers means that we're gonna be selling for three variables where each entry represents a variable. So only if all the variables zero would this set of factors be literally independent. So to do that, we just set up, um, an augmented matrix like this with all three vectors contaminated with a column maker of zeros. And you perform elementary rule operations. And once you do it, I'm not gonna do all of them here. But, uh, since I did them beforehand, once you do the final reel operations, you got to negative 68 and zero for the first row. So it stays. Are that stays the same? Yeah. I'm sorry. I noticed this wrong down wrong. Says positive to their, um, the second row will become zero negative five. And, um h plus 16. It means you're on the side. And the third row is just gonna be all zeros here. Ah, we don't even need thio completely solve this matrix into echelon form as we know that we see a pivot here and a pivot in this columns of these two columns all the pivot. But the third column does not have a problem. Position, position. Um, so the third variable will be a free variable, which means it can be any value from the number line, which would cause the other two variables to change because there would be written in terms of the free variable when you completely solve this system. Therefore, since, um, since the system has more than just a trivial solution more solutions with just a trivial solution, there is no value of h of age that makes this system linearly independent. Therefore, it's, um, it's linearly dependent for all age.

Problem. We're given age, which is son and dimensional. Subspace, subspace off. Hey, which is also and dimensional. And where will we want to show ages people to be so first to Christine to notice that that stuff space age as the basics with the in vectors and those in vectors has to be the new independent. So already down here. So that is Asia has a basis. Oh, and electors, because age it's in dimensional and more over those vectors has to be leaner. Independent, no. Are we nearly independent? Being dependent now spelled it wrong. Independent Dent. Okay, so Oh, I forgot to say that these factors has to be, you know, the independent in B because H e's a subspace off. All right, so now we can apply bassist zero, you know, a textbook, these this serum. So it says by our basic faces, the room. So they, the vectors will constitute well constitute a basis for yes. Well, so factors constituted he two, right? So that means issue. Because why? Because Asian each and b r o r both spend by the same set of vectors. So there's no other way to find a dis joint Asian V to make this happen. So this is our conclusion. We just have a vectors, that kind that span, uh, both H and visa. These two spaces has to be his two spaces have to be.

In this example, we have three different vectors that are provided where the third vector also depends on the value of H here. One question we could ask about this set of vectors is when with the offset be linearly independent, for example, for example, with that value depend on h. Was it always be linearly dependent, or will it always be linearly dependent to determine the answer to such questions? We can start off with the Matrix A where a has calm is formed from the vectors given to be V one, V two and V three. So this would be equal to one negative 14 in the first column. Three negative 57 for calm to and then negative 15 and h so for this given matrix, we're going to roll reduce to echelon form and that will tell us where the pivots are. First we start here and we're going to eliminate the entry Negative one and positive for below. So let me copy row one to start out, then add row one to row to replacing Row two. That gives us zero negative two and four, then multiply row one by negative four and added to Row three will obtain zero negative five and H plus four. So our new pivot position is found on the two to entry here in the Matrix, and we're just going to eliminate the entry below this time since we just require echelon form. So let me copy both row ones and two. So there's our Row one and row to Let's first Divide row to buy the quantity. Negative two will obtain 01 negative, too. Now our operation is going to be to multiply row two by positive five so that we can add it to Row three and obtain a zero here. Upon performing, that operation will obtain zero zero and H minus six. So let's look at the situation so far for pivots. Calm one and calm to are automatically pivot columns and calm. Three. Could be a pivot column, but we require H minus six to not be zero. So if H is not equal to positive six, then A has a pivot in every column. What this would mean is then, if we consider the Matrix equation a X equals zero vector. This equation has no free variables When h is not equal to positive six. And we know that whenever we're dealing with no free variables, we have linearly independent columns. So we're ready to state our our conclusion will say when h is not equal to six. The columns of a which are these vectors from the start of the problem, are linearly independent.

Okay. Now, in order to find a basis for the subspace of R. For that spanned by these four vectors, we want to get rid of any of vectors. Any of these vectors that are repetitive because they're dependent. So for example, we see that V4 is actually V one plus we can have before B three, V one plus a third of the three. So we can't have the four B three, M. V one all be in the basis because they're dependent. Now, let's see for V two, is there a way That we can write V2 as a combination of the other vectors? Let's see. We can have two times 3, one minus 2/3 B three. So that means that we can't have V 1, 2 and three all in the basis. Now can V to be expressed With just the 1? No. So that means that one example of a basis we can use is V one V two. We can also have V one, V 3 B. A basis. We just can't have Me Too NV three being the basis with V one Or V one, V 3 and V four together. So these are two possible bases.


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