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21. Given the subspace V = Spanllz,-3,4) (-1,S,3)}orR' Determine whether the vector(3,- 1,11) lies in that = subspace and justify your answer b) Find a basis...

Question

21. Given the subspace V = Spanllz,-3,4) (-1,S,3)}orR' Determine whether the vector(3,- 1,11) lies in that = subspace and justify your answer b) Find a basis for and concludedim()

21. Given the subspace V = Spanllz,-3,4) (-1,S,3)}orR' Determine whether the vector(3,- 1,11) lies in that = subspace and justify your answer b) Find a basis for and concludedim()



Answers

(a) find $n$ such that rowspace $(A)$ is a subspace of $\mathbb{R}^{n}$, and determine a basis for rowspace $(A) ;$ (b) find $m$ such that colspace $(A)$ is a subspace of $\mathbb{R}^{m},$ and determine a basis for colspace $(A)$. $$A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 5 & 6 & 7 \\ 9 & 10 & 11 \end{array}\right]$$

Let's find a basis for the road space of the following Matrix The Matrix, which all call P and has the falling in trees. 11 negative, three and two in the first row and 34 night of 11 and seven in the second row. Now we can use the following to fax to find using a quick method. Ah, basis for the road space of P. So the first fact is, if two matrices AP and P prime our row equivalent than that implies that there ro spaces are the same. The Rose basic p is equal to the road space of a P prime. And the second fact the second fact is that when we ro reduce a matrix to row echelon form, then low echelon for matrices have roads that are linearly independent. So row echelon form matrices row echelon form maitresse ease have rose that are linearly independent that are linearly independent. So if we could just roll reduce this matrix two off matrix that is a row echelon form of P. Then we'll get a new matrix that has the same road space of P and has roads that are linearly, literally independent. So we take the road space of this new matrix that is a Rochon form of the Matrix P. Then we'll have a set of vectors that is equal, whose span is equal to the aerospace of P, and that is and that set of vectors will be linearly independent. So we will have found a basis for the road space of peace unless lower due to this matrix, so positive video and try to Rover do sis matrix. Some said, assuming you've given it to try. So this matrix ISRO equivalent to the falling matrix, which you can check is in row echelon form or is in a row echelon form of the matrix p. So we can robot uses metric to the following matrix 11 negative 32 zero one negative to one. And the reason this matrix isn't row echelon form is because each pivot position has zeros directly below it. In this in the column that that pivot position is in and each perfect position is to the right off the pivot position in the row above it. So because of these two facts, we know that this matrix has a row space that the sit that's the same as the Aerospace of P. And we know since this matrix is in a row echelon form of the Matrix P, then this matrix has rows that are linearly independent. It has row vectors of their linearly, independent from each other. So therefore we immediately have a basis for the roast piece of P. It's namely the set containing the falling row vectors. It's the second team in the row Vector one comma, one common negative three comma two and the other vector in this set is zero comma, one common negative two comma one. This is a basis for the road space of P. Now the row space of P is the span of its row vectors. But we can also think of it as the span of its basis vectors and a stand of these two basis vectors is all in your combinations of these two vectors right here. These two vectors right here and all the new combinations of these two vectors are vectors with four real number injuries. So therefore we can say that the road space of P rose space of P since each vector in the road space of P must have four Rio number entries we consider the roast recipe is a self space off our four, the set of all vectors containing four real number entries. So to find a basis for the column space of P, we would first ro reduce the Matrix p ro reduce the Matrix P to a row echelon forms of overdue P to a row echelon form. And then next we would locate the pivot columns off the pivot columns of the low echelon form matrix to locate pivot columns of Roe Echelon Form matrix. And then third. We would take the corresponding columns in P, and we would use those to form our basis for the column space of P. And I'll explain what I mean by that in a second. So we would take corresponding Collins in P, the non road reduced matrix, the original Matrix. So what? I mean by taking the corresponding columns and p, let's say that I found out that the pivot columns of the low echelon form of the Matrix P turned out to be, say, columns one and three. Then that means that comes one and three of P. Combs one and three of P will form those column vectors. The column. Vector one in the column Vector three were formed a basis for the calm space of P. So looking at are row echelon form of the Matrix p This this guy right here is a row echelon form of the Matrix p we have that which is very nice, will rewrite it down here. So 11 negative 32 and 01 negative to one so we can locate the pivot columns of this matrix. It's this column right here and this column right here and these two columns the first and second column of P of our original Matrix piece of Let's be right, it Right below that, we read of Matrix P right below this row echelon form of the Matrix P. So 11 negative. 3211 negative 32 and 34 Negative 11 7234 Negative 11 7 So, since the first and second columns of the row Ashlan form of P, they we found out that those two are the pivot columns of this Lois want form. That means that the first and second columns off P the first and second Collins of P form those column because these contractors right here form a basis for the calm space of P. So a basis for the common space of P is the following set of factors of all. Instead of column vectors one comma three and one comma four. Now notice what we did. We wrote reduced the matrix p This matrix right here to a row echelon form of the matrix p to locate the pivot calms off the matrix p because the pivot columns of a row echelon form of the matrix p are exactly the pivot calms off the matrix p. And since we know that the span of all of the column vectors of P is exactly the coms basic p or we could also think of it as the span of the basis vectors off P. Excuse me of the basis sectors of the column Space of P. We know that this pain is all linear combinations of these two vectors right here in these two vectors and in fact, all linear combinations of these two vectors will have to real number components. So this span in which is equal to the column space of P, it's gonna be a subspace off our two, which is the center of all vectors having to Rio number components

Suppose we were given the following matrix, which I'll call P and this majors is equal to the following Negative 31 and seven. And suppose we were asked to find ah basis for the calm space of this matrix. Well, if we were asked to find a basis for the column space of this matrix, we wouldn't need to find a linearly independent sent s whose span who span is equal to the calm space off P Army troops. Now, take a minute and find the calm space of P writing it as the span of some set of vectors. So I'm assuming you've had a go at it. So the common space of P is the span of all off the column vectors of P. But P P is a three by one matrix. It has three rows and one column, so it only has one column vector. So the calm space of P is the span off the following set of vectors the span of the Vector three or excuse me? Negative. 31 and seven. Now notice that this set of vectors right here the scent of only one vector is linearly independent because we can say that we'll pause the video and see if you can determine why this is linearly independent. I'm assuming you've tried. So this set of vectors is linearly independent because the Onley solution the only see value that makes the equation see time the vector negative 31 and seven equal to the zero vector zero vector in our 30 vector with three zeros in it as it's entries, the only see value that makes this equation true is one sees equal to zero. All of the sea values make this statement false. So because of this fact, this implies of this set of vectors is linearly independent. And moreover, we know that the set of vectors that contains only one vector has a span that's equal to the column Space off P. So what can we say then? Check this out. We found a linearly independence that s whose span is equal to the confidence of peace. So we can say this set containing only one vector The vector Negative 31 and seven. This set containing only one vector is a basis for the calm space of P. And check this out. We know that the car in space of P is all linear combinations of this vector and all in your combinations of this vector are vectors that contain three components. So therefore we can say that is a subspace. The calm space of P is a subspace off our three since our three is the set of all vectors, the contained three rial number components. Now, to find a basis for the roast base of P, we would again have to find a linearly independent set s whose span who span is equal to the road space of P. No, remember that the roast base of P is the span of all of the row vectors of P. So check this out the span off all of the row vectors of P is the span of the following vectors the one dimensional vector Negative three the one dimensional Vector one and the one dimensional Vector seven And what I mean by one dimensional as these is that these vectors Onley contain one component. Now let me ask you, is this set of vectors right here linearly independent? Well, the answer is no. Well, why Sanford l is because, for example, I can write 71 of the veterans. In this set of vectors, I can write seven as a linear combination. Of the other two vectors in the set, for instance, seven is equal to with seven times one plus zero times negative three. So I can remove seven from this step of vectors and get a new set of actors get a new set of vectors, namely the set of vectors, Negative three and one who span is the same as my original set of vectors. So just like this set of vectors has a span that's equal to the road space of P. I can also say that this other vectors has a span that's equal to the road space of P and similarly, similarly, I can remove negative three because I can write Negative Three is a linear combination of the other vectors in this set and the only other vectoring that set us one so I can write negative three as, of course, negative three times one so I can remove negative three from the set of vectors without changing the span of this set of actors. In other words, I get a new set of vectors by removing negative three namely, the set of vectors containing only the 11 dimensional Vector one. And this new set of vectors has a stain span is the original set of vectors. So this span of this set of vectors has a span is also equal to the road space of P. Now, follow up question is this new set of vectors Ah, linearly independent set of vectors. And the answer is yes again because the only solution see to the equation C times the one dimensional vector one equaling the one dimensional zero vector zero. The only see value that makes this equation true is when c is equal toe one excuse me, etc. Is equal to zero. So therefore because this is the only value that makes this a statement true, This set of veterans right here is linearly independent. So check this out. We found a linearly independent set of vectors, mainly their own. The set of vectors containing only the one vector one and that set of vectors has a span equal to the roast base of piece of this, we can say, is a basis for the road space of P. And finally, what can we say where does the road space of P live, that is, Is it a subspace off our one? And the answer is yes, because the road space of P is all linear combinations off these 31 dimensional vectors. So every single vector in the aerospace of P is a one dimensional vector that has as its only one component off Rio number. So therefore it must live in the space off all vectors containing one component, and that one component is a real number. So therefore we can say that the road space of P lives in or that is to say, is a subspace of the good old real numbers a k a. R one.

We'll get something. This problem were given subspace. That is a B C D. Where? A minus three B. Let's see. Zero. So by this equation, we know that see can be represented as freebie minus a So that means C can be determined by being a so our vector a B c D can be re read. Could be right. We can rewrite this vector. Um, terms off, eh? Hey, freebie mind, say, and the so. I noticed that A, B and D R V variables here. Well, c can be represented as, uh, A and B. So keep going. We can separate the specter so that it will be 10 91. I'm a skater. Eight and zero. 13 zero. Time Specter beak. And for the vector d off. Sorry. Should be scaler instead of Victor. Um, times of skated be And poor skater d we have There was 00 times skater d All right, so our pieces will be these three vectors. 10 negative. 10 013 and 0001 All right, So this is our faces and our dimension of this subspace. Well, the three

So in this problem still, we are given a subspace. We need to find the basis and the dimension. That's a plus B to a hey minus B and negative. Okay, well, which may be our real numbers. Okay. Um, so again, we need to separate this vector purse. So we could, Although all the interests that contain a burst, that is a to a Sorry how? Just rewrite this too, eh? Three a zero. Right. Because in the purse, entry would contain that contains one A and, uh, second African things to a they're entering. Continue contains three a. So we couldn't plus find some e and nothing. Nothing in second entries that zero on. Don't be on connected. Okay, so this gives two vectors. 12 three. Sorry. Just rewrite the three 30 and the second vector one. You negative 11 So these two vectors are linearly independent. So this is the basis for the subspace. And you mentioned be two


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