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2) Find basis for the Hull space ker( A; of the mnatrixBasedVOUI findiug, determine the rank of AJ...

Question

2) Find basis for the Hull space ker( A; of the mnatrixBasedVOUI findiug, determine the rank of AJ

2) Find basis for the Hull space ker( A; of the mnatrix Based VOUI findiug, determine the rank of AJ



Answers

Find basis for ken(T) and range(T)

So we are given a linear transformation. The maps are cube To argue such that the linear transformation is the projection off the input vector onto a vector we given us tu minus one on one. Yeah. So this is a factor. What is the definition of projection Off a vector on toe? Another vector. So the definition is the projection off you onto V is given by u dot We divided by we don't we into the activity. Now let's take CCR input victories in our cube. So? So I can take us x one x two x three So I'll take yours. Um, us x one expects. I'm sorry. So us x one x two x d Now what is we given to minus 11 So let's find the projection first. So the projection is nothing. Bert the doctor worked off x one x two x three with tu minus 11 divided by what is magnitude off V square. Because we dot with more, Be square. So basically it's too square plus minus one Who will square plus one square Okay into my vector v The fact of these already given to us which is two minus 11 This quantity is a scale are right. So let's let's calculate it. The protection off you on TV is nothing but two x one minus x two x three So this is the doctor of definition rates. Corresponding elements should be multiplied and added It's some of the products basically and divided by the denominator is four plus one plus 16 times off. Tu minus 11 Right. So this is the linear transformation. Obviously, this is a linear transformation. Now we need to find the care. Neh lofty The dimension off, Colonel off TV call. It s nationality off. So let's find a colonel off. So what is the definition of Colonel? The cardinal definition is you off you zero vector. So the set off all such you such that be off you is a zero vector. We call it us the Colonel. Lofty Linux transformation. Okay, so let's check for which factors we get. Zero actors and output. So what is that? This is my tea off you. Right? So basically, tonight is a city off you. So the off you equate to zero. So in the off you is equated equated to zero so tee off you equal to zero vector. That should imply Onley. This quantity should be zero because six Uh, sorry, not six. This factor cannot be zero to minus one. Anyone, because it's a non zero vectors divided by six is also an incinerator. So the only possibility for tea off you would be zero factories to x one minus x two plus X three is equal to zero. So a small trick. So what we can do is I could write x one s one x 10 x three Okay, x two is what from this equation. What is x two from this equation to into x one plus one into x three on How can I write x 30 into x one plus one in tow? Extra so I can write my rector x one x two x three as x one times a vector 1 to 0 plus x three times a vector 011 This is a span off two vectors. So basically mine Careful off the linear transformation is nothing but span off which factors the two linearly independent vectors 1 to 0 island zero double one. So 120 on zero double one. So this is the capital of the linear transformation. And these were linearly independent vectors. How many are there too? So my diamond shin off Colonel Off T is how much to and this is mine. Ality off the linear transformation. So I got in. The lady has to Now what is? What do you mean by Naledi? Naledi is the dimension off Karenin right on. What you mean by rank Rank is basically the demands of the image, so rank off the linear transformation is the dimension off image dimensional image. So what do you what image? We already know the image. This is my image. But whatever input vector you give, this will be a scaler on by six is there so that is also a scaler. So basically mighty off any input vector to you is some Lambda Times The vector v So lambda is a scaler A friendly lambda is the scale are so what is my basis for real image? The basis for image space will be simply the vector We simply the vector we There's only one vector on your vacuous Lee One vector is linearly independent. So what should be the dimension off. One vector on back to it is a non zero vector, right? The dimension of zero victories anyway, zero. That is a different question. But this is a non zero vectors, so no, no issues. So what is the dimensional off this image? It's one. So the dimensional image is one which is nothing. Well, rank off t so my rank off t is a call to one. That's it. So rank of tea is one banality of these two. And what is my colonel? Off T The basis of Canon Lofty. It is this one. They sign this because cattle off these spanned by these two vectors.

Problem Number 39. Consider C one B one factor last see toe be to better plus C three B three vector is equal to the euro factor and this is wine. So here, see Juan is not equal to zero and I don t want better is the same as C one B one better and w two vector is corresponding to see to veto vector glass c three these three vector So we can say that deadly Warren Russ Heavy too is equal toe zero vector From here we say that the one everyone better and that b two vector Oh are lini early independent which contradict the serum which contradict that Be a Tim Fine points at that point time. Therefore we want to be with a zero piping prediction eso see through and see three people to zero rob one and e one better a sort of V two Vector and the three vector are linearly independent boy assumption or do you want back through? And people vector on the three Vector are linearly independent

Hello and welcome to Problem lieutenant. Chapter four. Section four you were asked to show that the given vectors are not a basis for a V. And the rest to find a basis for give them those vectors. So just looking at it we uh we have an identity that is relevant to this problem is that co sign of two X. Is equal to uh co sign squared X minus sine squared X. And as um V three is equal to the 1 -72 is not linearly independent. So no part B. Then asked well if that's not um literally independent. Can we get two of these that are literally independent? And that means um can we generate something such that um is of the Form zero is equal to a What time is the 1st 1? Co sign square decks? It must be some sign Squared x plus c. co sign of two x. That spends um the vector space. And from this we can again play this identity by doing this we'll get uh a plus C. Cosine squared X. But this um this science cortex will then disappear. So from this we know that science cortex is not in the span is not needed for the basis and it's just coastlines cortex and co sign up to X. And that concludes the problem

So here we are given that we have our transformation T. Which maps from our fourth? Our fourth is a linear transformation where we have T. Of X one X two X three X four is equal to um The vector here X three X four X one plus X two and X one minus X two. So um for a want to find the image of our vector V. Which is equal to the vector 10 negative 12 Okay so here we have the X one is equal to one X two is equal to zero X three is equal to negative one, X four is equal to two. So we have our transformation T. Of 10 negative 12 is equal to the vector here negative 12 And then one plus zero and one minus one gives the vector here. Okay um so so yeah so there is our answer right? That is um the image of the party. So there is our image. Okay and then for be going to find a basis for the kernel of tea and a range of tea. So the definition of the kernel of tea is the set um of wells of X one X two X three X four which live inside of our four. Such that um the transformation here maps to the zero vector of all zeroes. Okay um so we have a transformation T. Of excellent X two X three X four is equal to the zero vector implies that X three goes to zero X four goes to zero X one for six to goes to zero and X 1 96 2 goes to zero. Um So we have the kernel of tea right? Is just equal to all zeros. Um So therefore the kernel of T. Um has no basis, Right? Because um here Colonel of Tea has only um zero element and then the dimension of the kernel of T. Is zero. So therefore kernel of T. Has no basis. Okay um so now to find the range of T. Want to find um a standard matrix representation of tea. So for this we use um the standard basis vectors in R. Four. So here are our standard basic factors, right? E. One, E two, E. Three and E. Four which just has one in he is the first position, 2nd 3rd or fourth position. Fourth position was upward. Zeros every rose. So there are are our four basic factors. And then um we have okay the transformation so T uh V one is into this factor here. T. F E two TFT three. And we C. T. Of before. Okay so um then the standard matrix of the linear transformation is given here. So A is equal to do the transformation and we get this four by four matrix here. And then we're going to want to use echelon form on our matrix. A. So to do so we first swap um Rose one in row three. So where one goes to go three and go to what goes to go four to give this matrix here. And then we're doing well. Um are two minus R. One becomes our new R. Two. And then we um divide our two by negative two to give the matrix here. Which then we see in um row echelon form. And well um we have 490 Rose. So therefore the rank of A. Is four. Therefore the dimension of the range of T. Is equal to four. Um In other words the range space has four literally independent vectors. So now the defend the basis of the range of T. Well the basis is that it's given by our product is here. So this becomes a basis of the um of the range of tea. So um in other words, we can say that the columns of A are a basis of the range of tea. So we can stay here that the columns the columns of of our matrix A. Are a basis um of the range of T. All right, take care.


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