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(5i) 743i2.59 Js-JLet R be a ring that contains at least two elements Suppose for each nonzero a € R there is & unique b € R such that aba =4 Show ...

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(5i) 743i2.59 Js-JLet R be a ring that contains at least two elements Suppose for each nonzero a € R there is & unique b € R such that aba =4 Show that R has unityThe following is a part of the proof; determine whether its true Or false: We claim that ab is a unity for a # 0, b # 0. Let c e R. Then aba = a implies ca caba. Cancelling a we get € = cab. Then ab is unity-TrueFalse

(5i) 743i2.5 9 Js-J Let R be a ring that contains at least two elements Suppose for each nonzero a € R there is & unique b € R such that aba =4 Show that R has unity The following is a part of the proof; determine whether its true Or false: We claim that ab is a unity for a # 0, b # 0. Let c e R. Then aba = a implies ca caba. Cancelling a we get € = cab. Then ab is unity- True False



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For the arbitrary vectors $\mathbf{u}=\langle a, b\rangle, \mathbf{v}=\langle c, d\rangle,$ and $\mathbf{w}=\langle e, f\rangle$ and the scalars $c$ and $k,$ prove the following vector properties using the properties of real numbers. $$\mathbf{u}-\mathbf{v}=\mathbf{u}+(-\mathbf{v})$$

And this problem, we're talking about the properties of sets and specifically the relations between sets. We're talking about if a set or relation is reflexive, transitive and symmetric, and of course anti symmetric. The only thing in this problem is that we're not doing this proof based, we have to do this with diagrams. So what does that mean? We're going to have to analyze this problem using matrices. So let's first write our first matrix will call this AM sabar the matrix with respect to our relation. And our matrix would look like this. Who would have these entries? 010111011 So what does this mean? Well, are is going to be reflexive if each element in the diagonal of our matrix is one. And clearly we can see that is not the case. We have this first entry as zero. So our relation is not reflexive. And now let's square our matrix. So we basically take the first matrix and multiply it by that same matrix. When we do that we would get our square matrix equals this 11113 to 1 to two. So what does this mean? Well, the zero entries and our original matrix are non zero in our square matrix. So our isn't not to reflect part of me are is not transitive and I'm moving on to symmetry. How would we do this with matrices? We would have to transpose our original matrix. So are transposed matrix which we denote by this M. R. Raise to this. Tea is the exact same as our original matrix. So we can say that this is definitely symmetric because are transposed matrix is the same as our original now because our relation is symmetric, it's probably not anti symmetric. But we can check it proof based again, this is not anti symmetric because if we have some arbitrary matrix M. Sub I J equals zero, or an arbitrary matrix M sub J I equal to zero. Those don't exist for every I not equal to J. So we're not going to see Auntie Cemetery with with respect to this relation. So I hope this problem helped you understand how we can interpret the properties of the relations on sets, specifically using diagrams and matrices, as opposed to a proof based approach.

To check this subspace. What we need to do is we need to verify three exams. The first axiom is zero, vector should be belonging to a subspace. W. I mean the set W If you won't, you too belongs to W. Then we should verify that you want to see. You too also belongs to W. What is the closure property? Third, if you belongs to W then land up. You should also belongs to W Where lambda. Is it number from the field? Okay, if all these three exams satisfied, then we call W is a subspace of the factor space week. No, these set off all environ mattresses over the field kit. All right now W is the set of all symmetric matrices. So what do you mean by a symmetric matrix? The matrix was transposed the same as it's A So that's called a symmetric matrix. All right. Now if is symmetric then that's null. Matrix belongs to W. Yes, of course. Because the non matrix transpose his narrative so yes, it's true. Second axiom some of two symmetric matters is A and B. Supposed to be a symmetric B is also symmetric. Then a place be whole transpose is equal to the property of transport. It is a transport transport. But he transposes A. Because asymmetric transpose is B. Because B symmetric diet implies a place to be. He's also symmetric so that implies equals B belonged to W. So second exam is very fine. Now coming to Torrance if is symmetric lambda E. We'll transport the property of transport is lambda and a transport is equal to the lambda. A. Because the symmetric it is lambda lambda. He will transpose islam day. So that means lambda is also symmetric so it belongs to doug. So all the three exams are verified. So yes. Set of all symmetric matrices is a subspace. The second one upper triangular mattresses. So what do you mean by an upper triangular matics? All the elements below the main diagnosed should be zeroes, for example A B C 000 123 So this is an upper triangular matrix below this. All the elements are zeros of is a pet Wrangler. He's a pet Wrangler and he's also a strangler. Then it let's be is it a veranda? Of course, yes. Because these zeros get started with the other metrics in the same place. For example 123 4045006 If you add with 178 089 00 10. So what do you guys? 000000000 So some of the upper triangular mattresses. Also an upper triangular markets. So this is very frank. Null matrix is not trying biometrics because null metrics by default, all the elements below the main diagnosis anyway, zeros. In fact all the elements are real zeros. So it belongs to going mathematics, The Time magazine. If you want to play any number with the matrix scale are multiple, zero into any number of zero. So these zeros will still be retained. So lambda you Orlando A is also an upper triangular matrix. So yes, the set of a particular mattress is a subspace. The third diagonal matrix, basically, we can just individually say that the diagonal matrices from the subspace because lower triangular mattresses also from subspace. Because what do you mean by lower strangler? All the elements about domain that loves you with the same reasoning as a crime. And what is the diagonal matrix? It is both the lower triangle ring up strangler. Since both our Wrangler and Wrangler mattresses from subspace set of diagonal mattresses. Also from subsidies. Simple reason. Right. And skill harm metrics is a special case of diagonal matrix, scalar matrix means all the diagnosed elements should be zero, sorry, All the diagnostic elements should be same and other diagnosed elements other elements should be basically zeros. So this is a special case of diarrhea magics. And since diagnosed mattresses from subspace, the set of the subset of it also forms the suspects. Alright, so this is a prison. So

This question covers topic relating to linear and zebra and the span of uh basis of uh vector space. So you can see here we have the set as a finite and had the route that the the linear combinations of all vectors of S. Is actually there's a set of fee. Okay, so how do we do that? First of all? You refine the uh you can say like we can pick a finite set um the subset of S. And that obviously ideas upset of the vector space V. And because it's the it's an element in sign as actually inside V. So that ai can be rewritten as B J E J. What do you Hj? So E J is um it can be finite or infinite is a basics the basis for V. Okay, so we um so as you can see here um a I can be rewritten as a linear combination of these spaces. Right? And now if we pick any vector, like for example, I pick victor, A and A. Is A and finally linear combination of ai right? So for example, and for I ai right from one, for example, from one to let's say end and maybe uh M. Okay. And so we can re written that some as and for nighttime submission of changing from one to N. B to J. J. And death is just a linear combinations. Uh The finite clinical combination of or the vector E. Uh J. Right? So if you want to write it down precisely, it's just the summation from I from one to M. And information from che from one to end on. For I B J E K. Right? And if you put out the so ehh Okay. So now if you put out each A out put E. J. As you can okay, you can interchange this summation to change this to end and you change it to em. And and for I. B. J. E. J. Right? Okay. So now just uh J equal to one to end of E. J. Um, submission I from 1 to 1 for I beta J. Okay, So this is just a real number, right? So it's a linear combination, so that means I belong to the span of ass. Or so that means the span of S. Is upset of fee.

I want to prove U minus p is equal to U plus minus fee for factors. You and factors be so we have you might this mean is equal to to be minus and now, using our definition of vector subtraction, we have this equal to hey minus d c he dynasty. We also noticed I sense a c and e t r real numbers he had. This is equal to a plus. Negatives, bus negativity. But then this is just addition. So we have This is equal to you. Okay, Plus negatives. But we also noticed that this negative is nothing but a negative one. So we have a baby negative one times negative one time, Steve. And using our definition escape their multiplication, we can bring them negative one out. So we have a baby, plus maybe one times and bringing this up here we have. This is equal to replacing a B C d. Lift you and B, we have new plus negative one times. And this is equal to for us. Thank you. Be so following a chain of inequalities for the qualities we have. Therefore U minus fee is equal. You did


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