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H) Verify the dimension theorem for T....

Question

H) Verify the dimension theorem for T.

H) Verify the dimension theorem for T.



Answers

Let $H$ be an $n$ -dimensional subspace of an $n$ -dimensional vector space $V .$ Show that $H=V$ .

Hi here in this given problem we have we have to find dimensions of E five B by tT means the time rate of change of magnetic floods linked through our coil. No, as we know, using Faraday's laws of electro magnetic induction. No, no, this be five by DT in magnitude. The time rate of change of look slim through alcohol in magnitude. That is equal to IMF indios in the coil. The CMF industry is actually nothing but potential and potentially may be defined as work john but he would charge now for the work done. The dimensions will be M L two E minus two. But the dimensions of charge E. There is for MPO. So finally, the dimensions of this D five B i D T. Really beginning by I am and two E minus three a minus one. Which is the answer for this given problem here. Thank you.

This question asked us to use part one of the fundamental theorem of calculus to find the derivative. What we know is that given the integral, we note that the integral is H of X, therefore off of T. The function is inverse tan. Are are 10 of tea, which you can also write as 10 to the negative one of tea if you're using a calculator. Therefore, the derivative H of X would be negative arc tanne of instead of tea were now plugging in X because we're looking for the derivative h prime of X one over axe divided by X squared. This is the

Hello there. So for this exercise, we need to find the dimension of the formal spaces. The first one is a R four, where Well, this is actually the algebra of linear operators on our four, and we know that there is a result. There is a theory. Um, actually that say that the they mentioned of the Al Jazeera's of some linear space is equal to the to any square if the dimension of B is equals to end. So basically, this is saying that there is equal to the square of the dimension of P. Okay, so for this first exercise, we know that the they mentioned for the space are for Okay, So the dimension for our four is equal to four. So that implies that the dimension of a our four equals two for a square which is equals to 16. So this is the space of this algebra the next a space that we need to calculate The they mentioned is the space a So here we got. So the next one, it's eight or p two team. This is the space of polynomial of degree two. So let's remember that B two of t is you need only vary eight polynomial of degree two at most. So the basis for this a space of polynomial is given by the one T and T Square. So you can notice that this is the basis for this kind of polynomial is you can find. You can easily understand why because a linear combination of these elements will generate any polynomial. And these elements here are linearly independent. So there's a basis for the point on meals and the degree of this point is equal to three. So did I mentioned of B two t of t is equal to three and therefore the they mentioned of the space A of the algebra generated by this vector Space is equal to three square which is equal to mine. And the last one is the algebra generated by the linear operators of the vector space and of matrices of two times three. We know that they they mentioned of the space of matrices to buy three is equal to six. So that implies that the dimension of this algebra it's going to be six square which is equal to 36


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