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Let $H$ be the set of all vectors of the form $\left[\begin{array}{c}{2 t} \\ {0} \\ {-t}\end{array}\right] .$ Show that $H$ is a subspace of $\mathbb{R}^{3}$ . (Use the method of Exercise $9 . )$

$v=\left[\begin{array}{c}{2} \\ {0} \\ {-1}\end{array}\right]$

Calculus 3

Chapter 4

Vector Spaces

Section 1

Vector Spaces and Subspaces

Vectors

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and this problem, we're gonna be solving problem number 10 of section 4.4. Um, so the problem assets, if the given set are all vectors of this given definition, Where to? Where? We have a column vector of two teas there on negative to t. Here's some space of our three, which basically just means that, um, it spans are three. So for a given set, to be a subspace of another set, we need to fulfil three conditions. One it has to contain the zero vector to it has to be close under addition and three, it has be closed on her multiplication. So, um, if we just take a vector V at its base form to zero negative to t zero and negative t and we take t to Tia's any scaler values. So if we take t to be zero, we got two times 00 negative zero equal, which is 000 which is the zero vector. So the zero vector exists in this set so we can go on to the second condition. Clothes under a scaler are closing tradition. So if you take two vectors in this set where they multiply by a different scale our values So we can have V be to t zero negative t and you be to see zero negative c and we add these two together we get V plus you times are sorry. It would just be replace you. Oh, sorry. We Plus you would equal to to t plus to see you're a plus zero and negative t minus Siebel That is just the same thing as adding adding the scaler multiples together t plus c times this vector 201 which are just 20 negative one which is just the coefficients of the initial vectors. So this means that the vector remains the same but the scale of multiple just changes. It just adds and compounds. That means that we're still on the same line with varying scaler values and we're still within the same set. So it has closed under Skeletor, Earth is closed under addition and lastly, it has to be closed in her scaler multiplication. So if you just do, um, if you just do a scaler scene times the vector V, we got see Time's see Time's to t zero negative t That is just C Times, T times two 20 negative one Which again, is still this the same coefficients of the set? Um, So we still stay on the same line again? Because I just just the scale of all you were multiplying this by changes. See, times to you just get bigger. But we're still scaling the same line or vector. So we stay on the same line. Um, therefore, this is also closed on her skin. Her multiplication therefore, um h is a so space is a subsidies o r three

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