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In order for a collection v to be a vector space over the complex number c, the vector addition and scalar multiplication operations defined on the space must satisfy certain properties.
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And there are 10 of them, and let us first go through them.
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Suppose x and y belongs to the vector space v, then x plus y should also belong to v.
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Next, for three vectors, x, y, z, belonging to v x plus y summed first and then with z must be equal to y plus z summed first and then adding with x plus x plus from the left a 3 for x y belonging to the vector space x plus y equal to y plus x so multiplication must be commutative next for every x belonging to the vector space there should be the additive identity element 0 vector such that x plus 0 gives us back x.
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For every vector x belonging to the vector space, there must be an element minus x also belonging to the vector space, such that x added with minus x gives us the zero vector or the identity vector, the additive identity.
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Next, property m1.
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For x belonging to the vector space and alpha belonging to the complex numbers, alpha x also belongs to the vector space.
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Next property, for a vector x belonging to the vector space and alpha beta belonging to the complex numbers, alpha beta, first multiplied with each other, then multiplied with the vector, is equal to alpha, the vector beta x.
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Next property m3 for x and y belonging to vector space and alpha belonging to the complex numbers alpha times the vector x plus y must be equal to the sum of the two vectors alpha x and alpha y property m4 for x belonging to the vector space and alpha beta belonging to complex numbers the complex number alpha plus beta multiplying the vector x must be equal to the sum of the vectors alpha x and beta x the last property there must be the multiplicative identity one such that for every vector belong x belonging to the vector space one multiplying with this vector x must give us back the vector x itself now let us see in the various parts of the question if the given situation or if the given elements satisfy this properties of a vector space.
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So in part a, we have a vector, we have a space which consists of all polynomials of degree less than n.
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And let us define x, y, z to be the vectors in this space and let alpha and beta be complex scalers.
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So according to the definition, the vector x, y, z to be.
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Given by a 0 plus a 1 x etc till a n minus 1 x capital n minus 1 similarly we define the y with the various coefficients coefficients denoted as b0 b1 b n minus 1 and similarly the z with the coefficient c 0 c1 till c n minus 1 now for this one let us check all our properties the first property, whether it satisfies or not, so let us evaluate x plus y and collecting the coefficients, we see that the coefficients are given by a0 plus b0, then a1 plus b1, a2 plus b2, etc.
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So a0 plus b0, a1 plus b1, these are also complex numbers and therefore x plus y does give us a vector which belongs to the space v.
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So a1 property is satisfied.
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Let us evaluate first x plus y summed first and then added with z and this is the expression that we get.
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Next we evaluate x adding with the vector y plus z.
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Y plus z as evaluated first and if we do it sequentially we get this answer and clearly these two answers are equal and so a2 is satisfied.
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Next let us come to a3.
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Now, in a3, we have to prove that x plus y equal to y plus x.
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So let us evaluate x plus y.
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We get this expression and then let us evaluate y plus x and we get this expression there equal.
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So a3 is satisfied.
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Next, let us compute a4.
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Now, this is the condition about the additive identity 0.
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The zero element in this space is with all the co -covalued.
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Coefficients set to 0.
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So the zero vector is 0 plus 0x plus 0x squared etc till 0 x to the power and minus 1.
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So adding this 0 vector to x results in x itself.
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So if 4 is satisfied.
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Next there should be an additive inverse minus x.
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So here the additive inverse is defined like this with a minus sign in front of the coefficients.
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If we add this minus x to x, we find that it gives us the 0 vector.
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Next, next property, alpha times the x vector where alpha is a complex number gives us this expression.
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So all this alpha a not, alpha a1, alpha a2, etc.
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Then sets are complex numbers.
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Therefore, it does belong to the vector space and property m1 is satisfied.
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Property m2 let us first evaluate alpha beta and then multiply scalar do perform the scalar multiplication on x we find this expression and then alpha multiplying with the vector beta x also gives the same expression so m2 is satisfied now m3 we have to prove alpha times x plus y is equal to alpha x plus alpha y so first let us evaluate alpha times x plus y we get this expression.
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Alpha x plus alpha y also gives the same expression.
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So m3 is satisfied.
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Next m4.
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Let us first evaluate alpha plus beta times x.
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We get this expression.
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This is the same as alpha x plus beta x.
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So m4 is satisfied.
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Next the question of identity element in the sense of multiplication.
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So we see that if we take the number one, then it multiplies with x to give us back.
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Here all the coefficient gets multiplied with 1.
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1 is just a complex number and that's the multiplicative identity.
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And so m5 is also satisfied.
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So then for this part the answer is yes it is a vector space.
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So all polynomials whose degrees less than capital n forms a vector space.
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Next, let us come to part b, which tells us that here v is the set of polynomials under degree n that are even functions.
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Now here, the arguments that we have provided for part a applies, but with the coefficients that have odd subscripts set to 0...