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maksim [4K]
3 years ago
12

3. Let U and V be subspaces of a vector space W. Prove that their intersection UnV is also a subspace of W

Mathematics
1 answer:
kenny6666 [7]3 years ago
7 0

Answer:  The proof is done below.

Step-by-step explanation:  Given that U and V are subspaces of a vector space W.

We are to prove that the intersection U ∩ V is also a subspace of W.

(a) Since U and V are subspaces of the vector space W, so we must have

0 ∈ U and 0 ∈ V.

Then, 0 ∈ U ∩ V.

That is, zero vector is in the intersection of U and V.

(b) Now, let x, y ∈ U ∩ V.

This implies that x ∈ U, x ∈ V, y ∈ U and y ∈ V.

Since U and V are subspaces of U and V, so we get

x + y ∈ U  and  x + y ∈ V.

This implies that x + y ∈ U ∩ V.

(c) Also, for a ∈ R (a real number), we have

ax ∈ U and ax ∈ V (since U and V are subspaces of W).

So, ax ∈ U∩ V.

Therefore, 0 ∈ U ∩ V and for x, y ∈ U ∩ V, a ∈ R, we have

x + y and ax ∈ U ∩ V.

Thus, U ∩ V is also a subspace of W.

Hence proved.

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olganol [36]

The product of the given two matrices comes out to be \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Here we are given the 2 matrices as follows-

\left[\begin{array}{ccc}7&-2\\-6&2\end{array}\right] \left[\begin{array}{ccc}1&1\\3&3.5\end{array}\right]

To find the product of 2 matrices, the number of columns in the first matrix must be equal to the number of rows in the second matrix.

Here since both of the matrices are 2 × 2, their product is possible.

Now, to find the product, we need to multiply each element in the first row by each element of the 1st column of the second matrix and then find their sum. Similarly, we do this for all rows and columns.

Therefore,

\left[\begin{array}{ccc}(7*1)+(-2*3)&(7*1)+(-2*3.5)\\(-6*1)+(2*3)&(-6*1)+(2*3.5)\end{array}\right]

= \left[\begin{array}{ccc}(7)+(-6)&(7)+(-7)\\(-6)+(6)&(-6)+(7)\end{array}\right]

= \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Thus, the product of the given two matrices comes out to be \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Learn more about matrices here-

brainly.com/question/13595915

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The altitude of the triangle ABC, segment DA, has been
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Answer: sin(B)= h/c

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Area= 1/2acsin(B)

Step-by-step explanation: i got the answers frfr on edge smh yall know how that is

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which expression would be easier to simplify if you used the commutative property to change the order of the numbers​
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Step-by-step explanation:

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What is the range of the function y = x 2?
Leno4ka [110]

Answer:

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

Step-by-step explanation:

Given the function

y=x^2

  • We know that the range of a function is the set of values of the dependent variable for which a function is defined.

\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)

\mathrm{If}\:a

\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v

a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)

f\left(x\right)\ge \:0

Thus,

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

5 0
3 years ago
2. The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4.5. Suppose we going t
stiv31 [10]

Answer:

P(\bar X>76)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 76 we got:

z =\frac{76-75}{\frac{4.5}{\sqrt{81}}}= 2

And if we use the complement rule and the normal standard distribution or excel we got:

P(Z>2) = 1-P(Z

And we can find this using the ti 84 with the following code:

1-normalcdf(-1000, 76, 75, 0.5 )

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable who represent the score of golfers. We select a sampel size of 81.

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want to find this probability:

P(\bar X>76)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 76 we got:

z =\frac{76-75}{\frac{4.5}{\sqrt{81}}}= 2

And if we use the complement rule and the normal standard distribution or excel we got:

P(Z>2) = 1-P(Z

And we can find this using the ti 84 with the following code:

1-normalcdf(-1000, 76, 75, 0.5 )

6 0
3 years ago
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