Let
![\{\varphi_i~|~i\in\mathbb N,1\le i\le k\}](https://tex.z-dn.net/?f=%5C%7B%5Cvarphi_i~%7C~i%5Cin%5Cmathbb%20N%2C1%5Cle%20i%5Cle%20k%5C%7D)
be a set of orthogonal vectors. By definition of orthogonality, any pairwise dot product between distinct vectors must be zero, i.e.
![\varphi_i\cdot\varphi_j=\begin{cases}\|\varphi_i\|^2&\text{if }i=j\\0&\text{if }i\neq j\end{cases}](https://tex.z-dn.net/?f=%5Cvarphi_i%5Ccdot%5Cvarphi_j%3D%5Cbegin%7Bcases%7D%5C%7C%5Cvarphi_i%5C%7C%5E2%26%5Ctext%7Bif%20%7Di%3Dj%5C%5C0%26%5Ctext%7Bif%20%7Di%5Cneq%20j%5Cend%7Bcases%7D)
Suppose there is some linear combination of the
![\varphi_i](https://tex.z-dn.net/?f=%5Cvarphi_i)
such that it's equivalent to the zero vector. In other words, assume they are linearly dependent and that there exist
![c_i\in\mathbb R](https://tex.z-dn.net/?f=c_i%5Cin%5Cmathbb%20R)
(not all zero) such that
![\displaystyle\sum_{i=1}^kc_i\varphi_i=c_1\varphi_1+\cdots+c_k\varphi_k=\mathbf 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Ekc_i%5Cvarphi_i%3Dc_1%5Cvarphi_1%2B%5Ccdots%2Bc_k%5Cvarphi_k%3D%5Cmathbf%200)
(This is our hypothesis)
Take the dot product of both sides with any vector from the set:
![v_j\cdot\displaystyle\sum_{i=1}^kc_i\varphi_i=c_1\varphi_j\cdot\varphi_1+\cdots+c_k\varphi_j\varphi_k=\varphi_j\cdot\mathbf 0](https://tex.z-dn.net/?f=v_j%5Ccdot%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Ekc_i%5Cvarphi_i%3Dc_1%5Cvarphi_j%5Ccdot%5Cvarphi_1%2B%5Ccdots%2Bc_k%5Cvarphi_j%5Cvarphi_k%3D%5Cvarphi_j%5Ccdot%5Cmathbf%200)
By orthogonality of the vectors, this reduces to
![c_j\|\varphi_j\|^2=0](https://tex.z-dn.net/?f=c_j%5C%7C%5Cvarphi_j%5C%7C%5E2%3D0)
Since none of the
![\varphi_i](https://tex.z-dn.net/?f=%5Cvarphi_i)
are zero vectors (presumably), this means
![c_j=0](https://tex.z-dn.net/?f=c_j%3D0)
. This is true for all
![j](https://tex.z-dn.net/?f=j)
, which means only
![c_i=0](https://tex.z-dn.net/?f=c_i%3D0)
will allow such a linear combination to be equivalent to the zero vector, which contradicts the hypothesis and hence the set of vectors must be linearly independent.
Answer:
1
Step-by-step explanation:
Answer:
3 + 4x
Step-by-step explanation:
Let "a number" be x.
Therefore "four times a number" is 4 multiplied by x which is 4x.
Now add 3 for "3 plus".
3 + 4x
If you would like to write x = 1/3 * y in general form, you can do this using the following steps:
<span>The general form of the equation is: </span>ax + by + c = 0.
x = 1/3 * y
x - 1/3 * y = 0
The correct result would be <span>x - 1/3 * y = 0.</span>
Answer: Constant
No matter what the input x is, the output f(x) is going to be -4. Therefore, the output is constant.
Note: f(x) can be interchanged with y. So we can say y = -4.