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Dependence, Linear Transformations (Intro to Linear Transformations…

- - - - Think of this as vectors
        
        In this case we don't need the scalars to be all zero
        
        One vector is twice the other
        
        Not all linearly dependent sets are only multiples of each other #
  - - - The only way adding these two vectors results in a zero vector is if the weights are all zero
  - - - If there's only one vector then we analyze only
        xv=0
        
        So we only get the trivial case when v\(\neq\) 0 since that means x must be zero for the left side to be equal to the right side
        
        Therefore the zero vector is the only situation that is linearly dependent.
        
        If v is the zero vector then x can be a nonzero term and the left side will still equal the right side aka non trivial solution aka linearly dependent
        
        In short
        When \(v\neq0\) linear independent
        When v=zero vector linear dependent
      - Remember we're only interested in the weight can be a nonzero and xv still equal 0
    - - Think of this one visually. With only two vectors how can the combination result in zero
        
        Only if they're multiples of each other
        
        If they're multiples of each other we're able to use weights that'll cause the combination to result in zero
- - - - This can still happen if the vectors are not multiples of zero
        
        Vector w in this case is a linear combination of u and v
        
        Even though they're not multiples you can get a zero vector
        
        One example
        u+v-w =0
  - - - Not an augmented matrix
        \( \begin{pmatrix} 1 & 2 & 3&4\\ 7&0&1&1\\ 6&9&5&8\\ \end{pmatrix} \)
        
        \( \begin{pmatrix} 1 & 2 & 3&4\\ 0&-14&-20&-27\\ 0&-3&-13&-16\\ \end{pmatrix} \)
        
        \( \begin{pmatrix} 1 & 2 & 3&4\\ 0&3&13&16\\ 0&0&\frac{122}{3}&\frac{143}{3}\\ \end{pmatrix} \)
        
        This isn't an augmented matrix so if we wrote it out
        \(x_1+2x_2+3x_3+4x_4\\ 3x_2+13x_3+16x_4\\ \frac{122}{3}x_3+\frac{144}{3}x_4\)
        
        If you notice \(x_4\) is a free variable and we could use this to get a non trivial case for Ax=b
    - - In this case there are 5 vectors and only 3 entries
        p>n
    - - This means if there aren't more vectors than entries we can't say it's independent or dependent based off this theorem
  - - - But if one vector is the zero vector then we can make its weight non zero while all the other weights are zero for the vector and we'll end up with the zero vector
        
        \(1v_1+0v_2+...0v_n=0\)
        If \(v_1\) is the zero vector
- - - - When we multiply vector x and u by matrix A we get a new vector.
        Note that even the dimensions change (x goes from 4 by 1 to 2 by 1)
  - - - The set in \(R^n\) is called the domain. The set in \(R^m\) is called the codomain
        
        The vector T (x) in \(R^m\) is called an image
        
        The set of all images is called the range of T
  - - - This means find which vector with two entries after being multiplied with A results in b
        
        Ax=b
        \( \begin{pmatrix} 1 & -3 \\ 3&5\\ -1&7\\ \end{pmatrix} \) \( \begin{pmatrix} x_1 \\ x_2\\ \end{pmatrix} \) = \( \begin{pmatrix} 1 & -3&3\\ 3&5&2\\ -1&7&-5\\ \end{pmatrix} \)
        
        Reducing the augmented matrix we get x=\( \begin{pmatrix} 1.5 \\ -.5\\ \end{pmatrix} \)
    - - No since there were no free variables meaning that we have a unique solution
    - - This is essentially asking when we multiply A tou what is the result? In other words, how does this transformation look like
        
        \( \begin{pmatrix} 1 & -3 \\ 3&5\\ -1&7\\ \end{pmatrix} \) \( \begin{pmatrix} 2 \\ -1\\ \end{pmatrix} \) = \( \begin{pmatrix} 1*2 + -3*-1 \\ 3*2+5*-1\\ -1*2+7*-1\\ \end{pmatrix} \) = \( \begin{pmatrix} 5 \\ 1\\ -9\\ \end{pmatrix} \)
        
        Sou is transformed from {2,-1} to {5,1,-9}
    - - This is asking does there exist a vector that one we multiply with A results in c
        
        In other words is Ax=c consistent?