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Show that two given vectors span the real plane

Given vectors A = (2,1) and B = (1,3) show that every vector C = (c_1, c_2) in \mathbb{R}^2 is of the form C = xA + yB for some x,y \in \mathbb{R}. Express x and y in terms of c_1 and c_2.

Proof. (Note: I’m doing this without any real linear algebra since we don’t get to that until Volume 2. If you do know linear algebra then you would show the two vectors A and B are linearly independent and you’d be done.) Since

    \[ C = xA+ yB + (2x+y, x+ 3y) = (c_1, c_2) \]

we have

    \begin{align*}  c_1 &= 2x+y \\  c_2 &= x + 3y. \end{align*}

This implies

    \[ x = c_2 - 3y \quad \implies \quad c_1 = 2c_2 - 6y + y. \]


    \begin{align*}  y &= \frac{1}{5} \left( 2c_2 - c_1 \right) \\  x &= \frac{1}{5} \left( 3c_1 - c_2 \right). \end{align*}

This also shows that any vector in \mathbb{R}^2 can be obtained as xA + yB since given C = (c_1, c_2) we compute x and y by the formulas above. \qquad \blacksquare

One comment

  1. Alan Gonzaga says:

    There is an error on the first line
    Should be an “=” in the place of one of the “+”

    There is C = xA + yB “+”
    Should be C = xA + yB “=”

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