# Linear transformations using matrices

Let $$y_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \qquad y_2 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \qquad\text{and}\qquad y_3 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ and let $L$ be be a linear operator on $\mathbb R^3$ defined by $$L(c_1 y_1 + c_2 y_2 + c_3 y_3) = \left(c_1 + c_2 + c_3\right) y_1 + \left(2c_1+c_3\right) y_2 - \left(2c_2+c_3\right) y_3$$

Find a matrix representing $L$ with respect to the order pair basis $\{y1,y2,y2\}$.

I am not sure how to do this problem.

I know to find matrix $A$ you have to do the process $$\left[y_1 y_2 y_3 \mid L(y_1) \, L(y_2) \, L(y_3) \right] \sim [I \mid A]$$ But I am not sure how to find $L(y_1), L(y_2), L(y_3)$.

\begin{align}&Ly_1:=y_1+2y_2\\{}\\ &Ly_2:=y_1-2y_3\\{}\\&Ly_3:=y_1+y_2-y_3\end{align}\implies [L]=\begin{pmatrix}1&1&1\\2&0&1\\0&\!\!-2&\!\!-1\end{pmatrix}

Hint to deduce the above:

$$Ly_1=L(1\cdot y_1+0\cdot y_2+0\cdot y_3):=\;etc.$$

• Wait but how did you deduce that $L(y_1)=y_1+2y_2$ Feb 6, 2015 at 21:36
• @FernandoMartinez Read carefully the last line, and use linearity of $\;L\;$ and, of course, its definition . Feb 6, 2015 at 21:38
• Ok I think I see you have $L(1y_1)$ and $1=c_1$ and thus you apply the transformation formula. Getting $y_1+2y_2$ Feb 6, 2015 at 21:40
• @FernandoMartinez Exactamente. Feb 6, 2015 at 21:41

Order the right Hand side of the equation for $L(c_1y_1+c_2y_2+c_3y_3)$ by the coefficients in $c_1,c_2,c_3$. On the left Hand side you can use the linearity of $L$; for example $L(c_1y_1) = c_1 L(y_1)$. Compare the arguments on the left Hand side and the right Hand side; equality must be true for arbitrary coefficients $c_1,c_2,c_3$. You will get the mapped vectors $L(y_1),L(y_2),L(y_3)$ in Terms of $y_1,y_2,y_3$. Now it is simple to find the corresponding matrix $L$; remember: $(L(y_1),L(y_2),L(y_3)) = L * (y_1,y_2,y_3)$.

We want to have $${A} \begin{bmatrix} c_1 \\ c_2 \\c_3 \end{bmatrix}= \begin{bmatrix} c_1+c_2+c_3 \\ 2c_1+c_3 \\-2c_2 - c_{3} \end{bmatrix}$$ then: $A$ must be: $${A} = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 0 & -2 & -1 \end{bmatrix}$$