# Show that t is a linear transformation given the equation

I am not sure what the procedure is for solving these type of problems. Do I need to put the equations in reduced row echelon form? Do I need to solve for

$T \begin{bmatrix}1 \\ 0 \end{bmatrix}$ and $T \begin{bmatrix}0 \\ 1 \end{bmatrix}$

Any help in the procedure would be greatly appreciated.

Consider the function $T\begin{bmatrix} x_1\\ x_2\\x_3 \end{bmatrix} = \begin{bmatrix} x_1 - 2x_2 +7x_3\\ x_2 +4x_3 \end{bmatrix}$

How would you solve for one-to-one and onto after this?

I tried reduced row echelon form:

$\begin{bmatrix} 1 & -2 & 7 \\ 0 & 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 15 \\ 0 & 1 & 4 \end{bmatrix}$

Then I was thinking:

\begin{align*} x_1 &= -15x_3 & x_2 &= -4x_3 & x_3 &= free \end{align*}

Then I tried this:

$T\left[\begin{matrix} 1 \\ 0 \end{matrix} \right] = 1 \left[\begin {matrix} 1 \\ 0 \end{matrix} \right]+ 0 \left[\begin{matrix} -2 \\ 1 \end{matrix} \right]+ 0 \left[\begin{matrix} 7 \\ 4 \end{matrix} \right]$.

$T\left[\begin{matrix} 0 \\ 1 \end{matrix} \right] = 2 \left[\begin {matrix} 1 \\ 0 \end{matrix} \right]+ 1 \left[\begin{matrix} -2 \\ 1 \end{matrix} \right]+ 0 \left[\begin{matrix} 7 \\ 4 \end{matrix} \right]$.

Where does this matrix addition transformation rule come from? I see it in my textbook a lot but don't see an explanation anywhere. When do you use this rule?

$T\left[\begin{matrix} x_1 + y_1 \\ x_2 + y_2 \end{matrix} \right] = \left[\begin {matrix} x_1 + y_1 + x_2 + y_2 \\ x_2 + y_2 \end{matrix} \right]$.

• You only need to show that $T$ is linear, not one-to-one or onto. For what its worth, however, $T$ is not one-to-one due to the dimensions of its domain and codomain. – joeb Mar 14 '17 at 23:58
• @joeb can you explain the procedure for this please? – cokedude Mar 15 '17 at 0:02

An easy way is to write $T\left[\begin{matrix} x _1 \\ x_2 \\ x_3 \end{matrix} \right] = x_1\left[\begin{matrix} 1 \\ 0 \end{matrix} \right]+x_2\left[\begin{matrix} -2 \\ 1 \end{matrix} \right]+x_3\left[\begin{matrix} 7 \\ 4 \end{matrix} \right]$.
Thus $$T\left( \left[\begin{matrix} x _1 \\ x_2 \\ x_3 \end{matrix} \right]+\left[\begin{matrix} y _1 \\ y_2 \\ y_3 \end{matrix} \right]\right) = T\left( \left[\begin{matrix} x _1 +y_1\\ x_2+y_2 \\ x_3+y_3 \end{matrix} \right]\right)$$ $$= (x_1+y_1)\left[\begin{matrix} 1 \\ 0 \end{matrix} \right]+(x_2+y_2)\left[\begin{matrix} -2 \\ 1 \end{matrix} \right]+(x_3+y_3)\left[\begin{matrix} 7 \\ 4 \end{matrix} \right]$$ $$= \left(x_1\left[\begin{matrix} 1 \\ 0 \end{matrix} \right]+x_2\left[\begin{matrix} -2 \\ 1 \end{matrix} \right]+x_3\left[\begin{matrix} 7 \\ 4 \end{matrix} \right]\right) + \left(y_1\left[\begin{matrix} 1 \\ 0 \end{matrix} \right]+y_2\left[\begin{matrix} -2 \\ 1 \end{matrix} \right]+y_3\left[\begin{matrix} 7 \\ 4 \end{matrix} \right]\right)$$ $$= T\left( \left[\begin{matrix} x _1 \\ x_2 \\ x_3 \end{matrix} \right]\right)+T\left(\left[\begin{matrix} y _1 \\ y_2 \\ y_3 \end{matrix} \right]\right).$$
Prove that $T(c\mathbf{x}) = cT(\mathbf{x})$ analogously.
• By definition a linear transformation $S : V \to W$ between vector spaces over a common field (say the real field $\mathbb{R}$) is a function that satisfies the following two conditions: 1) that $S(\mathbf{x}+\mathbf{y}) = S(\mathbf{x}) + S(\mathbf{y})$ for any vectors $\mathbf{x,y} \in V$, and 2) that $S(c\mathbf{x}) = cS(\mathbf{x})$ whenever $c \in \mathbb{R}$ is a scalar and $\mathbf{x} \in V$ is a vector. If you can prove $S$ has these two properties, you're done. Also, you were probably downvoted because you didn't show your own attempts or explain why you were confused. – joeb Mar 15 '17 at 0:20
• Knowing the behavior of $T$ on a basis help you tremendously only if you already know that $T$ is a linear transformation - if you don't know $T$ is a linear transformation, its behavior on a basis tells you pretty much nothing. Moreover, reduced row echelon form doesn't really apply to this situation (reduced form helps one understand linear transformations, yes, but again, you don't know that $T$ is linear - you're trying to prove this). – joeb Mar 15 '17 at 1:43