13 is a linear transformation and .......Determine T

• MHB
Gold Member
MHB
Suppose that $T: \Bbb{R}^3 \rightarrow \Bbb{R}^3$ is a linear transformation and
$$T \begin{bmatrix} 1 \\1 \\0 \\ \end{bmatrix} = \begin{bmatrix} 1 \\2 \\1 \\ \end{bmatrix}, \quad T \begin{bmatrix} 1 \\0 \\1 \\ \end{bmatrix} = \begin{bmatrix} 1 \\0 \\2 \\ \end{bmatrix}, \quad T \begin{bmatrix} 0 \\1 \\0 \\ \end{bmatrix} = \begin{bmatrix} 2 \\2 \\3 \\ \end{bmatrix}.$$
Determine $T \begin{bmatrix} 1 \\2 \\3 \\ \end{bmatrix}$

ok this should be easy... but.. the examples were not that close to this

Gold Member
MHB
Suppose that $T: \Bbb{R}^3 \rightarrow \Bbb{R}^3$ is a linear transformation and
$$T \begin{bmatrix} 1 \\1 \\0 \\ \end{bmatrix} = \begin{bmatrix} 1 \\2 \\1 \\ \end{bmatrix}, \quad T \begin{bmatrix} 1 \\0 \\1 \\ \end{bmatrix} = \begin{bmatrix} 1 \\0 \\2 \\ \end{bmatrix}, \quad T \begin{bmatrix} 0 \\1 \\0 \\ \end{bmatrix} = \begin{bmatrix} 2 \\2 \\3 \\ \end{bmatrix}.$$
Determine $T \begin{bmatrix} 1 \\2 \\3 \\ \end{bmatrix}$

ok this should be easy... but.. the examples were not that close to this
Well, the most direct method would be to simply solve for T. But since T is linear there is another way.

Can you build $$\displaystyle \left [ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right ]$$ out of a linear combination of $$\displaystyle \left [ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right ]$$, $$\displaystyle \left [ \begin{matrix} 1 \\ 0 \\ 1 \end{matrix} \right ]$$, and $$\displaystyle \left [ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right ]$$?

-Dan

Gold Member
MHB
ok this probably is not exactly what it is supposed to be but

$T \begin{bmatrix} 1 \\1 \\0 \end{bmatrix} =\begin{bmatrix} 1 \\2 \\1 \end{bmatrix}, \quad T \begin{bmatrix} 1 \\0 \\1 \end{bmatrix} = \begin{bmatrix} 1 \\0 \\2 \end{bmatrix}' \quad T \begin{bmatrix} 0 \\1 \\0 \end{bmatrix} = \begin{bmatrix} 2 \\2 \\3 \end{bmatrix}$.

$=T\left[\begin{array}{c} 1&1&0\\ 1&0&1\\ 0&1&0 \end{array}\right]$

or possibly
$\left[\begin{array}{c}x_1+x_2 \\x_1+x_3\\x_2\end{array}\right]$

Last edited:
Gold Member
MHB
ok this probably is not exactly what it is supposed to be but

$T \begin{bmatrix} 1 \\1 \\0 \end{bmatrix} =\begin{bmatrix} 1 \\2 \\1 \end{bmatrix}, \quad T \begin{bmatrix} 1 \\0 \\1 \end{bmatrix} = \begin{bmatrix} 1 \\0 \\2 \end{bmatrix}' \quad T \begin{bmatrix} 0 \\1 \\0 \end{bmatrix} = \begin{bmatrix} 2 \\2 \\3 \end{bmatrix}$.

$=T\left[\begin{array}{c} 1&1&0\\ 1&0&1\\ 0&1&0 \end{array}\right]$

are we trying to build $Ax=B$
What I am trying to point you toward is
$$\displaystyle \left [ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right ] = a \left [ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right ] + b \left [ \begin{matrix} 1 \\ 0 \\ 1 \end{matrix} \right ] + c \left [ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right ]$$

so
$$\displaystyle T \left [ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right ] = a T \left [ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right ] + b T \left [ \begin{matrix} 1 \\ 0 \\ 1 \end{matrix} \right ] + c T \left [ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right ]$$

Note that this can only be done if T is linear.

-Dan

Gold Member
MHB
then?
$\left[\begin{array}{c}Ta+Tb \\Ta+Tc\\Tb\end{array}\right]$

i think I am getting confused by looking at too many examples

how would we know if T is linear?

Gold Member
MHB
how would we know if T is linear?
Suppose that $T: \Bbb{R}^3 \rightarrow \Bbb{R}^3$ is a linear transformation

then?
$$\displaystyle T \left [ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right ] = a T \left [ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right ] + b T \left [ \begin{matrix} 1 \\ 0 \\ 1 \end{matrix} \right ] + c T \left [ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right ]$$
and
$$T \begin{bmatrix} 1 \\1 \\0 \\ \end{bmatrix} = \begin{bmatrix} 1 \\2 \\1 \\ \end{bmatrix}, \quad T \begin{bmatrix} 1 \\0 \\1 \\ \end{bmatrix} = \begin{bmatrix} 1 \\0 \\2 \\ \end{bmatrix}, \quad T \begin{bmatrix} 0 \\1 \\0 \\ \end{bmatrix} = \begin{bmatrix} 2 \\2 \\3 \\ \end{bmatrix}.$$

As for
$\left[\begin{array}{c}Ta+Tb \\Ta+Tc\\Tb\end{array}\right]$
$Ta$ does not makes sense for $a\in\mathbb{R}$ because $T:\mathbb{R}^3\to\mathbb{R}^3$.

Gold Member
MHB
ok apparently I'm not understanding the steps
not sure what I should be asking