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

[karush]

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
I presume we could start with the middle one.

 

[topsquark]

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
I presume we could start with the middle one.

Well, the most direct method would be to simply solve for T. But since T is linear there is another way.

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

-Dan

 

[karush]

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

 

[topsquark]

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
[math]\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 ] [/math]

so
[math]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 ] [/math]

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

-Dan

 

[karush]

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?

 

[Evgeny.Makarov]

how would we know if T is linear?

Suppose that $T: \Bbb{R}^3 \rightarrow \Bbb{R}^3$ is a linear transformation

then?

[math]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 ] [/math]

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 $T:\mathbb{R}^3\to\mathbb{R}^3$.

 

[karush]

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