MHB What are the linear operators S and T?

[karush]

let $T:\Bbb{R}^2\rightarrow\Bbb{R}^2$ and $S:\Bbb{R}^2\rightarrow\Bbb{R}^2$ be defined by
$S\left[\begin{array}{c}x \\ y \end{array}\right]=\left[\begin{array}{c}3x+y \\ x+2y \end{array}\right],\qquad
T\left[\begin{array}{c}x \\ y \end{array}\right]=\left[\begin{array}{c}2y \\ 3x \end{array}\right]$
Find $S+T, \quad 3S+4T, \quad ST, \quad TS $

so
$S+T=\left[\begin{array}{c}3+1 \\ 1+2 \end{array}\right]
+\left[\begin{array}{c}3+0\\0+2 \end{array}\right]$
on $T$ $R_1 \leftrightarrow R_2$ before procede if ok...

 

[topsquark]

let $T:\Bbb{R}^2\rightarrow\Bbb{R}^2$ and $S:\Bbb{R}^2\rightarrow\Bbb{R}^2$ be defined by
$S\left[\begin{array}{c}x \\ y \end{array}\right]=\left[\begin{array}{c}3x+y \\ x+2y \end{array}\right],\qquad
T\left[\begin{array}{c}x \\ y \end{array}\right]=\left[\begin{array}{c}2y \\ 3x \end{array}\right]$
Find $S+T, \quad 3S+4T, \quad ST, \quad TS $

so
$S+T=\left[\begin{array}{c}3+1 \\ 1+2 \end{array}\right]
+\left[\begin{array}{c}3+0\\0+2 \end{array}\right]$
on $T$ $R_1 \leftrightarrow R_2$ before procede if ok...

The problem, as written, is a bit ambiguous.

Are you trying to find [math](S + T) \left [ \begin{matrix} x \\ y \end{matrix} \right ] = \left [ \begin{matrix} 3x + y \\ x + 2y \end{matrix} \right ] + \left [ \begin{matrix} 2y \\ 3x \end{matrix} \right ] [/math]

or S + T as an operator (which would be a 2 x 2 matrix.)

-Dan

 

[Evgeny.Makarov]

$S+T=\left[\begin{array}{c}3+1 \\ 1+2 \end{array}\right]
+\left[\begin{array}{c}3+0\\0+2 \end{array}\right]$

You can't write $S+T=\ldots$ because $S+T$ is a function and it must accept an argument. $(S+T)\left[\begin{array}{c}x \\ y\end{array}\right]=\ldots$ is OK.

 

[karush]

The problem, as written, is a bit ambiguous.

Are you trying to find [math](S + T) \left [ \begin{matrix} x \\ y \end{matrix} \right ] = \left [ \begin{matrix} 3x + y \\ x + 2y \end{matrix} \right ] + \left [ \begin{matrix} 2y \\ 3x \end{matrix} \right ] [/math]

or S + T as an operator (which would be a 2 x 2 matrix.)

-Dan

the first 3 lines was the given problem
possible my notation was off

 

[karush]

You can't write $S+T=\ldots$ because $S+T$ is a function and it must accept an argument. $(S+T)\left[\;{array}{c}x \\ y\end{array}\right]=\ldots$ is OK.

$(S+T)\left[\begin{array}{c}x \\ y\end{array}\right]
=\left[\begin{array}{c}3+1 \\ 1+2 \end{array}\right]
+\left[\begin{array}{c}3+0\\0+2 \end{array}\right]
=6-0=6$
hopfully

 

[Evgeny.Makarov]

$(S+T)\left[\begin{array}{c}x \\ y\end{array}\right]
=\left[\begin{array}{c}3+1 \\ 1+2 \end{array}\right]
+\left[\begin{array}{c}3+0\\0+2 \end{array}\right]$

So if $f(x)=3x$ and $g(x)=4x$, then $f(x)+g(x)=3+4$? Where did $x$ go? You need simply to add the results of $S$ and $T$ componentwise.

$\left[\begin{array}{c}3+1 \\ 1+2 \end{array}\right]
+\left[\begin{array}{c}3+0\\0+2 \end{array}\right]=6-0=6$

An order pair if numbers (a column) can never equal a single number.

 

[topsquark]

so thus
$(S+T)\left[\begin{array}{c}x \\ y\end{array}\right]=
\left[\begin{array}{c}3+1 \\ 1+2 \end{array}\right]
+\left[\begin{array}{c}3+0\\0+2 \end{array}\right]=
\left[\begin{array}{c}6+1\\1+4\end{array}\right]$

Where do the x and y values keep going? You don't have values for them, do you?

Many of your answers are not going to be simple numbers. You are dealing here with operators that act on vectors.

Here, we've already established that (post 2.)
[math](S + T) \left [ \begin{matrix} x \\ y \end{matrix} \right ] = \left [ \begin{matrix} 3x + 3y \\ 4x + 2y \end{matrix} \right ] [/math]

This is what I suspect the question is getting at... How do you use the operator S + T on a vector [math]\left [ \begin{matrix} x \\ y \end{matrix} \right ] [/math].

Look at the derivation and make sure you understand where things are going wrong. I'm not sure why you are having the problems you are having so please make sure to try deriving this on your own and let us know where things start going kablooey.

-Dan

 

[karush]

i thot matrix only was about coefficients

 

[topsquark]

i thot matrix only was about coefficients

They can be used that way. Would it help if I said that we can construct S?

[math]S = \left ( \begin{matrix} 3 & 1 \\ 1 & 2 \end{matrix} \right )[/math]

So we get
[math]S \left [ \begin{matrix} x \\ y \end{matrix} \right ] = \left ( \begin{matrix} 3 & 1 \\ 1 & 2 \end{matrix} \right ) ~ \left [ \begin{matrix} x \\ y \end{matrix} \right ] = \left [ \begin{matrix} 3x + y \\ x + 2y \end{matrix} \right ] [/math]

We can construct a matrix for T also. But the point of knowing that these are linear operators is that we can often skip deriving the matrix for the operator, which can save a considerable amount of time. That's what you are doing in trying to learn this.

-Dan