Finding the Standard Matrix of a Linear Transformation

[Drakkith]

Homework Statement

Let $T:ℝ^3→ℝ^2$ be the linear transformation defined by $\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\mapsto \begin{bmatrix} x_1 + x_2 + x_3\\ 0 \end{bmatrix}$.

i. Find the standard matrix for $T$.

Homework Equations

The Attempt at a Solution

For this problem I was able to guess that the standard matrix is $\begin{bmatrix} 1&1&1 \\ 0&0&0\end{bmatrix}$ (at least I think it is), but what I'd really like to understand is how to find it without simply guessing.

I've tried something like:
$A \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix} x_1+x_2+x_3\\0\end{bmatrix}$
$A=\begin{bmatrix} x_1+x_2+x_3\\0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}^{-1}$

But that doesn't work since the right side can't undergo matrix multiplication. I'm not sure what else to do.

 

[fresh_42]

Remember the janitor method: prepare the tools, inspect the task, work.

Preparation: Matrix multiplication is row $r_i$ times column $c$:
Inspection: Write $r_i c = (t_{i1},t_{i2},t_{i3}) \cdot (x_1,x_2,x_3)^\tau$ for $i = 1,2$
The job is now to solve this system of two linear equations for all combinations of $x_i$ possible.

This is usually done by $c=(1,0,0)\, , \,c=(0.1,0)\; , \; c=(0,0,1)$ to get the matrix entries $t_{ij}$.

 

[Drakkith]

Preparation: Matrix multiplication is row $r_i$ times column $c$:
Inspection: Write $r_i c = (t_{i1},t_{i2},t_{i3}) \cdot (x_1,x_2,x_3)^\tau$ for $i = 1,2$
The job is now to solve this system of two linear equations for all combinations of $x_i$ possible.

I'm sorry I'm not following you. Where did $t_{i1}$, $t_{i2}$, and $t_{i3}$ come from?

 

[fresh_42]

I'm sorry I'm not following you. Where did $t_{i1}$, $t_{i2}$, and $t_{i3}$ come from?

They are the entries of the matrix $T$, its rows: $row\,1 = r_1 = (t_{11},t_{12},t_{13})$ and $row\,2 = r_2 = (t_{21},t_{22},t_{23})$.
The column $c$ is the vector with the $x_i\, , \,c=(x_1,x_2,x_3)^\tau$ and the $\tau$ only means transposed, because I've written it as a row, but meant a column. So I had to transpose it. The given equation reads
$$
T \cdot x = \begin{bmatrix}t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}&\end{bmatrix} \cdot \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}t_{11}x_1+t_{12}x_2+t_{13}x_3 \\ t_{21}x_1+ t_{22}x_2+t_{23}x_3 \end{bmatrix} = \begin{bmatrix}r_1 \\ r_2\end{bmatrix} \cdot c = \begin{bmatrix}r_1 \cdot c \\ r_2 \cdot c\end{bmatrix}=\begin{bmatrix}x_1+x_2+x_3\\ 0 \end{bmatrix}
$$

 

[Drakkith]

They are the entries of the matrix $T$

Oh, so $T$ is the transform matrix then?

 

[fresh_42]

Oh, so $T$ is the transform matrix then?

Yes, usually the function and the matrix have the same letter. And the function is $T\, : \,\mathbb{R}^3 \longrightarrow \mathbb{R}^2$ with a matrix $T= \begin{bmatrix}t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}\end{bmatrix}$ in the standard basis.

Another way is directly substitute a vector $x=(1,0,0)^\tau$ in $Tx=\begin{bmatrix}1\\0 \end{bmatrix}$ and next $x=(0,1,0)^\tau$ and $x=(0,0,1)^\tau$. This gives also the searched values for the matrix entries $t_{ij}$.

 

[StoneTemplePython]

Another approach, that isn't emphasized as much, is to treat your matrix as representing a graph. You only need a tiny bit of graph theory for do this, but outside of markov and quantum stuff, people tend to ignore the matrix-graph representation I suppose.

The problem tells us that you have 3 starting states $x_1, x_2, x_3$ that each gets sent into one bucket with weighting one, and another bucket with weighting zero (or not sent to the second bucket at all, if you prefer). You can easily draw the circles and arrows of what this one time transition looks like, and then spend only a little time backfilling the associated matrix. (The idea of course is cleaner if the associated matrix is square, but I don't see a major issue with the non-square setup in this problem).

I happen to like drawing little pictures, so the graph theory approach has a lot of appeal to me, and maybe some others too.

 

[Drakkith]

I'm afraid I'm still not sure how to solve this.

$\begin{bmatrix}t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}x_1+x_2+x_3\\0\end{bmatrix}$

If I set up a system of equations, what goes on the right side of the equal sign for row 1? $x_1+x_2+x_3$? Or 1+1+1?
Neither make sense to me, because we have 3 variables but only 2 equations.

 

[fresh_42]

I'm afraid I'm still not sure how to solve this.

$\begin{bmatrix}t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}x_1+x_2+x_3\\0\end{bmatrix}$

If I set up a system of equations, what goes on the right side of the equal sign for row 1? $x_1+x_2+x_3$? Or 1+1+1?
Neither make sense to me, because we have 3 variables but only 2 equations.

We have the equation $Tx=x'$ with the matrix $T=(t_{ij})$ and the vectors $x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ and $x'=\begin{bmatrix}x_1+x_2+x_3\\0 \end{bmatrix}$.

If we write it out, i.e. $Tx$ in coordinates, then it gets
$$
T \cdot x = \begin{bmatrix}t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}&\end{bmatrix} \cdot \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}t_{11}x_1+t_{12}x_2+t_{13}x_3 \\ t_{21}x_1+ t_{22}x_2+t_{23}x_3 \end{bmatrix} =\begin{bmatrix}x_1+x_2+x_3\\ 0 \end{bmatrix}
$$
Now the crucial point is, that this has to hold for all values of $x_1,x_2,x_3$, because they represent any vector of the domain $\mathbb{R}^3$. This means, that we can substitute whatever we want for those values and can generate as many equations as we like. Most will be linearly dependent, so they won't necessarily lead to qualitatively different equations. E.g. replacing $(x_1,x_2,x_3)=(1,0,0)$ and $(x_1,x_2,x_3)=(2,0,0)$ will yield the same result. But the substitutions $(x_1,x_2,x_3)=(1,0,0)\; , \;(x_1,x_2,x_3)=(0,1,0)\; , \;(x_1,x_2,x_3)=(0,0,1)$ give us three times two independent equations to solve for the $t_{ij}$. Just plug them in and read out the values $t_{11}=1\; , \; t_{21}=0$, then $t_{12}=1\; , \; t_{22}=0$ and finally $t_{31}=1\; , \; t_{31}=0$.

 

[I like Serena]

Homework Statement

Let $T:ℝ^3→ℝ^2$ be the linear transformation defined by $\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\mapsto \begin{bmatrix} x_1 + x_2 + x_3\\ 0 \end{bmatrix}$.

We can read off the transformation matrix directly.

Matrix notation is really just a short hand.
We write:
$$\begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$
as a short hand for:
$$\begin{cases}ax+by\\ cx+dy\end{cases}$$
The matrix consists of the coefficients of a linear system.

In our case the coefficients of
$$\begin{bmatrix}
x_1 + x_2 + x_3\\ 0
\end{bmatrix}=
\begin{bmatrix}
1\cdot x_1 + 1\cdot x_2 + 1\cdot x_3\\ 0\cdot x_1 + 0\cdot x_2 + 0\cdot x_3
\end{bmatrix}$$
form the matrix that you've 'guessed'.

 

[Drakkith]

But the substitutions $(x_1,x_2,x_3)=(1,0,0)\; , \;(x_1,x_2,x_3)=(0,1,0)\; , \;(x_1,x_2,x_3)=(0,0,1)$ give us three times two independent equations to solve for the $t_{ij}$.

Fresh, where did you get those vectors from? And why are we plugging numbers into $x_i$? I thought we usually ignored them and only worried about their coefficients.

 

[Drakkith]

Fresh, where did you get those vectors from? And why are we plugging numbers into $x_i$? I thought we usually ignored them and only worried about their coefficients.

Thinking more on this, I believe that when we have a system of equations we usually have the coefficients of $x_i$ and we are trying to find each $x_i$. Here we want to find the coefficients. Is that correct?

 

[fresh_42]

Fresh, where did you get those vectors from? And why are we plugging numbers into $x_i$? I thought we usually ignored them and only worried about their coefficients.

We have a transformation, aka (linear) mapping, aka (linear) function $T\, : \, \mathbb{R}^3 \longrightarrow \mathbb{R}^2$. Our $T(x)$ here is the same as matrix of $T$, which I also denoted by $T$, applied to a three dimensional vector with real components, e.g. a vector $\vec{x}=(x_1,x_2,x_3) \in \mathbb{R}^3$. So $T(\vec{x}) = T \cdot \vec{x}\;^*)$.

$T$ is a function, which means it describes how elements from the domain are mapped to elements of the codomain, here vectors $\vec{x}=(x_1,x_2,x_3)$ to vectors $\vec{x}'=(x_1+x_2+x_3,0)$, the same as $f(x)=y=x^2$ maps $x \mapsto x^2$. And same as with the parabola $f(x)$ we can put in all kinds of values for $x$ to, e.g. draw the graph. Here we do it to find a description of $T$ in terms of a matrix.

___________
$^*)$ Before anyone complains that I didn't distinguish between the function and the matrix, yes, I know. I wanted to keep things easy here, as the topic is not the role of basis.

 

[fresh_42]

Thinking more on this, I believe that when we have a system of equations we usually have the coefficients of $x_i$ and we are trying to find each $x_i$. Here we want to find the coefficients. Is that correct?

Yes. It's like having $y=ax^2+bx+c=2x^2+4$ - find $a,b,c$. Same thing.

 

[Drakkith]

Yes. It's like having $y=ax^2+bx+c=2x^2+4$ - find $a,b,c$. Same thing.

Oh, okay.

So we have:
$T_{11}x_1+T_{12}x_2+T_{13}x_3= x_1+x_2+x_3$
$T_{21}x_1+T_{22}x_2+T_{23}x_3=0$

Solving the first:
$T_{11}x_1-x_1+T_{12}x_2-x_2+T_{13}x_3-x_3=0$
$x_1(T_{11}-1)+x_2(T_{12}2)+x_3(T_{13}-1)=0x_1+0x_2+0x_3$
$T_{11} - 1=0$, $T_{11} = 1$
$T_{12} - 1=0$, $T_{12} = 1$
$T_{13} - 1=0$, $T_{13} = 1$

The same process for the second yields:
$T_{21} = 0$
$T_{22}=0$
$T_{23}=0$

$T$ is thus: $T=\begin{bmatrix}1&1&1\\0&0&0\end{bmatrix}$

 

[fresh_42]

Yes. Maybe a bit complicated, but basically yes. I prefer the substitution method, as it directly yields the solution. For the componentwise comparison of the coefficients at the different $x_i$ we need that they are linear independent anyways, because we must give an argument, that a) $x_i \neq 0$ and b) that they don't eliminate themselves by addition. So we need the "any $x_i$" argument in both cases.

 

[Drakkith]

Yes. Maybe a bit complicated, but basically yes. I prefer the substitution method, as it directly yields the solution. For the componentwise comparison of the coefficients at the different $x_i$ we need that they are linear independent anyways, because we must give an argument, that a) $x_i \neq 0$ and b) that they don't eliminate themselves by addition. So we need the "any $x_i$" argument in both cases.

I'm still not sure how you did the substitution method.
Were those vectors chosen arbitrarily?

 

[fresh_42]

I'm still not sure how you did the substitution method.
Were those vectors chosen arbitrarily?

You have an equation $x_1(T_{11}-1)+x_2(T_{12}-2)+x_3(T_{13}-1)=0x_1+0x_2+0x_3$.

So firstly, we have to ask ourselves: Why can they be compared by each coefficient?

E.g. couldn't it be, that the $x_1$ part somehow swallows the $x_2$ part and only the $x_3$ part will be left. In such a case, we could only get something about the coefficient $T_{13}-1$ at $x_3$. Or any other constellation, in which they eliminate themselves. I mean $3+5=2+6$ doesn't mean $3=2$ and $5=6$. We have to rule out this possibility, in order to proceed with our comparison.

The solution to this is: No, because it has to be valid for any choices of constellations of the $x_i$ and therefore we can treat them like indeterminates (or variables). We are allowed to set $x_1=x_2=0$ and $x_3=1$, which leads to $0\cdot (T_{11}-1)+ 0\cdot (T_{12}-1) +1 \cdot (T_{13}-1)= T_{13}-1=0$. And similar for the other ones. I find this way of argumentation easier than any written sentences why $ax+by=cx+dy$ implies $a=c$ and $b=d$.

And secondly, if we have now $x_1(T_{11}-1)=0$, why can't be $x_1=0$ in which case we can't say anything about $T_{11}$.

Thing is, $x_1=0$ is allowed. But $x_1=1$ is also allowed. So we must not bother about the zero case, as we can treat $x_1$ again as indeterminate (or variable), i.e. we might as well substitute $x_1=1$ to get the result. Of course any other values unequal zero will do as well, but $1$ is usually easier. If it was something like $x_1 \dfrac{(T_{11}-1)}{7}$ I would not hesitate and substitute $x_1=7$. In case we avoid the substitution argument, we have to replace it by another argument, which allows us to rule out the $x_1=0$ case, because this one doesn't help.

 

[Drakkith]

You have an equation $x_1(T_{11}-1)+x_2(T_{12}-2)+x_3(T_{13}-1)=0x_1+0x_2+0x_3$.

So firstly, we have to ask ourselves: Why can they be compared by each coefficient?

E.g. couldn't it be, that the $x_1$ part somehow swallows the $x_2$ part and only the $x_3$ part will be left. In such a case, we could only get something about the coefficient $T_{13}-1$ at $x_3$. Or any other constellation, in which they eliminate themselves. I mean $3+5=2+6$ doesn't mean $3=2$ and $5=6$. We have to rule out this possibility, in order to proceed with our comparison.

Doesn't algebra already rule this out for us?

 

[fresh_42]

Doesn't algebra already rule this out for us?

Not automatically. E.g. if $ax+by=0$, then $a=b=0$ if we can treat $x,y$ as indeterminates (variables, free parameters, dimensions, whatever). But if e.g. $y=2x$ then all of a sudden, we have $ax+by=ax+2bx=(a+2b)x = 0$ and $a=b=0$ can not be concluded anymore. In this case it is only one among many possible solutions.

So the argument: My function $T\, : \,\mathbb{R}^3 \longrightarrow \mathbb{R}^2$ described by ... has to be valid for all possible points, vectors, elements of $\mathbb{R}^3$, therefore as well for those I deliberately choose, is in my mind more convincing than any abstract phrases about linear independent vectors.

The difference is, that the independent treatment of $x_1,x_2,x_3$ in $\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}$ is because they span three dimensions, i.e. are linearly independent at their different positions in the vector, can be treated as free variables because of it, whereas my substitution simply means to choose a certain basis of this space: $(1,0,0)\; , \;(0,1,0)\; , \;(0,0,1).$ And for vector spaces holds: what's true on the basis is true on the entire vector space. Of course only for linear statements, not things like length or so.

Thus substitution: "choose any $x_i$" and substitution by basis vectors: "choose a basis of $\mathbb{R}^3$" or componentwise comparison: "treat the $x_i$ as variables" are only different sides of the same coin.
However, the coin is necessary.

 

[Drakkith]

Ok I think I see what you're saying. Now, can you have a transform where you can't do my method above because $x_1...x_i$ aren't linearly independent?

 

[fresh_42]

Ok I think I see what you're saying. Now, can you have a transform where you can't do my method above because $x_1...x_i$ aren't linearly independent?

Not if we have $\vec{x}=\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}$ where each $x_i$ represents a separate dimension of $\mathbb{R}^3$. I'm only saying, that this circumstance is used and that considering them as free variables is the same as choosing a basis of three vectors for them.

But we could have a linear equation system with variables $x_i$ where it is not clear whether they represent a dimension each. E.g. the solutions to $Tx=\begin{bmatrix}3\\0 \end{bmatrix}$ define a plane in $\mathbb{R}^3$ given by $x_1+x_2+x_3=3$ or $x_3=-x_1-x_2+3$ where only $x_1$ and $x_2$ are free parameters (spanning the plane) and the third one is already determined by them (defining the position of this plane in space).

 

[Drakkith]

I see. That makes sense. Thanks fresh and everyone else. I think my understanding of transformations just increased by about 1000%.