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Given Linear Transformation $T: R^3->R^2 $ so that:
$$T \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} =\begin{bmatrix} 2 \\ -1 \end{bmatrix} , T \begin{bmatrix} 2 \\ -2 \\ 0 \end{bmatrix} =\begin{bmatrix} -2 \\ -2 \end{bmatrix} , T \begin{bmatrix} 2 \\ -1 \\ -2 \end{bmatrix} =\begin{bmatrix} -1 \\ 1 \end{bmatrix}$$
and Given Linear Transformation $S: R^2->R^3 $ so that:
$$S \begin{bmatrix} -1 \\ 2 \end{bmatrix} =\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} , S \begin{bmatrix} 1 \\ -1 \end{bmatrix} =\begin{bmatrix} 1 \\ -2 \\ -2 \end{bmatrix}$$

How can i find the representative matrix of $ST$ and the representative matrix of $TS$ with respect to the standard basis.?

Master Question
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  • $ST$ and $TS$ are the matrix representations of the operators composition $S\circ T$ and $T\circ $. You can find such operators and then compute their matrix representations. The wonderful thing is that it is equivalent to compute the matrix multiplications $ST$ and $TS$. I hope this comment helps you. – Dog_69 Jan 09 '18 at 22:13
  • @Dog_69 I still don't understand how can i find the representative matrix of $ST$ and the representative matrix of $TS$ with respect to the standard basis? – Master Question Jan 09 '18 at 22:24
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    In general, if $V,W$ are two finite dimensional vector spaces, $T:V\rightarrow W$ is a linear operator and ${v_i},{w_i}$ are bases of $V$ and $W$ resp., the matrix representation of $T$ on these basis is constructed setting the $i$th column as the vector $T(v_i)$ (expressed on the basis ${w_i}$, resulting to apply $T$ to the $i$th basis vector of $V$. – Dog_69 Jan 09 '18 at 22:32
  • @Dog_69 Can you explain it in clear answer for this question please? – Master Question Jan 09 '18 at 22:35

2 Answers2

1

Let $V,W$, $T:V\rightarrow W$ and $\{v_i\},\{w_i\}$ as above. The matrix representation of $T$ is now constructed as follows:

  • First, if $\mbox{dim }V=m$ and $\mbox{dim }W=n$ the matrix is $n\times m$ (because you applied it to $m\times1$ matrices to get $n\times 1$), and the size is independent of the choice of basis.

  • Now suppose

$$ T(e_1)=t^1_1w_1 + t^2_1w_2 + \cdots+ t^n_1w_n= \begin{pmatrix} t^1_1\\t^2_1\\ \vdots \\ t^n_1 \end{pmatrix} \tag{1} $$

Then

$$ T= \begin{pmatrix} t_1^1 & \cdot & \cdot & \cdots & \cdot \\ t^2_1 & \cdot & \cdot & \cdots & \cdot \\ \vdots & \vdots & \vdots & \ddots & \cdot\\ t^n_1 & \cdot & \cdot & \cdots & \cdot \end{pmatrix}. $$

  • Generically, for the $i$-th vector $e_i$ you will get

$$ T(e_i)=t^1i1w_1 + t^2_iw_2 + \cdots+ t^n_iw_n= \begin{pmatrix} t^1_i\\t^2_i\\ \vdots \\ t^n_1 \end{pmatrix} \tag{2} $$

and then

$$ T= \begin{pmatrix} t_1^1 & \cdots & t_i^1 & \cdots & \cdot \\ t^2_1 & \cdots & t_i^2 & \cdots & \cdot \\ \vdots & \vdots & \vdots & \ddots & \cdot\\ t^n_1 & \cdots & t_i^n & \cdots & \cdot \end{pmatrix}. $$

For finish, two remarks:

The (linear) operators $T$ and $S$ are independent of the choice of basis and more important, they are NOT matrices. Matrices are just REPRESENTATIONS, and, even the operator and its matrix are called equal, the are NOT equal. In this sense, $T\neq T$. For that reason, some authors call operators with calligraphic letters or add brackets to matrix representations: $[S],[T]$...

To compute product matrices (viewed as a representations) they must be in the SAME basis. For $[T]$, a right basis of $\mathbf R^3$ is $\{v_1=(1,2,0),v_2(2,-2,0),v_3(2,-1,-2)\}$, while the results $T(v_i)$ are supposed to be in the canonical basis $\{e_1=(1,0),e_2=(0,1)\}$. For $[S]$, a convenient basis of $\mathbf R^3$ is $\{w_1=(-1,1),w_2=(-1,2)\}$, while the results $S(w_j)$ are supposed to be in the canonical basis $\{e_1=(1,0,0),e_2=(0,1,0), e_3=(0,0,1)\}$. So, if you want to get the matrix representation $[ST]$ computing the matrix product $[S][T]$ you must to ensure both matrices are in the same basis.

Dog_69
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  • Firstly, thank you very much for your answer. Now, how can i ensure that both matrices are in the same basis? after all, for $[T]$ the vectors in his basis belongs to $R^3$ and for $[S]$ the vectors in his basis belongs to $R^2$.. – Master Question Jan 09 '18 at 23:34
  • Note that in eqs. (1) and (2), I'm writing the resulting vectors $T(v_1), T(v_i)$ as a linear combinations of $w_j$. For $S$, I would write $S(w_j)$ in terms of $v_i$. When I say two basis must match that is exactly what I want to say: you must to apply $T$ to some basis ${v_i}$ and write the result in terms of ${w_j}$, while for $S$, you must to apply it EXACTLY over THE SAME ${w_j}$ and write the result in terms of THE SAME basis ${v_i}$. – Dog_69 Jan 09 '18 at 23:43
  • I found that: $T = \begin{bmatrix}0& 1& 0\ -1 &0& -1.5 \end{bmatrix}$ and $S = \begin{bmatrix}2& 1\ -2 &0 \ -3 & -1 \end{bmatrix}$ . can you explain me please what is the next step? – Master Question Jan 10 '18 at 00:59
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    Now you have two possibilities. You can compose maps $S$ and $T$ and $T,S$ to find $T\circ S$ and $S\circ T$ resp. and then find the matrix representations of such maps (as before). But, as I have already said, there is a nice result which says that $[S\circ T] = [S][T]$, i.e, the matrix representation of the composition is the product of the matrices representations. So, all you have to do now is compute the matrix product$TS$ for the matrix representation of the map $T\circ S$ and $ST$ for the map $ST$. – Dog_69 Jan 10 '18 at 17:13
  • $S∘T$ it's $S*T$? – Master Question Jan 10 '18 at 17:16
  • It's the composition given by apply first $T$ and then $S$: $(S\circ T)(v)=S(T(v))$. Sometimes, people (included me) change the order and write $T\circ S$ to denote the map $(T\circ S)(v)=S(T(v))$. – Dog_69 Jan 10 '18 at 17:19
  • @MasterQuestion: It's a pleasure. – Dog_69 Jan 10 '18 at 17:51
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The following matrix

$$M_{T,B}= \begin{bmatrix} 2 &-2&-1 \\ -1&-2&1 \end{bmatrix}$$

represent the transformation T from the basis $\mathcal{B}$

$$v_{1,B}=\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, v_{2,B}= \begin{bmatrix} 2 \\ -2 \\ 0 \end{bmatrix} , v_{3,B} \begin{bmatrix} 2 \\ -1 \\ -2 \end{bmatrix}$$

to the standard basis $\mathcal{C}$.

Now you can change the basis and find the matrix $M_{T,C}$ in the standard basis.

Notably the matrix

$$N_B=[v_{1,B}\quad v_{2,B} \quad v_{3,B}]=\begin{bmatrix} 1&2&2 \\ 2&-2&-1 \\ 0&0&-2 \end{bmatrix}$$

represent the change of basis from the basis B to the standard basis thus

$$v_C=N_B\cdot v_B\iff v_B=N_B^{-1}\cdot v_C\iff v_B=N_C\cdot v_C$$

and $N_C$ represent the change of basis from the standard basis to basis B.

Thus, since the transformation T from the basis B to the standard basis is

$$w_C=M_{T,B}\cdot v_B$$

the transformation in the standard basis is given by

$$w_C=M_{T,B}\cdot N_C\cdot v_C$$

and the matrix $M_{T,B}\cdot N_C=M_{T,C}$ represent the transformation in the standard basis.

You can use the same method for S and then obtain $TS$ and $ST$ by simply matrix multiplication.

You should obtain

$$M_{T,C} = \begin{bmatrix}0& 1& 0\\ -1 &0& -\frac32 \end{bmatrix} \quad M_{S,C} = \begin{bmatrix}2& 1\\ -2 &0 \\ -3 & -1 \end{bmatrix}$$

Then

$$TS \to M_{TS,C}=M_{T,C}\cdot M_{S,C}= \begin{bmatrix}-2& 0\\ -7 &-3 \end{bmatrix}$$

$$ST \to M_{ST,C}=M_{S,C}\cdot M_{T,C}= \begin{bmatrix}-1& 2&\frac32\\ 0 &-2&0\\ 1 &-3&-2 \end{bmatrix}$$

user
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  • Can you explain why $M_{T,B}$ represent the transformation T from the basis B to the standard basis C? In addition, what is $M_{T,C}$ ? and how you know that $N_{B}$ represent the change of basis from the basis B to the standard basis? – Master Question Jan 09 '18 at 22:49
  • @MasterQuestion Of course, it is trivial: in the basis B the basis vectors have components $v_{1,B}=(1,0,0)$, $v_{2,B}=(0,1,0)$ , $v_{3,B}=(0,0,1)$ thus $M_{T,B}(1,0,0)=(2,-1)$ and so on. – user Jan 09 '18 at 22:52
  • Ok, can you explain what you mean while you said that "$M_{T,B}$ represent the transformation $T$ from the basis $B$ to the standard basis $C$"? it's not clear about the meaning of "the standard basis $C$". – Master Question Jan 09 '18 at 22:59
  • @MasterQuestion Take also a look hereat other my recent answer on this topic, you'll find all what you need. If you need further explanation I can explain further. https://math.stackexchange.com/questions/2565468/change-of-basis-am-i-doing-this-right/2565655#2565655 – user Jan 09 '18 at 23:04
  • https://math.stackexchange.com/questions/2560162/change-of-basis-for-linear-transformation-linear-algebra/2560175#2560175 – user Jan 09 '18 at 23:04
  • https://math.stackexchange.com/questions/708715/in-p-2-find-the-change-of-coordinates-matrix/2586553#2586553 – user Jan 09 '18 at 23:05
  • @MasterQuestion is it clear now? – user Jan 10 '18 at 06:48
  • i don't understand this part: $v_{C}=N_{B}*v_{B}$ , what is $v_{C}$ and what is $v_{B}$? – Master Question Jan 10 '18 at 10:32
  • $v_C$ are $v_B$ represent the (components of) same vector in the two different basis C "standard" and B. – user Jan 10 '18 at 10:51
  • Let us continue this discussion in chat. – Master Question Jan 10 '18 at 11:11
  • @MasterQuestion sorry I can't now – user Jan 10 '18 at 11:32