Linear transformation and matrix transformation

[HelpNeeder]

"Every linear transformation is a matrix transformation."
This statement is false.

But in the book Linear Algebra and Its Applications there is written:


However, doesn't this mean that every linear transformation is a matrix transformation?

Thank you in advance.

 

[lex]

https://math.stackexchange.com/questions/3673950/an-example-of-linear-transformation-not-matrix-transformation

 

[HelpNeeder]

Thank you, but I wanted to understand why this is not a contradiction.

 

[lex]

That is explained in the linked post. In David Lay's book, you can see he is talking about a linear transformation [MATH]T:\mathbb{R}^n \rightarrow \mathbb{R}^m[/MATH]Your statement has no such restriction. It refers to linear transformations between any vector spaces, not just [MATH]\mathbb{R}^n \text{ and } \mathbb{R}^m[/MATH](Examples are given in the linked post. You need to read it all the way to the bottom of the page.)

 

[HelpNeeder]

I understand it now. Thank you!

 

[HallsofIvy]

I would say that every linear transformation can be represented by a matrix, but I would not say that every linear transformation is a linear transformation since which matrix represents the linear transformation depends upon the basis for the vector space.

You can define a linear transformation without a basis but a specific matrix requires a specific basis.

 

[lex]

but I would not say that every linear transformation is a linear transformation

matrix?

I would say that every linear transformation can be represented by a matrix

See the final comment at the bottom of this post:
https://math.stackexchange.com/ques...near-transformation-not-matrix-transformation

 

[HallsofIvy]

matrix?

See the final comment at the bottom of this post:
https://math.stackexchange.com/ques...near-transformation-not-matrix-transformation

That says pretty much what I said.
"That said, there still is a way to "represent" T by a matrix."

The "T" referred to is the operation T(p)= p(0)x+ 3xp'(x) applied to the space or all polynomials of degree 3 or less.
That itself is NOT a matrix product but can be represented by one in a given basis. For example we can take as basis $\displaystyle v_1= 1$, $\displaystyle v_2= x$, $\displaystyle v_3= x^2$, and $\displaystyle v_4= x^3$.

Applying T to the basis vectors $\displaystyle T(v_1)= T(1)= 1x+ 3x(0)= x$, $\displaystyle T(v_2)= T(x)= 0x+ 3x(1)= 3x= 3v_2$, $\displaystyle T(v_3)= T(x^2)= 0x+ 3x(2x)= 6x^2= 6v_3$ and $\displaystyle T(v_4)= T(x^3)= 0x+ 3x(3x^2)= 9x^3= 9v_4$.

We can represent the basis "vector" 1 as $\displaystyle \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, x as $\displaystyle \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$, $\displaystyle x^2$ as $\displaystyle \begin{bmatrix} 0\\ 0 \\ 1 \\ 0 \end{bmatrix}$, and $\displaystyle x^3$ as $\displaystyle \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$.

Write $\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}$ as the matrix "representing" T.

Representing $\displaystyle T(v_1)= T(1)= x= v_2$ we have $\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}[\begin{bmatrix}1 \\ 0 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}a_{11} \\ a_{21} \\ a_{31} \\ a_{41}\end{bmatrix}= \begin{bmatrix}0 \\ 1 \\ 0 \\ 0\end{bmatrix}$ so we must have $\displaystyle a_{11}= 0$, $\displaystyle a_{21}= 1$, $\displaystyle a_{31}= 0$ and $\displaystyle a_{41}= 0$.

Representing $\displaystyle T(v_2)= T(x)= 3x= 3v_2$ we have $\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix} a_{21} \\ a_{22} \\ a_{23} \\ a_{24} \end{bmatrix}= \begin{bmatrix} 0 \\ 3 \\ 0 \\ 0 \end{bmatrix}$ so we must have $\displaystyle a_{21}= 0$, $\displaystyle a_{22}= 3$, $\displaystyle a_{23}= 0$, and $\displaystyle a_{24}= 0$.

Representing $\displaystyle T(v_3)= T(x^2)= 6x^3= 6v_3$ we have $\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}=$$\displaystyle \begin{bmatrix}a_{31}\\ a_{32} \\ a_{33}\\ a_{34}\end{bmatrix}=$$\displaystyle \begin{bmatrix}0 \\ 0 \\ 6 \\ 0 \end{bmatrix}$ so $\displaystyle a_{31}= 0$, $\displaystyle a_{32}= 0$, $\displaystyle a_{33}= 6$, and $\displaystyle a_{34}= 0$.

Representing $\displaystyle T(v_4)= T(x^3)= 9x^3= 9v_4$ we have $\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}\begin{bmatrix}0 \\ 0 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}a_{41} \\ a_{42} \\ a_{43}\\ a_{44}\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 9 \end{bmatrix}$ so that $\displaystyle a_{41}= 0$, $\displaystyle a_{42}= 0$, $\displaystyle a_{43}= 0$, and $\displaystyle a_{44}= 9$.

Putting that all together the matrix representing T, in this basis, is
\(\displaystyle \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & 9 \end{bmatrix}\(\displaystyle .\)\)

 

[lex]

@HallsofIvy
The final comment at the bottom of the page was: