2
$\begingroup$

Definition: Using the notation above, we call the $m \times n$ matrix $A$ defined by $A_{ij} = a_{ij}$ the matrix representation of $T$ in the ordered bases $\beta$ and $\gamma$.

The "notation above" to which the author refers is $T(v_j) = \sum_{i=1}^m a_{ij} w_i$. The author then remarks that the $j$-th column of $A$ is simply $[T(v_j)]_{\gamma}$

Here is one problem I am having trouble with: enter image description here

The question is: Define $T: M_2(\Bbb{R}) \to M_2(\Bbb{R})$ by $T(A) = A^T$. Compute $[T]_{\alpha}$.

From my understanding, $T(E_{11}) = T \left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \right) = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, $T(E_{12}) = T \left(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right) = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$, etc. And so the first column is $T(E_{11})$, the second column $T(E_{12})$, etc. Hence, the matrix representation should be

$$[T]_{\alpha} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$

But evidently the answer is

$$[T]_{\alpha} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}$$

Why is this the case? Moreover, how could either answer be an operator on $M_2(\Bbb{R})$ when the dimensions of the matrix are wrong?

Here is a related problem. Let $T : P_3(\Bbb{R}) \to P_2(\Bbb{R})$ by defined by $T(f(x)) = f'(x)$. According to my book, the matrix representation is

$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ \end{bmatrix}$$

but why wouldn't it be

$$\begin{bmatrix} 0 & 1 & 2x & 3x^2 \end{bmatrix}~~?$$

After all, $T(x^3)$ literally equals $3x^2$ and so should be the fourth column. Again, how could this matrix operate on $P_3(\Bbb{R})$.

In each problem, is the author implicitly using isomorphisms between $M_2(\Bbb{R})$ and $\Bbb{R}^4$, and $P_3(\Bbb{R})$ and $\Bbb{R}^4$? If so, this would be fallacious since isomorphisms haven't been introduced.

$\endgroup$
5
  • 1
    $\begingroup$ Relative to the basis $\alpha$, the matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is expressed as the vector $[a,b,c,d]^T\in\mathbb R^4$. Similarly, relative to the basis $\beta$, the polynomial $a+bx+cx^2+dx^3$ is represented by the vector $[a,b,c,d]^T\in\mathbb R^4$. $\endgroup$
    – amd
    May 19, 2017 at 0:40
  • $\begingroup$ @amd But are you not using some sort of isomorphism by doing that? From my understanding, elements in $M_2(\Bbb{R})$ are always "a square with numbers in each corner" (i.e., of the form $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$), no matter which basis is chosen (the only difference to be found between different basis is the numbers in those corners). No choice of basis for $M_2(\Bbb{R})$ could cause it to go from $\begin{bmatrix} ~ & ~ \\ ~ & ~ \\ \end{bmatrix}$ to $\begin{bmatrix} ~ \\ ~ \\ ~ \\ ~ \end{bmatrix}$... $\endgroup$
    – user193319
    May 19, 2017 at 7:54
  • $\begingroup$ ...unless you are using the isomorphism $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \mapsto \begin{bmatrix} a \\ c \\ b \\ d \\ \end{bmatrix}$ $\endgroup$
    – user193319
    May 19, 2017 at 7:56
  • $\begingroup$ Certainly, but one doesn’t need to introduce the concept of isomorphism to use these representations mechanically. I don’t know what text you’re using, but perhaps it presents things in this order to motivate the definition of isomorphism. $\endgroup$
    – amd
    May 19, 2017 at 7:59
  • $\begingroup$ Consider this as well: The only way to produce a $2\times2$ matrix when left-multiplying a $2\times2$ matrix by some other matrix is for this other matrix to also be $2\times2$. There is no such matrix that will produce the required transposition. The matrix that you came up with can’t possibly be correct, either. Left-multiplication of a $2\times2$ by this matrix is undefined, while the result of right-multiplication produces a matrix with the wrong shape. $\endgroup$
    – amd
    May 19, 2017 at 8:03

1 Answer 1

2
$\begingroup$

A matrix representation for a linear map describes how the transformation acts in the coordinate space (what you think as an implicit isomorphism is simply the definition).

If we fix a basis $u_1,u_2,\ldots,u_n$ of $U$ and a basis $v_1,\ldots,v_m$ of $V$ then a linear map $T\colon U\to V$ can be described as $y=[T]_ax$ where $x$ is the coordinate vector for $u$ and $y$ is the coordinate vector for the image of $u$, i.e. $$ u=\color{red}{x_1}u_1+\ldots+\color{red}{x_n}u_n,\quad Tu=\color{blue}{y_1}v_1+\ldots+\color{blue}{y_m}v_m. $$ Coordinate vectors can be organized as we like, for example, as a list $\color{red}{(x_1,x_2,\ldots,x_n)}$, but often it is convenient to set them up as a column vector in order to be able to refer to $[T]_a\color{red}x$ as a matrix multiplication. Thus, a matrix representation will be an $m\times n$ matrix.

Now in your example $T(A)=A^T$. We need to get the coordinate vector for $A$: $$ A=\begin{bmatrix}a & b\\c & d\end{bmatrix}=\color{red}a\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}+\color{red}b\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}+\color{red}c\begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}+\color{red}d\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}, $$ that is $$ \color{red}{x=\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}, $$ and similarly the coordinate vector for $A^T$ $$ A^T=\begin{bmatrix}a & c\\b & d\end{bmatrix}=\color{blue}a\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}+\color{blue}c\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}+\color{blue}b\begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}+\color{blue}d\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}, $$ that is $$ \color{blue}{y=\begin{bmatrix}a\\c\\b\\d\end{bmatrix}}. $$ Therefore, the matrix representation $[T]_a$ is the $4\times 4$ matrix that $\color{blue}y=[T]_a\color{red}x$. Note that the order of the basic vectors must be fixed.

The more systematic way to build $[T]_a$ is to map the basic vectors and organize the result column-wise. For example, $$ T(E_{11})=E_{11}=\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}=\color{blue}1\cdot E_{11}+\color{blue}0\cdot E_{12}+\color{blue}0\cdot E_{21}+\color{blue}0\cdot E_{22}, $$ hence, $$ [T]_a=\begin{bmatrix}\color{blue}1 & ? & ? & ?\\ \color{blue}0 & ? & ? & ?\\ \color{blue}0 & ? & ? & ?\\ \color{blue}0 & ? & ? & ? \end{bmatrix}, $$ and so on.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .