1
$\begingroup$

Given the function $f_{h}(x,y,z)=(x-z,y+hz,x+y+3z)$, what is the correct way to represent the matrix function in respect to the standard basis?

With the representation theorem, I would write the matrix in columns as: $$F_{h|S_3}=(f_h(e_1)|{S_3} \quad f_h(e_2)|{S_3} \quad f_h(e_3)|{S_3})=\begin{bmatrix}1 & 0 & 1\\ 0 & 1 & 1\\ -1 & h & 3\end{bmatrix}$$ But in my textbook it is written in rows as: $$F_{h|S_3}=(f_h(e_1)|{S_3} \quad f_h(e_2)|{S_3} \quad f_h(e_3)|{S_3})=\begin{bmatrix}1 & 0 & -1\\ 0 & 1 & h\\ 1 & 1 & 3\end{bmatrix}$$

What is the difference between them?

$\endgroup$
1
$\begingroup$

The correct way to represent the function $f_h$ in matrix form depends on the convention that you want to use to represent it. Let $(x,y,z) \in \mathbb{R}^3$. The matrix which you computed is useful for expressing $f_h$ as $$f_h(x,y,z) = \begin{bmatrix} x& y & z \end{bmatrix} \begin{bmatrix} 1& 0 & 1 \\ 0 & 1& 1\\ -1 & h & 3 \end{bmatrix}.$$ We can verify this by expanding the matrix product: \begin{align} \begin{bmatrix} x& y & z \end{bmatrix} \begin{bmatrix} 1& 0 & 2 \\ 0 & 1& 1\\ -1 & h & 3 \end{bmatrix} &= \begin{bmatrix} x \cdot 1 + y \cdot 0 + z \cdot (-1) \\ x \cdot 0 + y \cdot 1 + z \cdot h \\ x \cdot 1 + y \cdot 1 + z \cdot 3 \end{bmatrix} \\ &= \begin{bmatrix} x - z & y + hz & x + y + 3z \end{bmatrix}. \end{align} The result is a $1 \times 3$ matrix, which we can interpret as a row vector in $\mathbb{R}^3$. The matrix written in your textbook is useful for the following representation: $$f_h(x,y,z) = \begin{bmatrix} 1& 0 & -1 \\ 0 & 1& h\\ 1 & 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x-z \\ y + hz \\ x + y + 3z \end{bmatrix}.$$ The result is a $3 \times 1 $ matrix, which we can regard as a column vector in $\mathbb{R}^3$.

The two representations look different in the sense that the former yields a row vector and the latter a column vector, however they are the same in the sense that they can both be regarded as lists of three real numbers, i.e. as $(x-z,y+hz,x+y+3z) \in \mathbb{R}^3.$ The accepted answer here sums this idea up pretty well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.