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From S.L Linear Algebra:

Find the matrix associated with the following linear map:

$F: \mathbb{R}^n \rightarrow \mathbb{R}^n$ given by $F(X)=-X$

Considering that dimension of vector spaces are unknown, I've constructed a generalized method for the matrix associated with linear map given above.


Let $F: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a linear transformation given by $F(X)=-X$. Then $F(x_1,...,x_n)=-(x_1,...,x_n)$.

We shall find a $n \times n$ symmetric matrix $A=(a_{ij})$, such that $A(x_1,...,x_n)=-(x_1,...,x_n)$. With calculations, we can see that $\left ( (a_{11}x_1+...+a_{1j}x_{n}),...,(a_{i1}x_{1}+...+a_{ij}x_n) \right )=(-x_1,...,-x_n)$ and hence $\left ( ((1+a_{11})x_1+...+a_{1j}x_{n}),...,(a_{i1}x_{1}+...+(1+a_{ij})x_n) \right )=O$ ($O$ being a zero vector).

Considering that we are looking for basis of row space (image of linear transformation), all coefficients must be linearly independent within row vector, therefore kernel of all row vectors is trivial. We eventually get a $n \times n$ matrix with $a_{11},...,a_{ij}=-1$ (hence we get a "sparse" matrix which has $-1$ as diagonal elements):

$$\begin{pmatrix} -1_{11} & 0_{12} & ... & 0_{1j}\\ 0_{21} & -1_{22} & ... & 0_{2j}\\ ... & ... & ... & ... \\ 0_{i\,1} & 0_{i\,2} & ... &-1_{ij} \end{pmatrix}$$

I did index all elements of matrix to avoid confusion.


The generalized solution itself does seem quite confusing to me, it's much easier to find the matrix associated with linear transformation when dimension is known.

I'm looking for the basis of the matrix associated linear transformation above, which is the basis of row space of transformation (or the image).

Are there any mistakes in my solution? Is there a more simple and elegant way of finding a matrix associated with linear transformation where dimension of domain and codomain is unknown?

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    $\begingroup$ The dimensions are already given to you and it's $n$. $\endgroup$
    – cqfd
    Commented Jan 18, 2019 at 10:34
  • $\begingroup$ Are you referring basis? $\endgroup$
    – cqfd
    Commented Jan 18, 2019 at 10:35
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    $\begingroup$ How would you call the matrix for $F(X)=X$? Now, what do you do to a matrix when you just scale the vector, keeping the direction? $\endgroup$
    – orion
    Commented Jan 18, 2019 at 10:37
  • $\begingroup$ @ThomasShelby By certainity I implied exact integer as dimensions, e.g a linear map $\mathbb{R}^4 \rightarrow \mathbb{R}^2$ would be a $2x4$ matrix (and it would just be easier to write it). Now that I don't know the value of $n$, I created more "generalized" solution which seems little more complex than the solution that I would create for $2x4$ matrix. $\endgroup$
    – ShellRox
    Commented Jan 18, 2019 at 10:42
  • $\begingroup$ @orion My apologies, I'm unable to understand the comment. The goal is to find a matrix associated with linear transformation $\mathbb{R}^n \rightarrow \mathbb{R}^n$ defined by $F(X)=-X$ , which is $AX=-X$ (I need to find $A$). I did it by representing $A$ as $(a_{ij})$ and getting inner product for every row vector of $A=(a_{ij})$ by $X=(x_1,...,x_n)$ (and inner product is the element of $-X$). I'm trying to check the validity of my solution and whether or not there is a better, more elegant solution. $\endgroup$
    – ShellRox
    Commented Jan 18, 2019 at 10:48

1 Answer 1

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A matrix is determined by what it does to a basis. Let $\mathcal S=\{e_1,\dots,e_n\}$ be the standard basis. Then $Ae_i=-e_i, i=1,\dots, n$.

Thus $[A]_{\mathcal S}=\begin{pmatrix}-1&0&\dots&0\\0&-1&\dots&0\\\vdots&\ddots&\ddots&\vdots\\0&0&\dots&-1\end{pmatrix}$.

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  • $\begingroup$ Thank you for the answer. So the solution that I've presented above is valid, correct? Considering that we are looking for basis of row space, all elements must be linealry independent within the row space and hence all coefficients are zero, giving us $-1$ as a diagonal elements. $\endgroup$
    – ShellRox
    Commented Jan 18, 2019 at 11:51
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    $\begingroup$ I'm unfortunately unable to follow what you've done. Simply, the $i$th column is $Ae_i=-e_i$, where $e_i=(0,\dots,\underbrace{1}_{i-th place},\dots,0)$. $\endgroup$
    – user403337
    Commented Jan 18, 2019 at 12:13
  • $\begingroup$ I used the method given from this answer, in this case it was just a little more complex due to $n$ being variable. $\endgroup$
    – ShellRox
    Commented Jan 18, 2019 at 20:13
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    $\begingroup$ Right, well the mathematics appears to be ok. It's just that your wording is confusing: you talk of "linearly independent coefficients ", and btw the image is the column space, not the row space. $\endgroup$
    – user403337
    Commented Jan 18, 2019 at 20:18

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