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I've been struggling with the concept of finding the matrix of the operator, and need some help because I am preparing for an exam. I understand how to find the matrix of an operator/transformation when given a concrete example, but with abstract examples I get a little lost.

Problem Statement:Let $A$ be an $n\times n$ matrix, and let $V$ denote the space of $n$-dimensional row vectors. What is the matrix of the linear operator "right multiplication by $A$" with respect to the standard basis of $V$​?

So, I know that $A:V\rightarrow V$, and given a standard basis $B=(\textbf{v}_1, ..., \textbf{v}_n)$ of $V$, then any $\textbf{v}\in V$ can be written as $$\textbf{v}=a_1\textbf{v}_1+...+a_n\textbf{v}_n=\begin{bmatrix} a_1 & \dots & a_n \\ \end{bmatrix} \begin{bmatrix} - & \textbf{v}_1 & - \\ & \vdots & \\ - & \textbf{v}_n & - \end{bmatrix}$$

Now, am I on the right track here? Is $\begin{bmatrix} a_1 & \dots & a_n \\ \end{bmatrix}$ one of the row vectors of the matrix of $A$? We did a similar example in class with a transformation $T:V\rightarrow W$, and I was thinking to start breaking down the problem this way.

Also, when the author says standard basis of $V$ do they mean the standard basis $(\textbf{e}_1, ..., \textbf{e}_n)$? I was a little confused with that as well, because I think of $V$ being an abstract space, so i figure the standard basis could be different in $V$?

When I took Linear Algebra of real numbers, the concept of Transformations and change of bases was so clear to me, and now that we are abstracting more, I find myself getting very confused :(

Any help with this problem would be greatly appreciated!!!

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To answer the problem: Denote by $\mathbb F_n$ the vector space of row vectors with $n$ components coming from $\mathbb F$.

Let me call $L$ the linear transformation from $\mathbb F_n$ to $\mathbb F_n$. It looks like $L(x) =xA$, where $x\in \mathbb F_n$. Let $\{e_1,\ldots,e_n\}$ be the standard basis vector of $\mathbb F_n$. The ith column of the matrix rep of $L$ has coordinates given by $e_iA$. Since the basis we considered is the standard basis, this column is the transpose of the ith row of $A$. Hence, the matrix rep of $L$ is $A^T$.

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  • $\begingroup$ So you are saying that $A^{T}\textbf{x}=\textbf{x}^{T}A$? $\endgroup$ – yung_Pabs Oct 26 '15 at 4:47
  • $\begingroup$ That is incorrect. The left hand side is a column. The right hand side is a row. In practice, you do not equate $L(x)$ with the $Ax$ because they probably lie in different spaces. In general, $Ax$ gives you the coordinates. So what $A^Tx^T$ is giving you is the coordinates of $xA$. They are not really equal as vectors. $\endgroup$ – chhro Oct 26 '15 at 4:54
  • $\begingroup$ But I always consider the transpose of a matrix a row. So if I have $\textbf{x}^{T}A=\begin{bmatrix} x_{1} & \cdots & x_{n} \end{bmatrix} \begin{bmatrix} - & \textbf{a}_{1} & - \\ - & \vdots & - \\ - & \textbf{a}_{n} & - \end{bmatrix}$ Taking the transpose of this product would result in $A^{T}\textbf{x}=\begin{bmatrix} | & | & | \\ \textbf{a}_{1} & \cdots & \textbf{a}_{n} \\ | & | & | \end{bmatrix} \begin{bmatrix} x_{1}\\ \vdots \\ x_{n} \end{bmatrix} $ No? Which would give the result $\endgroup$ – yung_Pabs Oct 26 '15 at 23:37
  • $\begingroup$ I now understand that they aren't equal vectors, but we would have that $(\textbf{x}^T A)^{T}=A^{T}\textbf{x}$ $\endgroup$ – yung_Pabs Oct 26 '15 at 23:44

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