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I working on Axler's Linear Algebra Done Right and I found this problem. M(T) is defined to be a matrix representation of a linear map of T in $L(V,W)$ with respect to a given basis $v_1,...,v_n$ and $w_1,...,w_m$. Let $v \in V$ then $M(v)$ is a n-by-1 matrix of $a_1,...,a_n$ such that $v=a_1v_1+...+a_nv_n$. And $M(v_k)$ is computed with respect to $w_1,...,w_n$. Which means $M(v_k)$ is a m-by-1 matrix $(c_1 ... c_m)^T$ such that $c_1w_1+...+c_mw_m=v_k$. How could this be possible when $v_k$ is in V?

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  • $\begingroup$ I think you are talking about linear maps $T:V\rightarrow W$. Then it sounds like it really means $c_1w_1 + ... c_mw_m = T(v_k)$. Then it sounds like you really have $M(T)$ is a matrix with each row a $c$ vector. $\endgroup$ – Michael Sep 24 '18 at 13:13
  • $\begingroup$ So $M:L\rightarrow \mathbb{F}^{m,n}$ is a function that maps a linear transformation to a matrix, so a given linear transformation $T:V\rightarrow W$ gets mapped to a matrix $M(T)$. $\endgroup$ – Michael Sep 24 '18 at 13:21
  • $\begingroup$ So you mean this is a typo error in the book? The author really means $M(T(v_k))$ after all? $\endgroup$ – Michael D Nguyen Sep 24 '18 at 13:24
  • $\begingroup$ I would need to see the book. But if we define $M:L\rightarrow \mathbb{F}^{m,n}$, then writing $M(T(v_k))$ does not make sense because $T(v_k)$ is a vector in $w$, whereas the $M(\cdot)$ function needs a linear transformation as input, not a vector as input, i.e., $M(T)$ makes sense. $\endgroup$ – Michael Sep 24 '18 at 13:26
  • $\begingroup$ Either $M$ is the matrix associated to a particular linear map, once the isomorphism between $L(V,W)$ and $F^{m,n}$ has been fixed, and then $M(v_k)$ makes sense, or $M$ is the isomorphism, for any element $T\in L(V,W)$ you have a matrix $M(T)$, and then $M(T)(v_k)$ makes sense, but $M(v_k)$ does not $\endgroup$ – Jose Brox Sep 24 '18 at 13:27
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This really was a typo. Specifically, $\mathcal{M}(v_k)$ should be $\mathcal{M}(Tv_k)$ on lines 2, 4, 7, and 9 of page 85 of the third edition of my book Linear Algebra Done Right.

My apologies for the confusion caused by this error. I tried hard to eliminate all typos from the book, but a few slipped through.

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I worked it out! It's really a typo. I tried $M(T(v_k))$ and the equation fit perfectly.

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  • $\begingroup$ That does not make sense since the $M(\cdot)$ function needs a linear transformation as input. I think it means $M(T)e_k$ where $e_k=(0,0,...,1,0,0,...0)$. So $M(T)$ is a matrix and $M(T)e_k$ is the $k$th column of that matrix, which is hte $c$-vector associated with expressing $T(v_k)$ in terms of the basis vectors for $W$. $\endgroup$ – Michael Sep 24 '18 at 13:43
  • $\begingroup$ Yes. But later on, the author also add a definition for M mapping on a vector space. And thus there is this equation. But what doesn't make sense to me is that $v_k$ has to be expressed as a linear combination of a basis of W while $v_k$ is in V. $\endgroup$ – Michael D Nguyen Sep 24 '18 at 13:46
  • $\begingroup$ This is used to prove $M(Tv)=M(T)M(v)$ which expresses linear maps as matrix multiplication. $\endgroup$ – Michael D Nguyen Sep 24 '18 at 13:48
  • $\begingroup$ Then, it sounds like we really have a map $M:V\rightarrow \mathbb{F}^m$ . I wonder if the book distinguishes $M(\cdot)$ from $Mat(\cdot)$. $\endgroup$ – Michael Sep 24 '18 at 13:50
  • $\begingroup$ I found the book here: google.com/… what equation are you referring to? I do notice that equations (3.15) and and (3.13) use $\mathcal{M}(\cdot)$ informally to mean "the matrix of" but you can plug different types of inputs (vectors, maps, and so on) into it, so it does not formally define $\mathcal{M}:(domain)\rightarrow(range)$ as I expected. $\endgroup$ – Michael Sep 24 '18 at 14:07

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