Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading section 46 of Halmos' Finite Dimensional Vector Spaces. In this section Halmos poses two questions the first of which is:

Question 1: If $x$ is in a finite dimensional vector space $V$, write $x = \sum_i \xi_i x_i = \sum_i \eta_i y_i$, what is the relation between its coordinates $(\xi_1,\xi_2, \ldots \xi_n)$ with respect to the basis $X = (x_1, \ldots, x_n)$ and its coordinates $(\eta_1, \ldots ,\eta_n)$ in the basis $Y = (y_1 \ldots y_n)$?

After a few lines, Halmos defines the linear transformation $A$ by $A(x_i) = y_i$, $i=1,2, \ldots n$. From what I understand, suppose we have a basis vector $x_i$. Then this should correpond to the column vector

$$\left[\begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{array}\right]$$

in the basis $X$ where the $1$ is in the $i-th$ row of the vector.

Then if the matrix $A = a_{ij}$ is applied to this vector, we should have the $i-th$ column of the matrix. Let the $i-th$ column of the matrix be

$$\left[\begin{array}{c} a_{1i} \\ a_{2i} \\ \vdots \\ a_{ni} \end{array} \right].$$

However note that the $i-th$ column of the matrix is expressed in terms of the basis $X$, because Halmos states that $y_j = Ax_j = \sum_i a_{ij}x_i$. In other words, he has chosen to express each vector in the basis $Y$ as a linear combination of the basis vectors in $X$. Is there a particular reason of doing this? From my limited experience in linear algebra, if say we have a vector in a basis $X$, and we wish to express it in a basis $Y$, the thing to do would be to express $x_1$ as a linear combination of the $y_i's$ in $Y$, $x_2$, and so on. This is the opposite to what Halmos has done.

Is this difference anything significant to be aware of?

$\textbf{Edit : }$ I will write out the relevant bit that I am referring to in Halmos' book:

"Let $V$ be an $n$- dimensional vector space and let $X = (x_1, \ldots x_n)$ and $y=(y_1,\ldots y_n)$ be two bases in $V$. We may ask the following two questions:

Question 1. If $x$ is in $V$, $x = \sum_i \xi_ix_i = \sum_i\eta_i y_i$, what is the reation between its coordinates $xi_i$ with respect to $X$ and its coordinates $\eta_i$ with respect to $y$?

Question 2. If $(\xi_1, \ldots \xi_n)$ is an ordered set of $n$ scalars, what is the relation between the vectors $x = \sum_i \xi_ix_i$ and $y = \sum_i \xi_i y_i$?

Both these questions are easily answered in the language of linear transformations. We consider, namely the linear transformation $A$ defined by $Ax_i = y_i$. More explicitly:

$A(\sum_i \xi_i x_i) = \sum_i \xi_iy_i.$ "

share|cite|improve this question
View your two basis as two isomorphisms $x:K^ n\to V$ and $y:K^ n\to V$. Then you have $x(\xi)=y(\eta)$ iff $\xi=x^ {-1}(y(\eta))$. – Pierre-Yves Gaillard Aug 15 '11 at 10:53
@Pierre-Yves Gaillard Ok thanks. How does my matrix $A$ relate to this? – user38268 Aug 15 '11 at 11:28
To tell the truth, I focused on “Question 1”, and didn’t read carefully enough what you wrote next. Sorry. But do you agree that this answers “Question 1”? I’ll read more seriously the whole of your post, and if I have anything sensible to say about it (which I doubt), I’ll say it. – Pierre-Yves Gaillard Aug 15 '11 at 11:37
Je suis d'accord avec vous. Je suis curieux de savoir si la matrice que j'ai écrire $A$ a une connection avec les isomorphismes que vous avez dit. Par exemple pour écrire la matrice de la transformation linéare donnée par $x$, on forme la matrice dont colonnes sont les images des vecteurs $e_i$ dans $V$. Mais à mes yeus la matrice $A$ est la même matrice comme ce que j'ai dit avant. Je pense que je pourrais avoir tort.... – user38268 Aug 15 '11 at 12:25
Félicitations pour votre excellent français! Où l'avez-vous appris? - Le premier passage que je ne comprends pas dans votre question est: "the matrix $A = a_{ij}$". Vous aviez pourtant défini $A$ comme étant une transormation linéaire. Je vais essayer d'accéder au livre de Halmos et regarder le passage en question. – Pierre-Yves Gaillard Aug 15 '11 at 12:45
up vote 1 down vote accepted

The matrix which expresses basis vectors of one basis $X$ in terms of another basis $Y$ and the matrix which expresses vectors of $Y$ in terms of $X$ are just inverses to each other and which one you call $A$ and which one becomes $A^{-1}$ is a matter of notation or convention - you just have to pick the right one when doing an actual change of coordinates.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.