Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 know pretty well how to find the transformation matrix of a linear map (with respect to a basis). However, I am wondering whether it is also possible to do it the other way around?

This question arises because in one of my exercises in linear-algebra I first had to find the transformation matrix with respect to the canonical basis to the linear map $$F: \mathbb{R}_2[t] \to \mathbb{R}_2[t],\; s(t) \mapsto s(t) + s'(t) + ts''(t),$$ then I have to find the inverse map $F^{-1}$. The sample solution only provides the inverse of the transformation matrix, however, I want to know whether I can give an explicit function (using the inverse of the transformation matrix).

Thanks for any help in advance!

share|cite|improve this question
up vote 2 down vote accepted

Sure: if you have the matrix $A$ relative to the basis $\beta$, then the linear transformation that maps the coordinate vector $[\mathbf{v}]_{\beta}$ of $\mathbf{v}$ with respect to $\beta$ to $[A\mathbf{v}]_{\beta}$ has coordinate matrix $A$ with respect to $\beta$.

Or to be more explicit, if $$A = \left(\begin{array}{ccc} a_{11} & \cdots & a_{1n}\\ a_{21} & \cdots & a_{2n}\\ \vdots & \ddots & \vdots\\ a_{n1} & \cdots & a_{nn} \end{array}\right),$$ is the matrix relative to the basis $\beta=[\mathbf{v}_1,\ldots,\mathbf{v}_n]$, then the linear transformation defined by $$T(\mathbf{v}_j) = a_{1j}\mathbf{v}_1 + a_{2j}\mathbf{v}_2 + \cdots +a_{nj}\mathbf{v}_n$$ has coordinate matrix $A$ with respect to $\beta$.

In short: just "read off" what the linear transformation does to the $j$th vector of the basis by reading down the $j$th column of $A$.

share|cite|improve this answer
If I understand correctly, this is not quite what I was looking for. Let's take my exercise as an example. For the inverse function, we have the transformation matrix \begin{pmatrix} 1 & -1 & 4\\ 0 & 1 & -4\\ 0 & 0 & 1 \end{pmatrix} From this I can see that $F^{-1}(1) = 1, F^{-1}(x) = x-1, F^{-1}(x^2) = x^2-4x+4$ and due to linearity (is the inverse map linear too?), I can find $F^{-1}(s(t))$ for all $s(t) \in \mathbb{R}_2[t]$. However, I'm looking for a "more" explicit expression, something like $F^{-1}: s(t) \mapsto ...$, if you understand me... – Huy Jun 29 '11 at 20:17
@Huy: Yes, the inverse of an invertible linear transformation is necessarily a linear transformation. As for a formula, you can obtain it from linearity: $F^{-1}(a+bx+cx^2) = aF^{-1}(1) + bF^{-1}(x) + cF^{-1}(x^2) = (a-b+4c) + (b-4c)x + cx^4$. If you look carefully, the coefficients of the basis vectors in the image can be "read off" the rows of the inverse matrix. – Arturo Magidin Jun 29 '11 at 20:22
Because the coordinate vector of $a+bx+cx^2$ relative to the basis $[1,x,x^2]$ is $(a,b,c)^t$, so the image would be $$\left(\begin{array}{rrr}1 & -1 & 4\\0 & 1 & -4\\0&0&1\end{array}\right)\left(\begin{array}{c}a\\b\\c\end{array}\right)= a\left(\begin{array}{c}1\\0\\0\end{array}\right) + b\left(\begin{array}{r}-1\\1\\0\end{array}\right) + c\left(\begin{array}{r}4\\-4\\1\end{array}\right),$$and what's on the top line is the coefficient corresponding to $1$ (the first basis vector), what's on the middle line is the coefficient of $x$, and what's on the third line is the coefficient of $x^2$. – Arturo Magidin Jun 29 '11 at 20:40
@Huy: Sorry: I forgot the ping and it's too late to edit. Hopefully you'll see the reply now. – Arturo Magidin Jun 29 '11 at 20:46

Your Answer


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

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