Question: (a) Let $f: V \rightarrow W$ with $ V,W \simeq \mathbb{R}^{3}$ given by: $$f(x_1, x_2, x_3) = (x_1 - x_3, 2x_1 -5x_2 -x_3, x_2 + x_3).$$

Determine the matrix of $f$ relative to the basis $\{(0,2,1),(-1,1,1),(2,-1,1)\}$ of $V$ and $\{(-1,-1,0),(1,-1,2),(0,2,0)\}$ of $W$.

(b) Let $n \in \mathbb{N}$ and $U_n$ the vector space of real polynomials of degree $\leq n$. The linear map $f: U_n \rightarrow U_n$ is given by $f(p) = p'$. Determine the matrix of $f$ relative to the basis $\{1,t,t^{2},...,t^{n}\}$ of $U_n$.

My attempt so far: (a): First relative to the bases of $W$ I found the coordinates of an arbitrary vector: $\left( \begin{array}{r} a \\ b \\ c \end{array} \right) = x \left( \begin{array}{r} -1 \\ -1 \\ 0 \end{array} \right) + y \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) + z \left( \begin{array}{c} 0 \\ 2 \\ 0 \end{array} \right)$

$\begin{array}{l} a = -x + y \\ b = - x - y + 2z \\ c = 2y \end{array}$ or $\begin{array}{l} x = -a + \frac{1}{2}c \\ z = -\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c \\ y = \frac{1}{2}c \end{array}$

At this point I believe I have the linear combinations of the given basis in $W$ for an arbitrary vector, so next I take the vectors from $V$ and send them to $W$ using the given function:

$\begin{array}{l} f(v_1) = f(0,2,1) = (-1,-11,3) = (1 + \frac{3}{2})w_1 + \frac{3}{2}w_2 + (\frac{1}{2} - \frac{11}{2} + \frac{3}{2})w_3 \\ f(v_2) = f(-1,1,1) = (-2,-8,2) = (2+1)w_1 + w_2 + (1 - 4 +1)w_3 \\ f(v_3) = f(2,-1,1) = (1,8,0) = w_1 + (-\frac{1}{2} + 4)w_3 \end{array}$

or $\left( \begin{array}{rrc} \frac{5}{2} & 3 & 1 \\ \frac{3}{2} & 1 & 0 \\ -\frac{7}{2} & -2 & \frac{7}{2}\end{array} \right)$

Was I taking the correct steps? I didn't really do anything differently based on the fact that $V,W$ were isometric... Is there a particular significance or interpretation for the resulting matrix?

(b): Not really sure here...

$f(p) = p'$

would it make sense to write something like:

$f(1,t,t^{2},\dots, t^{n}) = (0,1,2t, \dots, nt^{n-1})$?

and if a basis for $(1,t,t^{2},\dots, t^{n})$ would be $A = \left( \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & \cdots & \cdots & 0 & 1 \end{array} \right)$

could i write:

$A' = \left( \begin{array}{ccccc} 0 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0 & 0 & 1 & 0 \end{array} \right)$?


Yes, your solution to (a) is correct assuming your computations for how to express the elements is correct. Though you don't need to find how to write an arbitrary vector in terms of the basis of $W$, you only need to find out how to write $f(0,2,1)$, $f(-1,1,1)$, and $f(2,-1,1)$ (the images of the vectors in the given basis for $V$) in terms of the given basis of $W$.

No, you never need to use the fact that $V$ and $W$ are isomorphic (we don't consider a metric here, so "isometric" is not appropriate here).

Your answer for (b), on the other hand, is not well done (or correct).

The basis for your vector space is $\{1,t,\ldots,t^n\}$. To find the matrix representation of $f$ relative to this basis, you need to find the image of each basis vector in the domain, and express it in terms of the basis vectors of the range. But remember: the vectors in $U_n$ are not tuples, they are polynomials. So $(1,t,\ldots,t^n)$ is not an element of $U_n$.

So, you would have to find each of $f(1)$, $f(t)$, $f(t^2),\ldots, f(t^n)$, and then express them in terms of the basis of the range, which happens to again be $\{1,t,t^2,\ldots,t^n\}$. So for instance, $f(t^2) = 2t$, so $$f(t^2) = 0\cdot 1 + 2\cdot t + 0\cdot t^2 + \cdots + 0\cdot t^n.$$ That means that the third column of the matrix (the one corresponding to the third basis vector, which is $t^2$) will be the transpose of $(0,2,0,\ldots,0)$, that is $$\left(\begin{array}{c} 0\\2\\0\\ \vdots \\ 0\end{array}\right).$$ Etc.

  • $\begingroup$ sorry, i'm a little confused by the exponent of "t" in the last line (if that even is an exponent and not some other notation). Where does that come from? $\endgroup$ – ghshtalt Nov 29 '10 at 20:53
  • $\begingroup$ @user3711: It stands for "Transpose"; it means that the column is not a row, but the column vector you get by turning that row in to a column. I'll edit to get rid of it, since it is easy to confuse with the $t$ from the polynomials. $\endgroup$ – Arturo Magidin Nov 29 '10 at 21:00

Here's a suggestion for seeing the bigger picture in your first question:

  1. The key idea below is that the column vectors of a matrix $M$ are the images of the standard ordered basis when the matrix $M$ is viewed as a transformation $M: \mathbb{R}^n \to \mathbb{R}^n$.

  2. It is easy to find a matrix of f with respect to the standard ordered basis of $\mathbb{R}^3$: it is given by a matrix with columns $\{1, 2, 0\}$, $\{0, -5, 1\}$, and $\{-1, -1, 1\}$. Call this matrix $C$.

  3. The matrix which transforms the given ordered basis of $V$ to the standard basis of $\mathbb{R}^3$ is given by writing the basis vectors as columns of this matrix. Similarly, this can be done for the basis of $W$. Let's call these matrices $A$ and $B$.

  4. You desire a matrix which represents $f$. Interpreting the matrices above as transformations, consider the following diagram:

$$\begin{array}{ccc} V & \to & \mathbb{R}^3\\ \downarrow & \qquad & \downarrow\\ W & \to & \mathbb{R}^3 \end{array}$$

where the horizontal maps are $A$ and $B$ and the right vertical map is $C$. Your desired matrix is the left vertical arrow. It is now clear that it is given by the matrix $B^{-1} C A$.

  • $\begingroup$ yes, thanks a lot for the explanation, it definitely helps to illustrate the bigger picture! $\endgroup$ – ghshtalt Nov 29 '10 at 21:03
  • $\begingroup$ +1, didn't think of that, but I should have. $\endgroup$ – rcollyer Jun 9 '11 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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