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Suppose $T:V\rightarrow V$ is a linear operator. Let a basis for $V$ be $B_1=\{e_1,e_2,\cdots,e_n\}$ .

Let $A$ be the matrix of $T$ relative to this basis.

This means : $T(e_k) = \sum_{i=1}^n a_{ik}~e_i$.

Now, my textbook says : Let us define two more matrices :

$E = [e_1,\cdots,e_n]$ and $E~' = [T(e_1), \cdots,T(e_n)]$

and then concludes that $E~'=EA$.

So, my question is : Aren't $A$ and $E'$ actually the same matrices? $A$ is obtained in the same manner : by taking the linear transformation of each column and writing it down in terms of the basis.

Could anyone please clear this confusion of mine.

Thank you very much for your help in this regard.

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    $\begingroup$ I think $E'=AE$ instead, since by definition $T(e_k)=Ae_k$. $\endgroup$ – KittyL Jan 31 '15 at 19:05
  • $\begingroup$ But Aren't $A$ and $E′$ actually the same matrices? $\endgroup$ – MathMan Jan 31 '15 at 19:06
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    $\begingroup$ $A$ is an $n\times n$ matrix, whereas $E'$ is not. $\endgroup$ – Alex Jan 31 '15 at 19:07
  • $\begingroup$ $E~' = [T(e_1), \cdots,T(e_n)]$ when expanded gives the same $A$? $\endgroup$ – MathMan Jan 31 '15 at 19:08
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    $\begingroup$ They are both $n\times n$, but they don't have to be the same. $[T(e_1), \cdots,T(e_n)]=[Ae_1, ..., Ae_n]$. The $e_k$'s don't have to be the standard basis. $\endgroup$ – KittyL Jan 31 '15 at 19:10
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Maybe an example would help sort things out. Consider the linear transformation $T(f(x)) = f(x+1)$ on the space of real polynomials of degree at most 3. In the basis $(1,x,x^2,x^3)$ we have $$ T(1) = 1, $$ $$ T(x) = x+1, $$ $$ T(x^2) = (x+1)^2 = x^2 + 2x + 1, $$ $$ T(x^3) = (x+1)^3 = x^3 + 3x^2 + 3x + 1. $$ Hence, in this basis, the matrix of $T$ is $$ A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ so that, writing $(e_1,e_2,e_3,e_4) = (1,x,x^2,x^3)$ to match your notation, we have $E = \begin{pmatrix} 1 & x & x^2 & x^3 \end{pmatrix}$ and \begin{align} EA &= \begin{pmatrix} 1 & x & x^2 & x^3 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 1 && 1 + x && 1 + 2x + x^2 && 1 + 3x + 3x^2 + x^3 \end{pmatrix} \\ &= \begin{pmatrix} 1 && (x+1) && (x+1)^2 && (x+1)^3 \end{pmatrix} \\ &= \begin{pmatrix} T(1) & T(x) & T(x^2) & T(x^3) \end{pmatrix} = E' \end{align}

The difference is that $E'$ is not an $n\times n$ matrix where you make the columns by picking off the coefficients of each of the $T(e_i)$, but instead each entry really is just $T(e_i)$.

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  • $\begingroup$ Got it. Thank you very much for the answer :) $\endgroup$ – MathMan Jan 31 '15 at 19:30
  • $\begingroup$ You're welcome! The matrix associated to a linear map is easy to get a little confused over, especially with indices all over, so I like to play with an easy example, like the one above, whenever coming to terms with something about them. $\endgroup$ – Alex Jan 31 '15 at 19:46
  • $\begingroup$ @Alex: Wow, aren't we seeing some Pascal's triangle there? And what would be the inverse of the matrix then? $\endgroup$ – orangeskid Feb 1 '15 at 1:34
  • $\begingroup$ @orangeskid, indeed that is Pascal's triangle appearing. The $i$-th column has the coefficients of $(x+1)^i$, and so you have $\binom{i}{0}, \binom{i}{1},\ldots , \binom{i}{i}, \ldots, \binom{i}{n}$ in place. For the inverse, you've got $(x+1)^i \mapsto x^i$, or equivalently $x^i \mapsto (x-1)^i$, so the coefficients are a 'negative Pascal' in some sense, $\binom{i}{0}, -\binom{i}{1},\ldots,(-1)^i\binom{i}{i}, \ldots (-1)^n\binom{i}{n}$. $\endgroup$ – Alex Feb 1 '15 at 18:06
  • $\begingroup$ @Alex: Oh, I see, I take the Pascal matrix you wrote and start changing signs on the shifted diagonals. I think I got it. And I see we can even multiply by a geometric sequence $1$, $\lambda$, $\lambda^2$,$\ldots$. So these are some maps on the polynomials. What are the transposes as linear maps? $\endgroup$ – orangeskid Feb 2 '15 at 2:42
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It is convenient to have operators act on vectors from the left, while scalars act on the vectors from the right. Take a linear operator $T$, $(e_1, \ldots, e_n)$ a basis. You express the images $Te_i$ in terms of this basis with the help of a matrix $A$, the matrix of the operator $T$ corresponding to base $(e_1, \ldots, e_n)$. The defining equation for $A$ is

$$(Te_1, \ldots, Te_n) = (e_1, \ldots, e_n) \cdot A$$

Let's see what happens to a vector $v$. Write $v$ in the basis $(e_1, \ldots, e_n)$. But write this in an efficient way: $$v = (e_1, \ldots, e_n) \cdot (x_1, \ldots x_n)^{t}$$

Apply $T$ and use that $T$ is linear ( yea, $T$ works from the left). We get

$$Tv = (Te_1, \ldots, Te_n)\cdot (x_1, \ldots x_n)^{t}$$

and using the previous equality:

$$Tv =((e_1, \ldots, e_n)\cdot A )\cdot(x_1, \ldots x_n)^{t}$$ or $$Tv =(e_1, \ldots, e_n)\cdot (A \cdot(x_1, \ldots x_n)^{t})$$

Therefore, the coordinates for the image $v' = Tv$ are

$$(x_1', \ldots x_n')^{t} = A (x_1, \ldots x_n)^{t}$$ that is

$$\left (\begin{array}{c} x'_1 \\x'_2\\ \ldots\\ x'_n\end{array} \right) = A \left (\begin{array}{c} x_1 \\x_2\\ \ldots\\ x_n\end{array} \right)$$

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