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Let's assume that a linear operator in V space in basis (e1, e2, e3, e4) has a matrix:

$A = \begin{pmatrix} 0&1&2&3\\5&4&0&-1\\3&2&0&3\\6&1&-1&7 \end{pmatrix}$

What look matrix of this operator will have in bases $(e_2, e_1, e_3, e_4)$ and $(e_1, e_1+e_2, e_1+e_2+e_3, e_1+e_2+e_3+e_4)$ ?

I see that this equation should have a look like:

$\begin{pmatrix} e_1\\e_2\\e_3\\e_4 \end{pmatrix}$ * $\begin{pmatrix} 0&1&2&3\\5&4&0&-1\\3&2&0&3\\6&1&-1&7 \end{pmatrix}$ = $\begin{pmatrix}e_2 \\ e_1 \\ e_3 \\ e_4\end{pmatrix}$

but what's next?

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2 Answers 2

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(a) The matricial equation of the operator $T$ in $B$ is $$Y=AX,\quad X=\begin{pmatrix}x_1\\ \vdots\\{x_n}\end{pmatrix}\text { coordinates of }x\text{ in }B,\;Y=\begin{pmatrix}y_1\\ \vdots\\{y_n}\end{pmatrix}\text { coordinates of }f(x)\text{ in }B.$$

(b) The change of basis matrix from $B=(e_2, e_1, e_3, e_4)$ to $$B'=(e_1, e_1+e_2, e_1+e_2+e_3, e_1+e_2+e_3+e_4)$$ is (transposing coefficients)$$P=\begin{pmatrix}{1}&{1}&{1}&1\\{0}&{1}&{1}&1\\{0}&{0}&{1}&1\\{0}&{0}&{0}&1\end{pmatrix}$$ (c) According to a well known theorem, the matricial equation of the operator $T$ in $B'$ is

$$Y'=\left(P^{-1}AP\right)X',\quad X'=\begin{pmatrix}x'_1\\ \vdots\\{x'_n}\end{pmatrix}\text { coordinates of }x\text{ in }B',\;Y'=\begin{pmatrix}y'_1\\ \vdots\\{y'_n}\end{pmatrix}\text { coordinates of }f(x)\text{ in }B'.$$

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  • $\begingroup$ So, in this case we should just find the resulting matrix from multiplying A on $P$ on the right and on $P^-1$ on the left? $\endgroup$ Commented Jan 20, 2017 at 10:30
  • $\begingroup$ Yes, that is right. $\endgroup$ Commented Jan 20, 2017 at 12:05
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Let's denote your standard basis ${e_1, e_2, e_3, e_4}$ by $V$ and your other basis ${e_1, e_1+e_2, e_1 + e_2 + e_3, e_1 + e_2 + e_3+e_4}$ by $W$.

You want to know, what happens, when you plug a vector $u$ defined in basis $W$ into your transformation. But since the Matrix $A$ just works with Basis $V$, you will first have to change $u$ from $W$ to $V$. After that, you will apply your matrix $A$. At last, you will have your result, but this reult ist (again) just given in terms of Basis $V$. So you change it again into $W$ to get you final result.

The Question is now, how to change basis. For that, you will have to take a look at your Basis itself. You have to check, what happens to the basis vectors, when you apply the change of basis. I.e. "How can y Basis vector of $W$ be expressed in terms of Basis vectors of $V$?"

This is a quite simple question, since the definition of $W$ tells you exactly that. The reverse operation is quite similar, but I will leave this up to you.

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