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Suppose $V$ is a vector space with two bases $\mathfrak{V}=\{v_1, \ldots, v_n\}$ and $\mathfrak{W}=\{w_1,\ldots,w_n\}$. $T:V \to V$ is a linear map and $M_\mathfrak{V}^\mathfrak{V}(T)=A$, $M_\mathfrak{W}^\mathfrak{W}(T)=B$ respectively denote the associated matrices of $T$ with respect to the bases $\mathfrak{V}$, $\mathfrak{W}$. Then, my linear algebra book says, the matrices $A$ and $B$ are conjugate i.e. there is a matrix $C$ such that $A=CBC^{-1}$. I can't figure out why this is the case. Any help will be appreciated. Thanks in advance.

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$C$ is the matrix which takes the ordered basis $\mathfrak{W}$ to the ordered basis $\mathfrak{V}$. – Kevin Dec 25 '11 at 2:15
up vote 2 down vote accepted

Each $w_i$ can be written uniquely as a linear combination of members of $\mathfrak{V}$: $$w_i=c_{1i}v_1+c_{2i}v_2+\cdots+c_{ni}v_n\;.$$ Let $$C=\pmatrix{c_{11}&c_{12}&c_{13}&\dots&c_{1n}\\ c_{21}&c_{22}&c_{23}&\dots&c_{2n}\\ c_{31}&c_{32}&c_{33}&\dots&c_{3n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ c_{n1}&c_{n2}&c_{n3}&\dots&c_{nn}}\;.$$

If $x=a_1w_1+a_2w_2+\cdots+a_nw_n$, i.e., if the coefficients of $x$ with respect to the basis $\mathfrak{W}$ are $\langle a_1,a_2,\dots,a_n\rangle$, then

$$C\pmatrix{a_1\\a_2\\a_3\\\vdots\\a_n}=\pmatrix{c_{11}a_1+c_{12}a_2+c_{13}a_3+\cdots+c_{1n}a_n\\ c_{21}a_1+c_{22}a_2+c_{23}a_3+\cdots+c_{2n}a_n\\ c_{31}a_1+c_{32}a_2+c_{33}a_3+\cdots+c_{3n}a_n\\ \vdots\\ c_{n1}a_1+c_{n2}a_2+c_{n3}a_3+\cdots+c_{nn}a_n}\;.$$

gives the coefficients of $x$ with respect to the basis $\mathfrak{V}$:

$$\begin{align*} x&=\sum_{i=1}^na_iw_i\\ &=\sum_{i=1}^na_i(c_{1i}v_1+c_{2i}v_2+\cdots+c_{ni}v_n)\\ &=\sum_{i=1}^n(c_{1i}a_iv_1+c_{2i}a_iv_2+\cdots+c_{ni}a_iv_n)\\ &=\sum_{i=1}^n\sum_{j=1}^nc_{ji}a_iv_j\\ &=\sum_{j=1}^n\sum_{i=1}^nc_{ji}a_iv_j\\ &=\sum_{j=1}^nv_j(c_{j1}a_1+c_{j2}a_2+\cdots+c_{jn}a_n)\\ &=(c_{11}a_1+c_{12}a_2+\cdots+c_{1n}a_n)v_1\\ &\qquad\qquad+(c_{21}a_1+c_{22}a_2+\cdots+c_{2n}a_n)v_2\\ &\qquad\qquad+(c_{31}a_1+c_{32}a_2+\cdots+c_{3n}a_n)v_3\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\vdots\\ &\qquad\qquad+(c_{n1}a_1+c_{n2}a_2+\cdots+c_{nn}a_n)v_n\;. \end{align*}$$

The matrix $C$ is a change-of-basis matrix, converting $\mathfrak{W}$-coordinates to $\mathfrak{V}$-coordinates. Clearly one can similarly construct the change-of-basis matrix for converting $\mathfrak{V}$-coordinates to $\mathfrak{W}$-coordinates, and since converting from one basis to the other and back again must produce the original coordinates, it follows that these two change-of-basis matrices must be inverses of each other. Thus, the change-of-basis matrix from $\mathfrak{V}$-coordinates to $\mathfrak{W}$-coordinates must be $C^{-1}$.

Now what happens when you calculate $CBC^{-1}x$ for some $x\in V$? If you start with its $\mathfrak{V}$-coordinates, multiplication by $C^{-1}$ gives you the same vector in $\mathfrak{W}$-coordinates, multiplication by $B$ gives you the $\mathfrak{W}$-coordinates of $T(x)$, and multiplication by $C$ finally converts these back to $\mathfrak{V}$-coordinates. These are the $\mathfrak{V}$-coordinates of $T(x)$, which could have been obtained directly by multiplying $x$ (in $\mathfrak{V}$-coordinates) by $A$. Thus, $CBC^{-1}x=Ax$ for every $x$, and hence $CBC^{-1}=A$.

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The reason is that we have an invertible matrix $X = \begin{pmatrix}v_1\vdots & \cdots & \vdots v_n\end{pmatrix}$ which maps the standard basis to $\mathfrak{V}$, and an invertible matrix $Y = \begin{pmatrix}w_1\vdots & \cdots & \vdots w_n\end{pmatrix}$ which maps the standard basis to $\mathfrak{W}$. Then $A=XTX^{-1}$, as $X^{-1}$ changes the basis to the standard basis, $T$ transforms the elements in the standard basis, then $X$ changes the basis back to $\mathfrak{V}$, and similarly $B=YTY^{-1}$. Thus $A = XTX^{-1} = X(YTY^{-1})X^{-1} = (XY)T(XY)^{-1}$.

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In Pierre Leroux's book Algèbre linéaire: une approche matricielle, I found this notation, which looks convenient to me:

If $T:V\to W$ is a linear map and $A\subset V,B\subset W$ are (finite ordered) basis, denote the corresponding matrix by $$ {}_BT_A. $$ Then the formula


holds whenever it makes sense.

In your case we have $T:V\to V$ and $A,B\subset V$. Denoting by $I$ the identity of $V$, we get $$ ({}_BT_B)=({}_BI_A)({}_AT_A)({}_AI_B) $$ and $$ ({}_BI_A)({}_AI_B)={}_BI_B. $$ As ${}_BI_B$ is the identity matrix, this shows that the matrices ${}_BI_A$ and ${}_AI_B$ are inverse.

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