Answer to Question #215530 in Linear Algebra for Komal

Question

T be a linear transformation on V and A and B be two ordered bases for V and A is the matrix of T relative to A and B is the matrix of T relative to B then prove that A and B are similar.

Answer 1

Let $\mathfrak{A}=(e_1, e_2,e_3)$, $\mathfrak{B}=(f_1, f_2, f_3)$ be two bases for the vector space V.

The fact that $A=(a_{ij})$ is the matrix of linear transformation T relative to the basis $\mathfrak{A}$ means that:

$Te_1=a_{11}e_1+a_{21}e_2+a_{31}e_3$

$Te_2=a_{12}e_1+a_{22}e_2+a_{32}e_3$

$Te_3=a_{13}e_1+a_{23}e_2+a_{33}e_3$

The fact that $B=(b_{ij})$ is the matrix of linear transformation T relative to the basis $\mathfrak{B}$ means that:

$Tf_1=b_{11}f_1+b_{21}f_2+b_{31}f_3$

$Tf_2=b_{12}f_1+b_{22}f_2+b_{32}f_3$

$Tf_3=b_{13}f_1+b_{23}f_2+b_{33}f_3$

Let C be the change-of-basis matrix:

$e_1=c_{11}f_1+c_{21}f_2+c_{31}f_3=\sum\limits_{i=1}^{3}c_{i1}f_i$

$e_2=c_{12}f_1+c_{22}f_2+c_{32}f_3=\sum\limits_{i=1}^{3}c_{i2}f_i$

$e_3=c_{13}f_1+c_{23}f_2+c_{33}f_3=\sum\limits_{i=1}^{3}c_{i3}f_i$

In general, $e_j=\sum\limits_{i=1}^{3}c_{ij}f_i$.

Then

$Te_j=\sum\limits_{i=1}^{3}c_{ij}Tf_i=\sum\limits_{i=1}^{3}c_{ij}\sum\limits_{k=1}^{3}b_{ki}f_k=\sum\limits_{k=1}^{3}f_k\sum\limits_{i=1}^{3}c_{ij}b_{ki}$

From the other side

$Te_j=\sum\limits_{i=1}^{3}a_{ij}e_i=\sum\limits_{i=1}^{3}a_{ij}\sum\limits_{k=1}^{3}c_{ki}f_k=\sum\limits_{k=1}^{3}f_k\sum\limits_{i=1}^{3}a_{ij}c_{ki}$

Comparing these two formulas, we obtain

$\sum\limits_{i=1}^{3}c_{ij}b_{ki}=\sum\limits_{i=1}^{3}a_{ij}c_{ki}$

In matrices: BC=CA or $B=CAC^{-1}$. This means that the matrices A and B are similar.