Linear Algebra - Isomorphism, Matrix of Linear Transformation Let T : R^{

Question

Linear Algebra - Isomorphism, Matrix of Linear Transformation
Let

T : R^{n} \to R^{n}

be a linear transformation. Let A be the standard matrix for T . Let

β

be an ordered basis for

R^{n}

and B is a matrix whose columns are vectors in

β

. Prove

{[T]}_{β} = B^{- 1} A B

.

Answer 1

Solution:
Let A be the standard matrix for T where $T : R^{n} \to R^{n}$ is a linear transformation. Let $β$ be an ordered basis for $R^{n}$ and B is a matrix whose columns are vectors in $β$ .
Define T(x)=Ax.
Let the vectors of $β b e v_{1}, v_{2}, \dots v_{n}$ .
Then,
$a_{1} v_{1} + a_{2} v_{2} + \dots + a_{n} V_{n} = A x$
$[v_{1} v_{2} . . . v_{n}] [\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{matrix}] = A x$
Further we have
$B [\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{matrix}] = A x (g i v e n : B = [v_{1} v_{2} . . . v_{n}])$
$[\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{matrix}] = B^{- 1} A x (R i g h t m u l t i p l y b y B^{- 1})$
Suppose the vector x is any vector from , then x represents any column of B.
Thus, we have $[\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{matrix}] = B^{- 1} A B$
Conclusion
The column matrix $[\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{matrix}]$ is coordinate vector with respect to B. that is ${[T]}_{β} = [\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{matrix}]$
Hence, it is proved that ${[T]}_{β} = B^{- 1} A B$ .

Linear Algebra - Isomorphism, Matrix of Linear Transformation Let T : R^{

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