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

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 $\beta$ be an ordered basis for ${R}^{n}$ and B is a matrix whose columns are vectors in $\beta$. Prove${\left[T\right]}_{\beta }={B}^{-1}AB$.
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Solution:
Let A be the standard matrix for T where $T:{R}^{n}\to {R}^{n}$ is a linear transformation. Let $\beta$ be an ordered basis for ${R}^{n}$ and B is a matrix whose columns are vectors in $\beta$.
Define T(x)=Ax.
Let the vectors of $\beta be{v}_{1},{v}_{2},\dots {v}_{n}$.
Then,
${a}_{1}{v}_{1}+{a}_{2}{v}_{2}+\dots +{a}_{n}{V}_{n}=Ax$
$\left[{v}_{1}{v}_{2}...{v}_{n}\right]\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\\ ⋮\\ {a}_{n}\end{array}\right]=Ax$
Further we have
$B\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\\ ⋮\\ {a}_{n}\end{array}\right]=Ax\left(given:B=\left[{v}_{1}{v}_{2}...{v}_{n}\right]\right)$

Suppose the vector x is any vector from , then x represents any column of B.
Thus, we have$\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\\ ⋮\\ {a}_{n}\end{array}\right]={B}^{-1}AB$
Conclusion
The column matrix $\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\\ ⋮\\ {a}_{n}\end{array}\right]$ is coordinate vector with respect to B. that is ${\left[T\right]}_{\beta }=\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\\ ⋮\\ {a}_{n}\end{array}\right]$
Hence, it is proved that ${\left[T\right]}_{\beta }={B}^{-1}AB$.