# Is the result of a matrix transformation equivalent to the that of that same matrix but orthonormali

pouzdrotf 2022-07-06 Answered
Is the result of a matrix transformation equivalent to the that of that same matrix but orthonormalized?
Say we have the transformation matrix $A$ of full rank such that ${A}^{-1}\ne {A}^{t},$, i.e., the matrix $A$ consists of linearly independent vectors which aren't orthogonal to each other, a vector v, and the orthonormalized transformation matrix ${A}^{\prime }.$ Is it true that $Av={A}^{\prime }v?$ And if not, is this unimportant?
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## Answers (1)

Karla Hull
Answered 2022-07-07 Author has 20 answers
One of the key features of the Gram-Schmidt process is that the vectors generated by it span the same $k$-vector subspace as the original vectors. Put into the language of matrices, this means that the column space of $A$ (or the range of $T$) is the same as the column space of ${A}^{\prime }$ (or the range of ${T}^{\prime }$). (Here, I assume that the linearly independent vectors we started with were the columns of $A.$.) Consequently, we have that

But that does not imply that $Av=T\left(v\right)={T}^{\prime }\left(v\right)={A}^{\prime }v$ for all $v\in V.$.
For instance, consider the $\mathbb{R}$ -vector space ${\mathbb{R}}^{2},$ i.e., the set of column vectors $\left(a,b{\right)}^{t}$ for $a,b\in \mathbb{R}.$ Observe that the vectors $v=\left(1,1{\right)}^{t}$ and $w=\left(1,2{\right)}^{t}$ are linearly independent (because the matrix with columns $v$ and $w$ has determinant 1). By Gram-Schmidt, we obtain an orthogonal basis ${v}_{1}=\left(1,1{\right)}^{t}$ and ${v}_{2}=\left(-\frac{1}{2},\frac{1}{2}{\right)}^{t},$ and we can convert this into an orthonormal basis by taking ${v}^{\prime }=\frac{{v}_{1}}{||{v}_{1}||}$ and ${w}^{\prime }=\frac{{v}_{2}}{||{v}_{2}||}.$
Using these as columns, we can form the $2×2$ matrix ${A}^{\prime }.$ We will denote by ${e}_{1}=\left(1,0\right)$ and ${e}_{2}=\left(0,1\right)$ the standard basis vectors of ${\mathbb{R}}^{2}.$ Observe that we have
$A{e}_{1}=v\ne \frac{{v}_{1}}{||{v}_{1}||}={v}^{\prime }={A}^{\prime }{e}_{1},$
and the analogous statement holds for ${e}_{2}$ in place of ${e}_{1}.$
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