# Linear Transformations and Bases

• Apr 28th 2012, 02:19 AM
Bernhard
Linear Transformations and Bases
I need some help or at least some assurance that my thinking on linear transformations and their matrix representations is correct.

I assume when we specify a linear transformation eg F(x,y, z) = (3x +y, y+z, 2x-3z) for example, that this is specified by its action on the variables and is not with respect to any basis.

However when we specify the matrix of a linear transformation T: V --> W that this is with respect to a basis in V and a basis in W

Of course if we have a linear transformation S: V -->V it could be that the two bases are the same.

If we are dealing in Euclidean space and no basis is mentioned regarding the matrix of a linear transformation, then I am assuming the standard bases are assumed.

Can someone either confirm I am correct in my thinking or point out the errors in my thinking?

Peter
• Apr 28th 2012, 11:28 AM
MathoMan
Re: Linear Transformations and Bases
Regarding your example F(x,y, z) = (3x +y, y+z, 2x-3z), consider the following.

Let $p=\left[\begin{array}{c} x \\ y\\ z \end{array} \right]$, and $A=\left[\begin{array}{ccc} 3 & 1 & 0\\ 0 & 1 & 1 \\ 2 & 0 & -3 \end{array} \right]$. Then $F(x,y,z)=F(p)=A\cdot p=\left[\begin{array}{c} 3x+y \\ y+z\\ 2x-3z \end{array} \right]$.

Obviously $A\in \mathbb{R}^{3\times 3}$ or $A:\mathbb{R}^3 \to \mathbb{R}^3$.

Standard canonical basis in $\mathbb{R}^3$, vectors $e_1=\left[\begin{array}{c} 1 \\ 0\\ 0 \end{array} \right]$, $e_2=\left[\begin{array}{c} 0 \\ 1\\ 0 \end{array} \right]$, $e_3=\left[\begin{array}{c} 0 \\ 0\\ 1 \end{array} \right]$ are mapped to $f_i=A e_i$, columns of the matrix $A$:
$f_1=\left[\begin{array}{c} 3 \\ 0\\ 2 \end{array} \right]$, $f_2=\left[\begin{array}{c} 1 \\ 1\\ 0 \end{array} \right]$ and $f_3=\left[\begin{array}{c} 0 \\ 1\\ -3 \end{array} \right]$.

Every two bases in vector space V are equivalent. Meaning that every vector can be given in either one of them as a linear combination of the vectors forming the basis, the difference is in coefficients used.

If nothing is specified then standard basis is assumed.
• Apr 28th 2012, 01:49 PM
Deveno
Re: Linear Transformations and Bases
in general, the matrix of a linear transformation is determined "only up to similarity" (if T is a linear transformation, and A is the matrix for T in the standard basis, then if P is the (necessarily invertible) "change-of basis" matrix from the standard basis to any other basis, the matrix for T in the NEW basis is: PTP-1).

this is why similarity classes are important in the study of matrices.

note that just WRITING a vector as (x,y,z) is actually implying a basis:

(x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1) = xe1 + ye2 + ze3.

that is, in the standard basis B = {e1,e2,e3} the "coordinates" of a vector (x,y,z) are just [x,y,z]B.

in Rn, particularly for n = 2 and n = 3, the normal "coordinate axes" are so firmly entrenched in our minds, we forget we PUT them there in the first place, there is no "x-axis" in actual space. so, for a rotation (like, for example, a ferris wheel), we "choose" an origin (the center of the wheel) to be able to describe its motion mathematically. it's sort of like choosing a language to give expression to an idea....the idea is just what it is, but we need WORDS to convey it.
• Apr 28th 2012, 06:45 PM
Bernhard
Re: Linear Transformations and Bases
Thanks for the help above

Much appreciated

Peter