Mathematics Prelims

June 25, 2009

Definition of Similarity Using Linear Transformations

Filed under: Algebra,Linear Algebra — cjohnson @ 12:24 pm

Let’s suppose we have a linear transformation \tau on \mathbb{R}^3 which performs the following:

\displaystyle \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] \mapsto \left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right]

\displaystyle \left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right] \mapsto \left[ \begin{array}{c} 4 \\ 5 \\ 6 \end{array} \right]

\displaystyle \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] \mapsto \left[ \begin{array}{c} 5 \\ 7 \\ 0 \end{array} \right]

Now, the matrix representation of this transformation with respect to the standard basis is clearly

\displaystyle \left[ \begin{array}{ccc} 1 & 4 & 5 \\ 2 & 5 & 7 \\ 3 & 6 & 0 \end{array} \right]

But suppose we were to use a different basis for \mathbb{R}^3, like say \zeta = \{ (2, 1, 0)^T, \, (1, 0, 1)^T, \, (3, 0, -1)^T \}.  We see that our transformation maps these basis vectors as follows:

\displaystyle \left[ \begin{array}{c} 2 \\ 1 \\ 0 \end{array} \right] \mapsto \left[ \begin{array}{c} 6 \\ 9 \\ 12 \end{array} \right]

\displaystyle \left[ \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right] \mapsto \left[ \begin{array}{c} 6 \\ 9 \\ 3 \end{array} \right]

\displaystyle \left[ \begin{array}{c} 3 \\ 0 \\ -1 \end{array} \right] \mapsto \left[ \begin{array}{c} -2 \\ -1 \\ 9 \end{array} \right]

Notice that with the vectors we have on both the left and the right above are the coordinates with respect to the standard basis.  We’d like to see what the matrix representing \tau looks like with respect to the \zeta basis, so let’s convert the vectors on the right to \zeta-coordinates.  Recalling that a change of basis is simply a system of equations where the columns of the coefficient matrix are the coordinates of the basis vectors (and the inverse of this matrix if we want to go the other way), we have that

\displaystyle \left[ \begin{array}{c} 6 \\ 9 \\ 12 \end{array} \right] = \left[ \begin{array}{c} 9 \\ 6 \\ -6 \end{array} \right]_{\zeta}

\displaystyle \left[ \begin{array}{c} 6 \\ 9 \\ 3 \end{array} \right] = \left[ \begin{array}{c} 9 \\ -\frac{3}{4} \\ -\frac{15}{4} \end{array} \right]_{\zeta}

\displaystyle \left[ \begin{array}{c} -2 \\ -1 \\ 9 \end{array} \right] = \left[ \begin{array}{c} 19 \\ \frac{7}{4} \\ - \frac{5}{2} \end{array} \right]_{\zeta}

So with respect to our \zeta basis, the representation of \tau is

\displaystyle \left[ \begin{array}{ccc} 9 & 9 & 19 \\ 6 & -\frac{3}{4} & \frac{7}{4} \\ -6 & -\frac{15}{4} & -\frac{5}{2} \end{array} \right]

We will denote this matrix as _\zeta[\tau]_\zeta where the right-most subscript means that inputs are in \zeta-coordinates, and the left-most subscript means the outputs are in \zeta-coordinates as well.  Note that we could calculate _\zeta[\tau]_\zeta by going from \zeta coordinates to standard coordinates, using the earlier matrix, then going back to \zeta-coordinates.  That is,

\displaystyle _\zeta[\tau]_\zeta = _\zeta[I]_e \, _e[\tau]_e \, _e[I]_\zeta

where _e[I]_\zeta refers to the \zeta-to-standard basis change of basis matrix.  For notational convenience, define the following

A := _\zeta[\tau]_\zeta

B := _e[\tau]_e

P := _e[I]_\zeta

Then the above becomes

A = P^{-1} \, B \, P

Any matrices A and B for which there exists a P with satisfying this equation are called similar matrices.  Note that two matrices are similar if and only if they represent the same linear transformation, but with respect to different bases.  Also note that every non-singular matrix represents a change of basis matrix.  Similarity forms an equivalence relation on the set of square matrices / the set of linear transformations from a finite dimensional vector space to itself.

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