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Section 4.2 Matrix Equivalence and Similarity

Subsection 4.2.1 Change of Bases and Transformations

Recall that if \(\mathcal{B}\) is one basis for a vector space and \(\mathcal{D}\) is another, then the transformations \(Rep_{\mathcal{B}\mathcal{D}}\) and \(Rep_{\mathcal{D}\mathcal{B}}\) change coordinates back and forth between them, and that \((Rep_{\mathcal{D}\mathcal{B}})^{-1}=Rep_{\mathcal{B}\mathcal{D}}\text{.}\) (See also Figure 3.4.5 and Figure 3.4.6.)

If \(T\) is a transformation of the vector space in terms of the basis \(\mathcal{B}\text{,}\) then

\begin{align*} \widehat{T}\amp=Rep_{\mathcal{D}\mathcal{B}}\circ T\circ Rep_{\mathcal{B}\mathcal{D}}\\ \amp=Rep_{\mathcal{D}\mathcal{B}}\circ T\circ (Rep_{\mathcal{D}\mathcal{B}})^{-1}, \end{align*}

is the same transformation except in terms of \(\mathcal{D}\text{.}\)  1 Why is the composition of the transformations written right to left? When this situation exists we say that the transformations \(T\) and \(\widehat{T}\) are similar (or in some settings we say they are conjugate).

Figure 4.2.1. Transformations Between and Within Bases
In your text (map representations, p.259 and similarity, p.388) you will see this written something like:

\begin{equation*} T=Rep_{\mathcal{B}\mathcal{B}}\left(T\right) \end{equation*}

and

\begin{align*} \widehat{T}\amp=Rep_{\mathcal{D}\mathcal{D}}\left(T\right)\\ \amp=Rep_{\mathcal{D}\mathcal{B}}\circ T\circ (Rep_{\mathcal{D}\mathcal{B}})^{-1}\\ \amp=Rep_{\mathcal{D}\mathcal{B}}\circ Rep_{\mathcal{B}\mathcal{B}}\left(T\right)\circ (Rep_{\mathcal{D}\mathcal{B}})^{-1}. \end{align*}

Subsection 4.2.2 Similar Matrices

Definition 4.2.2. Similar Matrices.

Two matrices \(A\) and \(B\) are said to be similar if there exists an invertible (non-singular) matrix \(P\) such that

\begin{equation*} B=P\, A\, P^{-1}. \end{equation*}

More generally, we say that two elements related in this way are conjugates.

Investigation 4.2.1. A Basic Example.

Suppose we have two matrices

\begin{equation*} A= \left( \begin{array}{rr} 1\amp 3 \\ -1 \amp-2\\ \end{array} \right),\ \mbox{and}\ B= \left( \begin{array}{rr} -10 \amp 7 \\ -13 \amp 9 \end{array} \right) \end{equation*}

and we want to see if they are similar. By definition this means we need to find \(P\) such that

\begin{equation*} A=P\, B\, P^-1 \end{equation*}

or equivalently

\begin{equation*} A\, P=P\, B. \end{equation*}

Let

\begin{equation*} P= \left( \begin{array}{rr} p_1 \amp p_2\\ p_3 \amp p_4 \end{array} \right) \end{equation*}

so that the equation above becomes

\begin{equation*} \left( \begin{array}{rr} p_1+3p_3\amp p_2+3p_4 \\ -p_1-2p_3 \amp -p_2-2p_4\\ \end{array} \right)= \left( \begin{array}{rr} -10p_1-13p_2\amp 7p_1+9p_2 \\ -10p_3-13p_4\amp 7p_3+9p_4 \\ \end{array} \right) \end{equation*}

leaving us with four equations and four unknowns. Here we write it as a homogeneous system:

\begin{align*} 11p_1+13p_2+3p_3\amp = 0\\ 7p_1+8p_2-3p_4\amp = 0\\ p_1-8p_3-13p_4\amp = 0\\ p_2+7p_3+11p_4\amp = 0 \end{align*}

or as a matrix equation

\begin{equation*} \left( \begin{array}{rrrr|r} 11 \amp 13 \amp 3 \amp 0 \amp 0\\ 7 \amp 8 \amp 0 \amp -3 \amp 0\\ 1 \amp 0 \amp -8 \amp -13 \amp 0\\ 0 \amp 1 \amp 7 \amp 11 \amp 0\\ \end{array} \right). \end{equation*}

With a little work this reduces to

\begin{equation*} \left(\begin{array}{rrrr|r} 1 \amp 0 \amp -8 \amp -13 \amp 0\\ 0 \amp 1 \amp 7 \amp 11 \amp 0\\ 0 \amp 0 \amp 0 \amp 0 \amp 0\\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{array}\right). \end{equation*}

So if we let \(p_3=-1\) and \(p_4=1\) we get \(p_1=5\) and \(p_2=-4\text{,}\) and

\begin{equation*} P= \left( \begin{array}{rr} 5 \amp -4\\ 1 \amp -1 \end{array} \right). \end{equation*}

Note that since \(p_3\) and \(p_4\) are free variables there are in fact infinitely many solutions. Finally, we will see in Subsection 4.3.1 that there is an easier way.

Here are some basic key observations:

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