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Discover Linear Algebra

Discovery guide 26.1 Discovery guide

Recall.

Square matrices \(A, B \) are called similar if there exists an invertible matrix \(P \) so that
\begin{equation*} \inv{P} A P = B \text{.} \end{equation*}
In this case, \(P \) is called the transition matrix that realizes the similarity of \(A \) and \(B \text{.}\)
In Chapter 25, we were interested in determining the conditions under which a square matrix is similar to a diagonal matrix. We called such a matrix diagonalizable. In this discovery guide, we’ll explore similarity more generally, both geometrically and algebraically.

Discovery 26.1.

(a) Symmetry.

The equality defining similar matrices seems one-directional: it would seem more appropriate to say that matrix \(A \) is similar to matrix \(B \), rather than saying that they are similar together, because \(A \) can be “transformed” into \(B \) via the transition matrix \(P \text{.}\)
Convince yourself that this distinction is not important by verifying that if \(A \) is similar to \(B \text{,}\) then \(B \) is also similar to \(A \text{:}\)
\begin{equation*} \inv{(\fillinmath{XXX})} B (\fillinmath{XXX}) = A \text{.} \end{equation*}

(b) Reflexivity.

Fill in the blanks to prove that every square matrix is similar to itself:
\begin{equation*} \inv{(\fillinmath{XXX})} A (\fillinmath{XXX}) = A \text{.} \end{equation*}

(c) Transitivity.

Suppose \(A, B \) are similar via transition matrix \(P \text{,}\) and that \(B \) is also similar to a third matrix \(C \) via transition matrix \(Q \text{,}\) so that \(\inv{Q} B Q = C\text{.}\)
Fill in the blanks to verify that \(A \) and \(C \) must also be similar:
\begin{equation*} \inv{(\fillinmath{XXX})} A (\fillinmath{XXX}) = C \text{.} \end{equation*}

Aside: Terminology.

Discovery 26.2. The geometry of similarity.

Consider the matrices
\begin{align*} A \amp = \begin{bmatrix} 1 \amp 3 \\ 3 \amp 1 \end{bmatrix} \text{,} \amp B \amp = \begin{abmatrix}{rr} 4 \amp 0 \\ 0 \amp -2 \end{abmatrix} \text{,} \amp P \amp = \begin{abmatrix}{rr} 1 \amp -1 \\ 1 \amp 1 \end{abmatrix} \text{,} \amp \inv{P} \amp = \frac{1}{2} \begin{abmatrix}{rr} 1 \amp 1 \\ -1 \amp 1 \end{abmatrix} \text{.} \end{align*}
You don’t need to verify this, but \(A \) and \(B \) are similar via transition matrix \(P \text{,}\) so that \(\inv{P} A P = B \text{.}\)
As usual, write \(\basisfont{S} \) for the standard basis \(\basisfont{S} = \{ \uvec{e}_1, \uvec{e}_2 \} \) of \(\R^2 \text{.}\) Also write \(\basisfont{B} \) for the basis \(\basisfont{B} = \{ \uvec{p}_1, \uvec{p}_2 \} \) of \(\R^2 \) formed by the columns of \(P \) (Statement 11 of Theorem 21.5.5).

(a)

On a set of \(xy \)-axes, plot the vectors \(\uvec{v} \) and \(A \uvec{v} \text{,}\) where \(\uvec{v} = \left[\begin{smallmatrix} 3 \\ -2 \end{smallmatrix}\right] \text{.}\)

(b)

Use \(\inv{P} \) to compute the coordinate vector \(\matrixOf{\uvec{v}}{B} \text{.}\)
Hint.
What are the columns of the transition matrix \(\ucobmtrx{B}{S} \text{?}\) Do you know another matrix in this activity that has those same columns? Then use Statement 3 of Proposition 22.5.4.

(c)

Our usual \(xy \)-coordinate system is really the \(\basisfont{S} \)-coordinate system, as
\begin{equation*} \begin{bmatrix} x \\ y \end{bmatrix} = x \uvec{e}_1 + y \uvec{e}_2 \end{equation*}
for every vector in the plane.
Let’s call the \(\basisfont{B} \)-coordinate system the \(wz \)-coordinate system, with \(w \) on the horizontal axis and \(z \) on the vertical axis. On a new set of \(wz \)-axes (don’t erase your \(xy \)-axes from before!), plot the vectors \(\matrixOf{\uvec{v}}{B} \) and \(B \matrixOf{\uvec{v}}{B} \text{.}\)

(d) Analyze.

When computing \(B \matrixOf{\uvec{v}}{B} \text{,}\) the \(4 \) in the upper left entry of \(B \) multiplied the \(w \)-component of \(\matrixOf{\uvec{v}}{B} \text{,}\) and the \(-2 \) in the lower right entry of \(B \) multiplied the \(z \)-component of \(\matrixOf{\uvec{v}}{B} \text{.}\)
Describe this pattern in geometric terms, by considering how the diagonal entries of \(B \) determined how the vector \(\matrixOf{\uvec{v}}{B} \) was transformed into the vector \(B \matrixOf{\uvec{v}}{B} \text{.}\)

(e)

Just as the standard basis vectors \(\uvec{e}_1, \uvec{e}_2 \) correspond to the \(x \)- and \(y \)-axes, respectively, the \(\basisfont{B} \)-basis vectors \(\uvec{p}_1, \uvec{p}_2 \) correspond to the \(w \)- and \(z \)-axes, respectively.
Plot vectors \(\uvec{p}_1, \uvec{p}_2 \) on your original set of \(xy \)-axes from Task a, and then extend each of them in both directions (maybe with dashed lines) to create a set of \(wz \)-axes superimposed on the \(xy \)-axes.

(f) Compare.

Try to determine if your geometric description of the transformation \(\matrixOf{\uvec{v}}{B} \mapsto B \matrixOf{\uvec{v}}{B} \) from Task d is consistent with the geometric transformation \(\uvec{v} \mapsto A \uvec{v} \) on your first diagram, but relative to the new superimposed \(wz \)-axes.

(g) Connect.

Use \(P \) to convert \(B \matrixOf{\uvec{v}}{B} \) from \(\basisfont{B} \)-coordinates back to \(\basisfont{S} \)-coordinates. Surprised?
Now look back at Task b to fill in the blank: \(P B \matrixOf{\uvec{v}}{B} = \fillinmath{XXX} \uvec{v} \text{.}\)

(h) Reflect.

We started with the assumption that \(A \) and \(B \) are similar matrices. Reflect on the reason you were asked to carry out this activity.

Discovery 26.3. The algebra of similarity.

By muliplying both sides on the left by \(P \text{,}\) a similarity relation \(\inv{P} A P = B \) becomes \(A P = P B \text{.}\)
For this activity, assume we are working with \(3 \times 3\) matrices.

(a)

Think of \(P \) as made up of three columns:
\begin{equation*} P = \begin{bmatrix} | \amp | \amp | \\ \uvec{p}_1 \amp \uvec{p}_2 \amp \uvec{p}_3 \\ | \amp | \amp | \end{bmatrix}\text{.} \end{equation*}
Using the pattern of (✶✶✶) in Subsection 4.3.7, write down an expression for the first column of the product matrix \(A P \text{.}\)

(b)

Now think of \(B \) as made up of three columns:
\begin{equation*} B = \begin{bmatrix} | \amp | \amp | \\ \uvec{b}_1 \amp \uvec{b}_2 \amp \uvec{b}_3 \\ | \amp | \amp | \end{bmatrix} \end{equation*}
(but for now stop thinking of \(P \) as a collection of columns). Again using the pattern of (✶✶✶) in Subsection 4.3.7, write down an expression for the first column of the product matrix \(P B \text{.}\)

(c)

Let’s explore your expression from Task b a little further. Suppose the first column of \(B \) is the vector \(\uvec{b}_1 = \left[\begin{smallmatrix} 5 \\ 3 \\ -1 \end{smallmatrix}\right] \text{.}\) Use the matrix-times-vector pattern from (✶✶) in Subsection 22.3.2 to express the first column of \(P B \) as a linear combination.

(d)

For \(A P = P B \) to be true, we must at least have the first columns of \(A P \) and \(P B \) equal. Set your expressions from Task a and Task c to be equal to help you fill in the following:
the coordinate vector of relative to the basis must be equal to .
Hint.
Your expression from Task c is a linear combination, and linear combinations tell us coordinates.

(e)

If we analyzed and compared the second columns of \(A P \) and \(P B \) in the same fashion, would we come to the same pattern as in Task d? What words in the pattern would you change? What if we analyzed and compared the third columns of \(A P \) and \(P B \) in the same fashion?
In each of the remaining discovery activities, assume that square matrices \(A,B \) are similar via transition matrix \(P \text{,}\) with \(\inv{P} A P = B \text{.}\)

Discovery 26.4.

Verify that the squares \(A^2 \) and \(B^2 \) are also similar.
Do you think the same is true for every pair of powers \(A^k \) and \(B^k \) (with the same exponent on both matrices)?
What about \(\inv{A} \) and \(\inv{B} \text{?}\)

Discovery 26.5.

Demonstrate that the transition matrix \(P \) transforms null space vectors of \(B \) into null space vectors of \(A \text{.}\)
What matrix will transform null space vectors of \(A \) into null space vectors of \(B \text{?}\)

Discovery 26.6.

Demonstrate that the transition matrix \(P \) transforms column space vectors of \(B \) into column space vectors of \(A \text{.}\)
What matrix will transform column space vectors of \(A \) into column space vectors of \(B \text{?}\)
Hint.
Recall that a vector \(\uvec{b} \) is in the column space of \(B \) if and only if the system \(B \uvec{x} = \uvec{b} \) has at least one solution.

Discovery 26.7.

We have seen that similar matrices have the same eigenvalues (Statement 3 of Proposition 25.6.1).
Demonstrate that the transition matrix \(P \) transforms eigenvectors of \(B \) for a particular eigenvalue into eigenvectors of \(A \) for that same (shared) eigenvalue.
What matrix will transform eigenvectors of \(A \) into eigenvectors of \(B \text{?}\)