Similar Matrices and Their Properties

Let \(A\) and \(B\) be \(n \times n\) matrices. Then the products \(AB\) and \(BA\) are both \(n \times n\) matrices. In most cases the products \(AB\) and \(BA\) are not equal.

However, for some pairs of \(n \times n\) matrices \(A\) and \(B\), we are able to find an invertible matrix \(P\) such that \(PB = AP\). This leads to the following definition.

The following theorem shows that similarity (\(\sim \)) satisfies reflexive, symmetric, and transitive properties.

item:reflexive It is clear that \(A\sim A\) (let \(P=I\)).

item:symmetric If \(A\sim B,\) then for some invertible matrix \(P\),

\begin{equation*} A=P^{-1}BP \end{equation*}
and so
\begin{equation*} PAP^{-1}=B \end{equation*}
But then
\begin{equation*} \left ( P^{-1}\right ) ^{-1}AP^{-1}=B \end{equation*}
which shows that \(B\sim A\).

item:transitive Now suppose \(A\sim B\) and \(B\sim C\). Then there exist invertible matrices \(P,Q\) such that

\begin{equation*} A=P^{-1}BP,\ B=Q^{-1}CQ \end{equation*}
Then,
\begin{equation*} A=P^{-1} \left ( Q^{-1}CQ \right )P=\left ( QP\right ) ^{-1}C\left ( QP\right ) \end{equation*}
showing that \(A\) is similar to \(C\).

Any relation satisfying the reflexive, symmetric and transitive properties is called an equivalence relation. Theorem th:similarityequivalence proves that similarity between matrices is an equivalence relation. Practice Problem prob:lessthan gives a good example of a relation that is NOT an equivalence relation.

As we will see later, similar matrices share many properties. Before proceeding to explore these properties, we pause to introduce a simple matrix function that we will continue to use throughout the course.

It is easy to see that \(\mbox {tr}(A + B) = \mbox {tr} A + \mbox {tr} B\) and that \(\mbox {tr}(cA) = c \mbox { tr}(A)\) holds for all \(n \times n\) matrices \(A\) and \(B\) and all scalars \(c\). The following fact is more surprising.

Write \(A = [a_{ij}]\) and \(B = [b_{ij}]\). For each \(i\), the \((i, i)\)-entry \(d_{i}\) of the matrix \(AB\) is given as follows: \(d_{i} = a_{i1}b_{1i} + a_{i2}b_{2i} + \dots + a_{in}b_{ni} = \sum _{j}a_{ij}b_{ji}\). Hence

\begin{equation*} \mbox {tr}(AB) = d_1 + d_2 + \dots + d_n = \sum _{i}d_i = \sum _{i}\left (\sum _{j}a_{ij}b_{ji}\right ) \end{equation*}
Similarly we have \(\mbox {tr}(BA) = \sum _{i}\left (\sum _{j}b_{ij}a_{ji}\right )\). Since these two double sums are the same, we have proven the theorem.

The following theorem lists a number of properties shared by similar matrices.

Let \(B = P^{-1}AP\) for some invertible matrix \(P\).

For th:properties_similar_det, \(\det B = \det (P^{-1}) \det A \det P = \det A\) because \(\det (P^{-1}) = 1/ \det P\) (Theorem th:detofinverse).

Similarly, for th:properties_similar_rank \(\mbox {rank} B = \mbox {rank}(P^{-1}AP) = \mbox {rank} A\), because multiplication by an invertible matrix cannot change the rank. To see this, note that any invertible matrix is a product of elementary matrices. Multiplying by elementary matrices is equivalent to performing elementary row (column) operations on \(A\), which does not change the row (column) space, nor the rank. It follows that similar matrices have the same rank.

For th:properties_similar_trace, we make use of Theorem th:trAB=trBA:

\begin{equation*} \mbox {tr} (P^{-1}AP) = \mbox {tr}[P^{-1}(AP)] = \mbox {tr}[(AP)P^{-1}] = \mbox {tr} A. \end{equation*}
As for th:properties_similar_char_poly,
\begin{align*} \det (B-\lambda I) &= \det \{P^{-1}AP-\lambda (P^{-1}P)\} \\ &= \det \{ P^{-1}(A-\lambda I)P\} \\ &= \det (A-\lambda I), \end{align*}

so \(A\) and \(B\) have the same characteristic equation. Finally, th:properties_similar_char_poly implies th:properties_similar_eig because the eigenvalues of a matrix are the roots of its characteristic polynomial.

Even though the properties in Theorem th:properties_similar cannot be used to show two matrices are similar, these properties come in handy for showing that two matrices are NOT similar.

The next theorem shows that similarity is preserved under inverses, transposes, and powers:

See Practice Problem prob:similarproperties.

Practice Problems

At the beginning of this section we mentioned that similarity of \(n \times n\) matrices is an equivalence relation.

An equivalence relation is a binary relation \(R\) on elements of a set \(S\) that has the following properties:

  • The reflexive property: \(x R x\) for every \(x \in S\)
  • The symmetric property: If \(x R y\), then \(y R x\) for every \(x,y \in S\)
  • The transitive property: If \(x R y\) and \(y R z\), then \(x R z\) for every \(x,y,z \in S\)

Let \(S\) be the set of all positive integers. We can show that the relation “less than” (symbolized by \(<\)) is NOT an equivalence relation on this set. To see this, note that “less than” is not reflexive, because \(s<s\) is not true for any positive integer \(s\).

1.
Is the relation “less than” symmetric? YesNo
2.
Is the relation “less than” transitive? YesNo
Another relation between matrices we have studied in this course is that two matrices can be “row equivalent”. Is the relation “row equivalent”
1.
reflexive? YesNo
2.
symmetric? YesNo
3.
transitive? YesNo

Use Theorem 7 to show that \(A\) and \(B\) are not similar. \(A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\), \(B = \begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix}\)

Use Theorem 7 to show that \(A\) and \(B\) are not similar. \(A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\), \(B = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}\)
Use Theorem 7 to show that \(A\) and \(B\) are not similar. \(A = \begin{bmatrix} 1 & 2 & -3 \\ 1 & -1 & 2 \\ 0 & 3 & -5 \end{bmatrix}\), \(B = \begin{bmatrix} -2 & 1 & 3 \\ 6 & -3 & -9 \\ 0 & 0 & 0 \end{bmatrix}\)

Show that \(\begin{bmatrix} 1 & 2 & -1 & 0 \\ 2 & 0 & 1 & 1 \\ 1 & 1 & 0 & -1 \\ 4 & 3 & 0 & 0 \end{bmatrix}\) and \( \begin{bmatrix} 1 & -1 & 3 & 0 \\ -1 & 0 & 1 & 1 \\ 0 & -1 & 4 & 1 \\ 5 & -1 & -1 & -4 \end{bmatrix}\) are not similar.

Let
\[A=\begin{bmatrix}1 & 2 & 0\\2 & -1 & -2\\0 & -2 & 1\end{bmatrix}\]
The eigenvalues of \(A\) are known to be \(\lambda _1=-3\), \(\lambda _2=3\) and \(\lambda _3=1\). Find the corresponding eigenspaces.
\[\mathcal {S}_{-3}=\text {span}\left (\begin{bmatrix}\answer {-1}\\\answer {2}\\1\end{bmatrix}\right )\]
\[\mathcal {S}_3=\text {span}\left (\begin{bmatrix}\answer {-1}\\\answer {-1}\\1\end{bmatrix}\right )\]
\[\mathcal {S}_1=\text {span}\left (\begin{bmatrix}\answer {1}\\\answer {0}\\1\end{bmatrix}\right )\]
We will refer to the eigenvectors you listed above as \(\vec {x}_1\), \(\vec {x}_2\) and \(\vec {x}_3\).
Form matrix \(P\) whose columns are the eigenvectors \(\vec {x}_1\), \(\vec {x}_2\), and \(\vec {x}_3\) you found in the previous part. Use technology to find \(P^{-1}\).
\[P^{-1}=\frac {1}{6}\begin{bmatrix}\answer {-1} & \answer {2} & \answer {1}\\\answer {-2} & \answer {-2} & \answer {2}\\\answer {3} & \answer {0} & \answer {3}\end{bmatrix}\]
Find the product \(P^{-1}AP\). What did you get? Which of the following is true?
\(A\sim P\) \(A\sim D\) for ANY diagonal matrix \(D\) \(A\sim D\) for the diagonal matrix \(D\) whose diagonal entries are the eigenvalues of \(A\)
Prove that if \(A\) and \(B\) are \(n\times n\) matrices and \(A\sim B\), then
1.
\(A^{-1} \sim B^{-1}\) (provided that the inverses exist),
2.
\(A^T \sim B^T\), and
3.
\(A^k \sim B^k\) for all integers \(k \geq 1\).

(See Theorem 12)

Let \(\lambda \) be an eigenvalue of \(A\) with corresponding eigenvector \(\vec {x}\). If \(B = P^{-1}AP\) is similar to \(A\), show that \(P^{-1}\vec {x}\) is an eigenvector of \(B\) corresponding to \(\lambda \).

Text Source

The text in this section is a compilation of material from Section 7.2.1 of Ken Kuttler’s A First Course in Linear Algebra (CC-BY) and Section 5.5 of Keith Nicholson’s Linear Algebra with Applications (CC-BY-NC-SA).

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, p. 362-364.

Many of the Practice Problems are Exercises from W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, pp. 298-310.