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Discover Linear Algebra

Jeremy Sylvestre

Section 26.5 Theory

Subsection 26.5.1 Properties of trace

We have introduced a new matrix operation in this chapter, so we will record some of its algebraic properties. Most of these are straightforward to verify from the definition of trace (of a square matrix) (see the Terminology and notation section for this chapter), and we leave that task to you, the reader.
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Proof of Statement 6.

Suppose that \(A,B\) are \(n \times n\) matrices. Set \(C = A B\) and \(D = B A\text{.}\) Since \(C\) is a product matrix, each of its entries is obtained from the product of a row of \(A\) with a column of \(B\text{.}\) In particular, the diagonal entries of \(C\) are:
\begin{equation*} \begin{array}{rcccccccc} c_{11} \amp = \amp a_{11} b_{11} \amp + \amp a_{12} b_{21} \amp + \amp \dotsb \amp + \amp a_{1n} b_{n1} \text{,} \\ c_{22} \amp = \amp a_{21} b_{12} \amp + \amp a_{22} b_{22} \amp + \amp \dotsb \amp + \amp a_{2n} b_{n2} \text{,} \\ \amp \vdots \\ c_{nn} \amp = \amp a_{n1} b_{1n} \amp + \amp a_{n2} b_{2n} \amp + \amp \dotsb \amp + \amp a_{nn} b_{nn} \text{,} \end{array} \end{equation*}
and \(\trace C\) is the sum of these sums. If you form a similar expression for each diagonal entry of \(D\) as a row of \(B\) times the corresponding column of \(A\text{,}\) you will find that, as a grid of products of matrix entries, it is simply the transpose of the grid above. Conclude that the total sum of each grid is the same.
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Subsection 26.5.2 Similarity is an equivalence relation

We will record here the result of Discovery 26.1 without proof, as we have already justified it in that activity.
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As discussed in Subsection 26.3.3, Theorem 26.5.2 implies that similarity partitions \(\matrixring_n(\C)\) into similarity classes. There are some special similarity classes, which explains our initial focus on diagonal form (Chapter 25) as the first “simple” form of interest.
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Proof idea.

All that is required is to prove that a scalar matrix \(k I\) can only be similar to itself. That is, if \(B \similar k I\) is true, then it must in fact be that \(B = kI\text{.}\) We leave the details to you, the reader.
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Remark 26.5.4.

Note that this proposition applies to both the zero matrix \(\zerovec = 0 I\) and the identity matrix \(I = 1 I\text{.}\)
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Subsection 26.5.3 Properties of similar matrices

Now we’ll state and prove the properties of similar matrices that can be transferred from one to the other via a transition matrix that realizes the similarity.
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Proof.

Assume \(B = \inv{P} A P\text{.}\)
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Case of positive \(k\).
We have
\begin{equation*} B^k = (\inv{P} A P)^k = (\inv{P} A P) (\inv{P} A P) (\inv{P} A P) \dotsm (\inv{P} A P), \end{equation*}
where the expression on the right has \(k\) factors of \((\inv{P} A P)\text{.}\) By associativity, rearrange the brackets:
\begin{equation*} B^k = \inv{P} A (P \inv{P}) A (P \inv{P}) A (P \dotsm \inv{P}) A P \text{.} \end{equation*}
Each factor of \(P \inv{P}\) is equal to the identity matrix, and the identity matrix has no effect in matrix multiplication. Thus,
\begin{equation*} B^k = \inv{P} A A A \dotsm A P, \end{equation*}
where there are \(k\) factors of \(A\text{,}\) since we started with \(k\) factors of \((\inv{P} A P)\text{.}\) We now have \(B^k = \inv{P} A^k P\text{,}\) showing that \(A^k\) and \(B^k\) are similar, as desired.
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If we assume that \(A\) is invertible, then so is \(B\) since it is equal to the product of invertible matrices \(\inv{P}\text{,}\) \(A\text{,}\) and \(P\) (Statement 4 of Proposition 5.5.5). If we instead assume that \(B\) is invertible, then we can make the same argument using expression \(A = P B \inv{P}\) to conclude that \(A\) is invertible as well.
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For the remainder of the cases of showing similarity of powers of \(A\) and \(B\text{,}\) assume that \(A\) and \(B\) are both invertible.
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Case of \(k = 0\).
By convention, both \(A^0 = I\) and \(B^0 = I\text{,}\) and the identity matrix is similar to itself.
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Case of \(k = -1\).
Demonstrating similarity of \(\inv{A}\) and \(\inv{B}\) is straightforward, and we leave it to you, the reader.
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Case of \(k \lt -1\).
Write \(k = - m\) for \(m\) a positive integer. If \(A\) and \(B\) are similar, then so are \(\inv{A}\) and \(\inv{B}\) by a previous case. But then so are positive powers of \(\inv{A}\) and \(\inv{B}\text{,}\) also by a previous case. In particular, the matrices
\begin{align*} \bigl(\inv{A}\bigr)^m \amp = A^{-m} = A^k \text{,} \amp \bigl(\inv{B}\bigr)^m \amp = B^{-m} = B^k \end{align*}
are similar.
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Proof idea for Statement 1.

Assume \(\uvec{x} = \uvec{x}_0\) is a solution to \(B \uvec{x} = \zerovec\text{,}\) so that \(B \uvec{x}_0 = \zerovec\) is true. Substitute the similarity relation \(B = \inv{P} A P\) into this vector equality and rearrange to demonstrate that \(\uvec{x} = P \uvec{x}_0\) is a solution to \(A \uvec{x} = \zerovec\text{.}\)
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Proof idea for Statement 2.

If vector \(\uvec{b}\) is in the column space of \(B\text{,}\) then the system \(B \uvec{x} = \uvec{b}\) is consistent. (See the discussion in Subsection 21.3.1.) So suppose \(\uvec{x} = \uvec{x}_0\) is a solution to this system, so that \(B \uvec{x}_0 = \uvec{b}\) is true. Substitute the similarity relation \(B = \inv{P} A P\) into this vector equality and rearrange to demonstrate that \(\uvec{x} = P \uvec{x}_0\) is a solution to the system \(A \uvec{x} = P \uvec{b}\text{.}\) Hence this system is consistent, which means that \(P \uvec{b}\) is in the column space of \(A\text{.}\)
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The following corollary follows from the proposition due to the fact that the transition matrix \(P\) is invertible.
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Proof idea.

We will outline the proof of equal nullities by considering maximally linearly independent sets of null space vectors; the proof of equal ranks can be similarly obtained by considering maximally linearly sets of column space vectors.
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We have already seen in Proposition 26.5.6 that if \(B = \inv{P} A P\text{,}\) then the transition matrix \(P\) transforms null space vectors of \(B\) into null space vectors of \(A\text{.}\) The symmetry of the similarity relation tells that \(\inv{P}\) will do the reverse: transform null space vectors of \(A\) into null space vectors of \(B\text{.}\) Applying Statement 2 of Proposition 21.5.1 with \(E = P\) to a maximal linearly independent set of null space vectors of \(B\text{,}\) or with \(E = \inv{P}\) to a maximal linearly independent set of null space vectors of \(A\text{,}\) we see that such maximal linearly independent sets must be the same size for \(A\) as for \(B\text{.}\) As dimension is the size of a basis, and a basis is a maximally linearly independent set (see Statement 2.a of Theorem 19.5.2) the dimensions of the two null spaces must be the same.
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Now we record some facts relating eigenvalues and eigenvectors between similar matrices. Note that we have already recorded (and proved) some of these facts in Subsection 25.6.1.
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Proof.

We have already stated and proved the statement about determinant as Statement 1 of Proposition 25.6.1.
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To prove that similar matrices have the same trace, we will use Statement 6 of Proposition 26.5.1. If \(B = \inv{P} A P\) for some invertible matrix \(P\text{,}\) then
\begin{align*} \trace B \amp = \trace (\inv{P} A P) \\ \amp = \trace \bbrac{\inv{P} (A P)} \\ \amp = \trace \bbrac{(A P) \inv{P}} \\ \amp = \trace \bbrac{A (P \inv{P})} \\ \amp = \trace (A I) \\ \amp = \trace A \text{.} \end{align*}
Note that in the above manipulations, we are not asserting that \(\inv{P} (A P)\) is equal to \((A P) \inv{P}\text{.}\) Instead, we are using the \(\trace (M N) = \trace (N M)\) pattern from Proposition 26.5.1, applied to \(M = \inv{P}\) and \(N = A P\text{.}\)
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Proof idea.

Note that we have already stated and proved the fact about eigenvalues and their multiplicities as Statement 3 of Proposition 25.6.1. So all that remains is to prove the proposed fact about eigenvectors.
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So assume \(B = \inv{P} A P\text{,}\) and suppose \(\uvec{x}_0\) is an eigenvector of \(B\) corresponding to eigenvalue \(\lambda\text{.}\) By definition, this means
\begin{equation*} B \uvec{x}_0 = \lambda \uvec{x}_0\text{.} \end{equation*}
Substitute the assumed similarity relation for \(A\) and \(B\) and rearrange to
\begin{equation*} A (P \uvec{x}_0) = \lambda (P \uvec{x}_0)\text{,} \end{equation*}
which says that \(P \uvec{x}_0\) is an eigenvector of \(A\) corresponding to the shared eigenvalue \(\lambda\text{.}\)
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Similarly to null spaces and column spaces of similar matrices above, the following corollary follows from the theorem due to the fact that the transition matrix \(P\) is invertible.
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Proof idea.

This is essentially the same as the proof of Corollary 26.5.7: if \(B = \inv{P} A P\text{,}\) then from Theorem 26.5.9 we know that the transition matrix \(P\) will transform vectors in the eigenspace \(E_\lambda(B)\) into vectors in the eigenspace \(E_\lambda(A)\) for any shared eigenvalue of \(A\) and \(B\text{,}\) and \(\inv{P}\) will do the reverse. Applying Statement 2 of Proposition 21.5.1 with \(E = P\) to a maximal linearly independent set of vectors in \(E_\lambda(B)\text{,}\) or with \(E = \inv{P}\) to a maximal linearly independent set of vectors in \(E_\lambda(A)\text{,}\) we see that such maximal linearly independent sets must be the same size for \(A\) as for \(B\text{.}\)
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