<Home>Teaching><MATH 115A (Spring 2023)>Worksheet 8

Worksheet 8

Problem 3

Let $A \in M_{n \times n}\p{\F}$ be similar to an upper-triangular matrix, and suppose that $A$ has distinct eigenvalues $\lambda_1, \ldots, \lambda_k$ with corresponding algebraic multiplicities $m_1, \ldots, m_k$ .

Prove that $\tr A = \sum_{i=1}^k m_i\lambda_i$ .
Prove that $\det A = \prod_{i=1}^k \lambda_i^{m_i}$ .

Solution.

This follows from two facts:

If $A$ and $B$ are similar, then they have the same eigenvalues.
The eigenvalues of an upper-triangular matrix are its diagonal entries.

Fact 1 follows from the fact that if $B = QAQ^{-1}$ , then

\det\p{B - \lambda I} = \det\p{Q\p{A - \lambda I}Q^{-1}} = \det\p{A - \lambda I},

i.e., $A$ and $B$ have the same characteristic equation. For Fact 2, if $A$ is upper-triangular with diagonal entries $a_1, \ldots, a_n$ , then because $A - \lambda I$ is also upper-triangular with diagonal entries $a_1 - \lambda, \ldots, a_n - \lambda$ , the characteristic equation of $A$ is

p_A\p{\lambda} = \det\p{A - \lambda I} = \prod_{i=1}^n \p{a_i - \lambda}.

Thus, the roots of $p_A$ are exactly the diagonal entries, so the eigenvalues of $A$ are also just the diagonal entries.

Note that $\tr\p{AB} = \tr\p{BA}$ from a simple calculation:
$\begin{aligned} \tr\p{AB} = \sum_{i=1}^n \p{AB}_{ii} &= \sum_{i=1}^n \sum_{j=1}^n A_{ij} B_{ji} \\ &= \sum_{j=1}^n \sum_{i=1}^n B_{ji} A_{ij} \\ &= \sum_{j=1}^n \p{BA}_{jj} \\ &= \tr\p{BA}. \end{aligned}$
Thus, if $A = QUQ^{-1}$ , where $U$ is an upper-triangular matrix, then
$\tr A = \tr\p{QUQ^{-1}} = \tr\p{Q^{-1}QU} = \tr U.$
But the diagonal entries of $U$ are the eigenvalues of $A$ counted with multiplicity, so
$\tr U = \sum_{i=1}^k m_i\lambda_i.$
Again, write $A = QUQ^{-1}$ , where $U$ is upper-triangular. Then the determinant of $U$ is just the product of the diagonal entries, which are the eigenvalues of $A$ counted with multiplicity. Thus,
$\det A = \det U = \prod_{i=1}^k \lambda_i^{m_i}.$

Problem 4

Let $V$ be a finite dimensional vector space over $\F$ , let $T \in \mathcal{L}\p{V}$ be invertible.

Prove that if $\lambda$ is an eigenvalue of $T$ , then $\lambda^{-1}$ is an eigenvalue of $T^{-1}$ .
Prove that the eigenspace of $T$ corresponding to $\lambda$ is the same as the eigenspace of $T^{-1}$ corresponding to $\lambda^{-1}$ .
Prove that if $T$ is diagonalizable, then $T^{-1}$ is diagonalizable.

Solution.

These all follow from the fact that $v$ is an eigenvector for $T$ if and only if it's an eigenvector for $T^{-1}$ . If $v$ is an eigenvector for $T$ with eigenvalue $\lambda$ , then

T\p{v} = \lambda v \implies v = \lambda T^{-1}\p{v} \implies T^{-1}\p{v} = \lambda^{-1} v.

Note that because $T$ is invertible, $\lambda \neq 0$ , so $\lambda^{-1}$ is well-defined.

Since $v$ is an eigenvector of $T$ , we know that $v \neq 0$ , so this precisely says that $v$ is an eigenvector for $T$ with eigenvalue $\lambda^{-1}$ . Applying the forward direction with $T, \lambda$ replaced with $T^{-1}, \lambda^{-1}$ and noting that $\p{T^{-1}}^{-1} = T$ and $\p{\lambda^{-1}}^{-1} = \lambda$ gives the reverse direction. (Alternatively, you can just do the same computation, but in reverse.)

Our computation above shows that if $\lambda$ is an eigenvalue for $T$ , then $\lambda^{-1}$ is an eigenvalue for $T^{-1}$ .
Recall that the eigenspace of $T$ corresponding to $\lambda$ is the set
$E_\lambda\p{T} = \set{v \in V \mid T\p{v} = \lambda v}.$
We showed that $T\p{v} = \lambda v$ if and only if $T^{-1}\p{v} = \lambda^{-1} v$ , and this implies $E_\lambda\p{T} = E_{\lambda^{-1}}\p{T^{-1}}$ .
Let $\beta = \set{v_1, \ldots, v_n}$ be an eigenbasis of $V$ for $T$ . Then $\beta$ is also a set of eigenvectors for $T^{-1}$ . Since $\beta$ was a basis of $V$ , this means that $\beta$ is also an eigenbasis of $V$ for $T^{-1}$ .