Ch05. Eigenvectors and Eigenvalues

5.2 The Characteristic Equation

Useful information about the eigenvalues of a square matrix $A$ is encoded in a special scalar equation called the characteristic equation of $A$.

Determinants

• Let $A$ be an $n \times n$ matrix, let $U$ be any echelon form obtained from $A$ by row replacements and row interchanges (without scaling), and let $r$ be the number of such row interchanges.
• Then the determinant of $A$, written as $\text{det }A$, is $(-1)^{r}$ times the products of diagonal entries $u_{11},\cdots,u_{nn}$ in $U$.
• If $A$ is invertible, then $u_{11},\cdots,u_{nn}$ are all pivots (because $A \sim I_n$ and the $u_{ii}$ have not been scaled to 1’s).
• Otherwise, at least $u_{nn}$ is zero, and the product $u_{11}\cdots u_{nn}$ is zero.
• Thus

Example 2

Compute $\text{det }A$ for $$A= \begin{bmatrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{bmatrix}$$ .

Solution:

The following row reduction uses one row interchange:

$$A \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 & -2 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & -2 & 0 \\ 0 & -6 & -1 \end{bmatrix} \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & -2 & -1 \\ 0 & 0 & -1 \end{bmatrix} = U_1$$

• So $\text{det }A$ equals $(-1)^1(1)(-2)(-1)=-2$.

The following alternative row reduction avoids the row interchange and produces a different echelon form.

• The last step adds $\frac{-1}{3}$ times row 2 to row 3: $$A \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 & -2 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 & 0 & \frac{1}{3} \end{bmatrix} = U_2$$
• This time $\text{det }A$ equals $(-1)^0(1)(-6)\left(\frac{1}{3}\right)=-2$, the same as before. $\blacksquare$

Theorem: The Invertible Matrix Theorem. (continued)

Let $A$ be an $n \times n$ matrix. Then $A$ is invertible if and only if:

19.The number 0 is not an eigenvalue of $A$.

20.The determinant of $A$ is not zero.

Theorem 3: Properties of Determinants

Let $A$ and $B$ be an $n \times n$ matrices.

1. $A$ is invertible if and only if $\text{det }A \ne 0$.
2. $\text{det }AB = (\text{det }A)(\text{det }B)$.
3. $\text{det }A^T = \text{det }A$.
4. If $A$ is triangular, then $\text{det }A$ is the product of the entries on the main diagonal of $A$.
5. Row operations and Determinants
   • A row replacement operation on $A$ does not change the determinant.
   • A row interchange changes the sign of the determinant.
   • A row scaling also scales the determinant by the same scalar factor.

The Characteristic Equation

• Theorem 3(1) shows how to determine when a matrix of the form $A-\lambda I$ is not invetible.

  $\text{det }(A-\lambda I) \ne 0$ 로 invertible여부 판별.

• The scalar equation $\text{det } (A-\lambda I)=0$ is called the characteristic equation of $A$.
• A scalar $\lambda$ is an eigenvalue of an matrix $A$ if and only if $\lambda$ satisfies the characteristic equation $$\text{det } (A-\lambda I)=0$$

Example 3

Find the characteristic equation of $$A = \begin{bmatrix} 5 & -2 & 6 & -1 \\ 0 & 3 & -8 & 0 \\ 0 & 0 & 5 & 4 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ .

Solution :

Form $A-\lambda I$, and use Theorem 3(4):

$$\begin{align} \text{det }(A-\lambda I) &= \text{det } \begin{bmatrix} 5-\lambda & -2 & 6 & -1 \\ 0 & 3-\lambda & -8 & 0 \\ 0 & 0 & 5-\lambda & 4 \\ 0 & 0 & 0 & 1-\lambda \end{bmatrix} \\ &= (5-\lambda)(3-\lambda)(5-\lambda)(1-\lambda) \end{align}$$

• The characteristic equation is $$(5-\lambda)^2(3-\lambda)(1-\lambda) = 0$$ or $$(\lambda-5)^2(\lambda-3)(\lambda-1) = 0$$
• Expanding the product, we can also write $$\lambda^4 -14 \lambda^3 + 68 \lambda^2 - 130 \lambda +75 = 0$$

• If $A$ is an $n \times n$ matrix, then $\text{det }(A-\lambda I)$ is a polynomial of degree $n$ called the characteristic polynomial of $A$.
• The eigenvalue 5 in Example 3 is said to have multiplicity 2 because $(\lambda-5)$ occurs two times as a factor of the characteristic polynomial.
• In general, the (algebraic) multiplicity of an eigenvalue $\lambda$ is its multiplicity as a root of the characteristic equation.

Similarity

• If $A$ and $B$ are $n \times n$ matrices, then $A$ is similar to $B$ if there is an invertible matrix $P$ such that $P^{-1}AP=B$, or, equivalently, $A=PBP^{-1}$.
• Writing $Q$ for $P^{-1}$, we have $Q^{-1}BQ=A$.
• So $B$ is also similar to $A$, and we say simply that $A$ and $B$ are similar.
• Changing $A$ into $P^{-1}AP$ is called a similarity transformation.

Theorem 4:

If $n \times n$ matrices $A$ and $B$ are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities).

Proof:

If $B=P^{-1}AP$ then, $$B-\lambda I = P^{-1}AP - \lambda P^{-1}P = P^{-1}(AP-\lambda P)= P^{-1}(A-\lambda I )P$$

• Using the multiplicative property (2) in Theorem (3), we compute $$\begin{align} \text{det}(B-\lambda I) &= \text{det}\left[ P^{-1}(A-\lambda I)P\right] \\ &= \text{det}(P^{-1})\text{det}(A-\lambda I)\text{det}(P) \tag{2} \end{align}$$
• Since $\text{det}(P^{-1})\text{det}(P) = \text{det}(P^{-1}P)= \text{det }I = 1$, we see from equation (2) that $\text{det}(B-\lambda I) = \text{det}(A-\lambda I)$.

Warnings

• The matrices $$\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \text{ and } \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$$ are not similar even though they have the same eigenvalues.
• Similarity is not the same as row equivalence (If $A$ is row equivalent to $B$, then $B=EA$ for some invertible matrix $E$). Row operations on a matrix usually change its eigenvalues.