MATH 240

Mon. November 4th, 2019


PCB=(PBC)1P_{C\leftarrow B}=(P_{B\leftarrow C})^{-1}


Eigenspace

Recall that λ\lambda is an eigenvalue of a n×nn\times n matrix AA if Ax=λxAx=\lambda x for some xx in Rn\R^n.

Note that this means (AλI)x=0(A-\lambda I)x=0 has non-trivial solutions.
The null space of AλIA-\lambda I is precisely the collection of xx for which (AλI)x=0(A-\lambda I)x=0.

Definition. The eigenspace of AA corresponding to the eigenvalue of λ\lambda of AA is the null space of AλIA-\lambda I.


Ex. Find a basis for the eigenspace corresponding to λ=2\lambda =2 for A=(416216218)A= \begin{pmatrix} 4&-1&6\\2&1&6\\2&-1&8 \end{pmatrix}.

A2I=(416216218)(200020002)=(216216216)\begin{aligned} A-2I&=\begin{pmatrix} 4&-1&6\\2&1&6\\2&-1&8 \end{pmatrix}-\begin{pmatrix} 2&0&0\\0&2&0\\0&0&2 \end{pmatrix}\\ &=\begin{pmatrix} 2&-1&6\\2&-1&6\\2&-1&6 \end{pmatrix} \end{aligned}

(216021602160)(216000000000)(1123000000000)(x1x2x3)=x2(1210)+x3(301)\begin{aligned} &\begin{pmatrix} 2&-1&6&0\\2&-1&6&0\\2&-1&6&0 \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 2&-1&6&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 1&-\frac{1}{2}&3&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}\\ \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}&=x_2\begin{pmatrix} \frac{1}{2}\\1\\0 \end{pmatrix}+x_3\begin{pmatrix} -3\\0\\1 \end{pmatrix} \end{aligned}

Thus {(1210),(301)}\{\begin{pmatrix} \frac{1}{2}\\1\\0 \end{pmatrix},\begin{pmatrix} -3\\0\\1 \end{pmatrix}\} is a basis for the required eigenspace.


Calculating Eigenvalues

Observation:
(AλI)x=0(A-\lambda I)x=0 has non-trivial solutions if det(AλI)=0\text{det}(A-\lambda I)=0. This comes from the Invertible Matrix Theorem. (Recall that eigenvectors are nonzero by definition, so if there are only trivial solutions then AA has no eigenvalues.)

Definition: The equations det(AλI)=0\text{det}(A-\lambda I)=0 is called the characteristic equation (for AA).

Theorem: A scalar λ\lambda is an eigenvalue if and only if λ\lambda satisfies the characteristic equation det(AλI)=0\text{det}(A-\lambda I)=0.


Ex. Find the eigenvalues of (357029001)\begin{pmatrix} 3&5&7\\0&2&9\\0&0&1 \end{pmatrix} by finding the zeroes of the characteristic equation.

AλI=(357029001)(λ000λ000λ)=(3λ5702λ9001λ)\begin{aligned} A-\lambda I&=\begin{pmatrix} 3&5&7\\0&2&9\\0&0&1 \end{pmatrix}-\begin{pmatrix} \lambda&0&0\\0&\lambda&0\\0&0&\lambda \end{pmatrix}\\ &=\begin{pmatrix} 3-\lambda&5&7\\0&2-\lambda&9\\0&0&1-\lambda \end{pmatrix} \end{aligned}

AλIA-\lambda I in this case is a triangular matrix, so its determinant is just the product of the entries on its diagonal.

det(AλI)=(3λ)(2λ)(1λ)=0λ=3, 2, 1\begin{aligned} det(A-\lambda I)&=(3-\lambda)(2-\lambda)(1-\lambda)=0\\ \lambda&=3,\ 2,\ 1 \end{aligned}

The eigenvalues of AA are 33, 22, and 11.
From this example we can infer that the eigenvalues of any triangular matrix are the entries on its diagonal.


Characteristic Polynomials

Observation:
The characteristic equation is always a polynomial. This is why it is also called the characteristic polynomial.

If some characteristic polynomial has degree 3, does it have an eigenvalue?
Yes, because eigenvalues occur at the real roots of their characteristic polynomial, and all polynomials of odd degrees have at least one real root by the Intermediate Value Theorem. (Note that polynomials of even degrees may not have any real roots.)

Definition. The (algebraic) multiplicity of an eigenvalue λ\lambda is its multiplicity as a root of the characteristic polynomial.
For instance, the characteristic polynomial (λ2)2(λ3)(\lambda -2)^2(\lambda -3) has two eigenvalues – 2, which has a multiplicity of 2, and 3, which has a multiplicity of 1.


Ex. Find the eigenvalues and their multiplicities from the characteristic polynomial λ64λ512λ4\lambda^6-4\lambda^5-12\lambda^4.

λ64λ512λ4=λ4(λ24λ12)=λ4(λ6)(λ+2)=0\begin{aligned} \lambda^6-4\lambda^5-12\lambda^4&=\lambda^4(\lambda^2-4\lambda-12)\\ &=\lambda^4(\lambda-6)(\lambda+2)\\ &=0 \end{aligned}

The eigenvalues are 0 (with multiplicity 4), 6 (with multiplicity 1), and -2 (with multiplicity 1).