Mathematics Guide

M2 Extra (I) ​

Characteristic Equation ​

Let A=(4123)A=\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}.
Suppose AX=λXAX=λX where λλ is a scalar and X=(xy)X=\begin{pmatrix} x \\ y \end{pmatrix} is a non-zero matrix.
Prove that ∣A−λI∣=0\begin{vmatrix} A-λI \end{vmatrix}=0.
This is called the characteristic equation of AA.

Solution

AX=λXAX−λX=O(A−λI)X=O(4−λ123−λ)(xy)=(00)\begin{aligned} AX &= λX \\ AX-λX &= O \\ (A-λI)X &= O \\ \begin{pmatrix} 4-λ & 1 \\ 2 & 3-λ \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{aligned}

This is a system of homogenous equations.
Since XX is non-zero, this system must have non-trivial solutions.
Therefore:
∣4−λ123−λ∣=∣A−λI∣=0\begin{vmatrix} 4-λ & 1 \\ 2 & 3-λ \end{vmatrix}=\begin{vmatrix} A-λI \end{vmatrix}=0

Eigenvalue and Eigenvector ​

Hence, find λλ and XX such that AX=λXAX=λX.
λλ is called an eigenvalue of AA.
XX is called an eigenvector of AA.

Solution

∣A−λI∣=0∣4−λ123−λ∣=0λ2−7λ+10=0λ=2 or 5\begin{aligned} \begin{vmatrix} A-λI \end{vmatrix} &= 0 \\ \begin{vmatrix} 4-λ & 1 \\ 2 & 3-λ \end{vmatrix} &= 0 \\ λ^{2}-7λ+10 &= 0 \\ λ &= 2 \text{ or } 5 \end{aligned}

If λ=2λ=2,

(2121)(xy)=(00)2x+y=0\begin{aligned} \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ 2x+y &= 0 \end{aligned}

We can take X=(1−2)X=\begin{pmatrix} 1 \\ -2 \end{pmatrix}.

We may also take (2−4)\begin{pmatrix} 2 \\ -4 \end{pmatrix} or (3−6)\begin{pmatrix} 3 \\ -6 \end{pmatrix} or ...
Any multiples of XX is also an eigenvector of AA.

If λ=5λ=5,

(−112−2)(xy)=(00)−x+y=0\begin{aligned} \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ -x+y &= 0 \end{aligned}

We can take X=(11)X=\begin{pmatrix} 1 \\ 1 \end{pmatrix}.

Hence, AA has two pairs of eigenvalues and eigenvectors:
λ1=2λ_1=2 and X1=(1−2)X_1=\begin{pmatrix} 1 \\ -2 \end{pmatrix}
λ2=5λ_2=5 and X2=(11)X_2=\begin{pmatrix} 1 \\ 1 \end{pmatrix}

They satisfy:
AX1=λ1X1AX_1=λ_1X_1
AX2=λ2X2AX_2=λ_2X_2

Diagonalization ​

Find some matrix DD and PP such that A=PDP−1A=PDP^{-1} where DD is diagonal.
This procedure is called the diagonalization of AA.

Solution

Let P=(x1x2y1y2)P=\begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix} and D=(d100d2)D=\begin{pmatrix} d_1 & 0 \\ 0 & d_2 \end{pmatrix}.

A=PDP−1AP=PDA(x1x2y1y2)=(x1x2y1y2)(d100d2)=(d1x1d2x2d1y1d2y2)\begin{aligned} A &= PDP^{-1} \\ AP &= PD \\ A\begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix} &= \begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix} \begin{pmatrix} d_1 & 0 \\ 0 & d_2 \end{pmatrix} \\ &= \begin{pmatrix} d_1x_1 & d_2x_2 \\ d_1y_1 & d_2y_2 \end{pmatrix} \end{aligned}

We can split this into two equations:

A(x1y1)=(d1x1d1y1)=d1(x1y1)A\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} d_1x_1 \\ d_1y_1 \end{pmatrix} = d_1\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}

A(x2y2)=(d2x2d2y2)=d2(x2y2)A\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} d_2x_2 \\ d_2y_2 \end{pmatrix} = d_2\begin{pmatrix} x_2 \\ y_2 \end{pmatrix}

In other words:
d1d_1 and d2d_2 are the eigenvalues of AA.
(x1y1)\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} and (x2y2)\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} are the eigenvectors of AA.

Hence:
d1=λ1=2d_1=λ_1=2 and (x1y1)=(1−2)\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}=\begin{pmatrix} 1 \\ -2 \end{pmatrix}
d2=λ2=5d_2=λ_2=5 and (x2y2)=(11)\begin{pmatrix} x_2 \\ y_2 \end{pmatrix}=\begin{pmatrix} 1 \\ 1 \end{pmatrix}

Therefore:
D=(λ100λ2)=(2005)D=\begin{pmatrix} λ_1 & 0 \\ 0 & λ_2 \end{pmatrix}=\begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}
P=(11−21)P=\begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}

Diagonalizing Other Expressions ​

Given that we have diagonalized A=PDP−1A=PDP^{-1}.
Diagonalize AnA^{n}, i.e. express AnA^{n} in the form of P(a diagonal matrix)P−1P(\text{a diagonal matrix})P^{-1}.

Solution

An=(PDP−1)(PDP−1)(PDP−1)...(PDP−1)=PDP−1PDP−1PDP−1...PDP−1=PDnP−1\begin{aligned} A^n &= (PDP^{-1})(PDP^{-1})(PDP^{-1})...(PDP^{-1}) \\ &= PD\cancel{P^{-1}P}D\cancel{P^{-1}P}D\cancel{P^{-1}}...\cancel{P}DP^{-1} \\ &= PD^nP^{-1} \\ \end{aligned}

Since Dn=(λ1n00λ2n)D^n=\begin{pmatrix} λ_1^n & 0 \\ 0 & λ_2^n \end{pmatrix} is also diagonal,
we successfully diagonalize AnA^n.

Diagonalize A−1A^{-1}.

Solution

A−1=(PDP−1)−1=(P−1)−1D−1P−1=PD−1P−1\begin{aligned} A^{-1} &= (PDP^{-1})^{-1} \\ &= (P^{-1})^{-1}D^{-1}P^{-1} \\ &= PD^{-1}P^{-1} \end{aligned}

where D−1=(λ1−100λ2−1)D^{-1}=\begin{pmatrix} λ_1^{-1} & 0 \\ 0 & λ_2^{-1} \end{pmatrix} is diagonal.

Remember (ABC)−1=C−1B−1A−1(ABC)^{-1}=C^{-1}B^{-1}A^{-1}.

Diagonalize (A−1)n(A^{-1})^n.

Solution

(A−1)n=(PD−1P−1)n=P(D−1)nP−1\begin{aligned} (A^{-1})^n &= (PD^{-1}P^{-1})^n \\ &= P(D^{-1})^nP^{-1} \end{aligned}

where (D−1)n=(λ1−n00λ2−n)(D^{-1})^n=\begin{pmatrix} λ_1^{-n} & 0 \\ 0 & λ_2^{-n} \end{pmatrix} is diagonal.

Diagonalize A+2IA+2I.

Solution

A+2I=PDP−1+P(2I)P−1=P(D+2I)P−1\begin{aligned} A+2I &= PDP^{-1}+P(2I)P^{-1} \\ &= P(D+2I)P^{-1} \end{aligned}

where D+2I=(λ1+200λ2+2)D+2I=\begin{pmatrix} λ_1+2 & 0 \\ 0 & λ_2+2 \end{pmatrix} is diagonal.

Diagonalize (A+2I)−1(3A5+I)(A+2I)^{-1}(3A^{5}+I).

Solution

(A+2I)−1(3A5+I)=(P(D+2I)P−1)−1(3PD5P−1+PIP−1)=(P(D+2I)−1P−1)(P(3D5+I)P−1)=P(D+2I)−1(3D5+I)P−1\begin{aligned} &(A+2I)^{-1}(3A^{5}+I) \\ &= (P(D+2I)P^{-1})^{-1}(3PD^{5}P^{-1}+PIP^{-1}) \\ &= \left(P(D+2I)^{-1}\cancel{P^{-1}}\right)\left(\cancel{P}(3D^{5}+I)P^{-1}\right) \\ &= P(D+2I)^{-1}(3D^{5}+I)P^{-1} \end{aligned}

where (D+2I)−1(3D5+I)=((λ1+2)−1(3λ15+1)00(λ2+2)−1(3λ25+1))(D+2I)^{-1}(3D^{5}+I)=\begin{pmatrix} (λ_1+2)^{-1}(3λ_1^{5}+1) & 0 \\ 0 & (λ_2+2)^{-1}(3λ_2^{5}+1) \end{pmatrix} is diagonal.

Since (D+2I)−1(3D5+I)(D+2I)^{-1}(3D^{5}+I) is calculated from diagonal matrices DD and II, the result must also be diagonal.

Observe the same function form:
(A+2I)−1(3A5+I)(A+2I)^{-1}(3A^{5}+I)
(D+2I)−1(3D5+I)(D+2I)^{-1}(3D^{5}+I)
(λ+2)−1(3λ5+1)(λ+2)^{-1}(3λ^{5}+1)

Function of Matrix ​

This leads to a powerful theorem about any function f(A)f(A) of a matrix AA:

Theorem

If A=PDP−1A=PDP^{-1} where DD is diagonal, then:

f(A)=Pf(D)P−1=P(f(λ1)00f(λ2))P−1f(A) = Pf(D)P^{-1} = P\begin{pmatrix} f(λ_1) & 0 \\ 0 & f(λ_2) \end{pmatrix}P^{-1}

We can thus diagonalize and evaluate any expression f(A)f(A).

Example

Evaluate (A+A−1)10(A+A^{-1})^{10}.

Solution

(A+A−1)10=P(D+D−1)10P−1=P((λ1+λ1−1)1000(λ2+λ2−1)10)P−1=... do the multiplication\begin{aligned} & (A+A^{-1})^{10} \\ &= P(D+D^{-1})^{10}P^{-1} \\ &= P\begin{pmatrix} (λ_1+λ_1^{-1})^{10} & 0 \\ 0 & (λ_2+λ_2^{-1})^{10} \end{pmatrix}P^{-1} \\ &= ... \text{ do the multiplication} \end{aligned}

General Formula for High Power ​

From the above results, we have the formula:
An=P(λ1n00λ2n)P−1A^n = P\begin{pmatrix} λ_1^n & 0 \\ 0 & λ_2^n \end{pmatrix}P^{-1} ...(1)

and:
(A−1)n=P(λ1−n00λ2−n)P−1(A^{-1})^n = P\begin{pmatrix} λ_1^{-n} & 0 \\ 0 & λ_2^{-n} \end{pmatrix}P^{-1} ...(2)

Note that we can get (2) simply by replacing nn with −n-n in (1).

Theorem

Any formula for AnA^n will also work for (A−1)n(A^{-1})^n simply by replacing nn with −n-n.

Similar Matrices ​

Let B=QAQ−1B=QAQ^{-1}. We say that such a matrix BB is similar to AA.
Diagonalize BB.

Solution

B=Q(PDP−1)Q−1=(QP)D(QP)−1\begin{aligned} B &= Q(PDP^{-1})Q^{-1} \\ &= (QP)D(QP)^{-1} \end{aligned}

where DD is diagonal.

Sequence ​

PQ ​