Introductory Linear Algebra/Eigenvalues and eigenvectors

Example. (Formula of power of a diagonal matrix) Let ${\displaystyle D={\begin{pmatrix}3&0\\0&-5\end{pmatrix}}}$. Then, ${\displaystyle D={\begin{pmatrix}3^{n}&0\\0&(-5)^{n}\end{pmatrix}}}$ for each positive integer ${\displaystyle n}$, since $${\displaystyle {\begin{pmatrix}3&0\\0&-5\end{pmatrix}}{\begin{pmatrix}3^{k}&0\\0&(-5)^{k}\end{pmatrix}}={\begin{pmatrix}3^{k+1}&0\\0&(-5)^{k+1}\end{pmatrix}},}$$ and we can prove the formula of power of a diagonal matrix by induction.

Example. Let ${\displaystyle P={\begin{pmatrix}1&1\\2&3\\\end{pmatrix}}}$ and ${\displaystyle D={\begin{pmatrix}3&0\\0&-5\end{pmatrix}}}$. Then, it can be computed that ${\displaystyle P^{-1}={\begin{pmatrix}3&-1\\-2&1\\\end{pmatrix}}}$. Let ${\displaystyle A=PDP^{-1}={\begin{pmatrix}19&-8\\48&-21\\\end{pmatrix}}}$. Then, $${\displaystyle {\begin{aligned}A^{n}&=(PDP^{-1})^{n}=\underbrace {(PD{\color {blue}P^{-1}})({\color {blue}P}DP^{-1})\cdots (PD{\color {brown}P^{-1}})({\color {brown}P}DP^{-1})} _{n\;PDP^{-1}{\text{'s}}}\\&=PD(\underbrace {\color {blue}P^{-1}P} _{\color {blue}I})DP^{-1}\cdots PD(\underbrace {\color {brown}P^{-1}P} _{\color {brown}I})DP^{-1}\\&=PD\underbrace {{\color {blue}I}D\cdots {\color {brown}I}D} _{n-1\;ID{\text{'s}}}P^{-1}\\&=P\underbrace {DD\cdots D} _{n\;D{\text{'s}}}P^{-1}\\&=PD^{n}P^{-1}\\&=P{\begin{pmatrix}3^{n}&0\\0&(-5)^{n}\end{pmatrix}}P^{-1}\qquad {\text{by above example}}\\&={\begin{pmatrix}1&1\\2&3\\\end{pmatrix}}{\begin{pmatrix}3^{n}&0\\0&(-5)^{n}\end{pmatrix}}{\begin{pmatrix}3&-1\\-2&1\\\end{pmatrix}}\\&={\begin{pmatrix}3^{n+1}-2(-5)^{n}&(-5)^{n}-3^{n}\\6(3^{n})-6(-5)^{n}&3(-5)^{n}-2(3^{n})\\\end{pmatrix}}\end{aligned}}}$$

Definition. (Diagonalizable matrix) A square matrix ${\displaystyle A}$ is diagonalizable if there exists an invertible matrix ${\displaystyle P}$ such that ${\displaystyle P^{-1}AP}$ is a diagonal matrix.

Remark. An equivalent condition is ${\displaystyle A=PDP^{-1}}$ for some diagonal matrix ${\displaystyle D}$ and invertible matrix ${\displaystyle P}$, which matches with the form in the motivation section. So, if a matrix is diagonalizable, we can compute its power conveniently.

Example. The matrix ${\displaystyle I_{n}}$ is diagonalizable, since there exists ${\displaystyle P=I_{n}}$ such that ${\displaystyle P^{-1}I_{n}P}$ is a diagonal matrix (which is ${\displaystyle I_{n}}$). Also, there exists ${\displaystyle P=I_{n},D=I_{n}}$ such that ${\displaystyle I_{n}=PDP^{-1}}$.

Exercise.

Select all diagonalizable matrix (matrices).

	A zero matrix.
	${\displaystyle 2I.}$
	A diagonal matrix.

Definition. (Eigenvector and eigenvalue) Let ${\displaystyle A}$ be a square matrix. A nonzero vector ${\displaystyle \mathbf {v} }$ is an eigenvector of ${\displaystyle A}$ if there exists a scalar ${\displaystyle \lambda }$ such that ${\displaystyle A\mathbf {v} =\lambda \mathbf {v} }$. In that case, ${\displaystyle \lambda }$ is an eigenvalue of ${\displaystyle A}$ corresponding to eigenvector ${\displaystyle \mathbf {v} }$.

Remark.

• ${\displaystyle A\mathbf {v} =\lambda \mathbf {v} }$ means that multiplying the vector ${\displaystyle \mathbf {v} }$ by the matrix ${\displaystyle A}$ is equivalent to multiplying it by a scalar (scaling of vector)
• the prefix eigen- has the meanings of 'own', 'proper', and 'characteristic'

Example. (Eigenvectors of identity matrix) Each vector ${\displaystyle \mathbf {v} \in \mathbb {R} ^{n}}$ is an eigenvector of ${\displaystyle I_{n}}$, since $${\displaystyle I_{n}\mathbf {v} =\mathbf {v} =1\cdot \mathbf {v} ,}$$ for each vector ${\displaystyle \mathbf {v} \in \mathbb {R} ^{n}}$, and their corresponding eigenvalues are all ${\displaystyle 1}$.

Exercise.

Select all correct statement(s).

	If ${\displaystyle \mathbf {v} }$ is an eigenvector of invertible matrix ${\displaystyle A}$, then it is also an eigenvector of ${\displaystyle A^{-1}}$.
	If ${\displaystyle \lambda }$ is an eigenvalue of ${\displaystyle A}$, then ${\displaystyle \lambda ^{n}}$ is an eigenvalue of ${\displaystyle A^{n}}$.
	Each vector ${\displaystyle \mathbf {v} \in \mathbb {R} ^{n}}$ is an eigenvctor of the zero matrix ${\displaystyle O_{n\times n}}$.
	Zero vector is an eigenvector of each square matrix.
	If there exists an eigenvector for a matrix, then there are infinitely many eigenvectors of that matrix.

Theorem. (Diagonalization) Let ${\displaystyle A}$ be an ${\displaystyle n\times n}$ matrix. Then, ${\displaystyle A}$ is diagonalizable if and only if ${\displaystyle A}$ has ${\displaystyle n}$ linearly independent eigenvectors. If ${\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}}$ are linearly independent eigenvectors of ${\displaystyle A}$ corresponding to the eigenvalues ${\displaystyle \lambda _{1},\ldots ,\lambda _{n}}$ (some of them are possibly the same), we can define an invertible matrix ${\displaystyle P}$ with columns ${\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}}$ and a diagonal matrix ${\displaystyle D}$ with diagonal entries ${\displaystyle \lambda _{1},\ldots ,\lambda _{n}}$ such that $${\displaystyle A=PDP^{-1}.}$$

Proof. In the following, we use ${\displaystyle {\begin{pmatrix}\mathbf {v} _{1}&\cdots &\mathbf {v} _{n}\end{pmatrix}}}$ to denote the matrix with columns ${\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}}$, in this order. $${\displaystyle {\begin{aligned}&&A&=PDP^{-1}\\&\Leftrightarrow &AP&=PD\underbrace {PP^{-1}} _{I}\\&\Leftrightarrow &A{\begin{pmatrix}\mathbf {v} _{1}&\cdots &\mathbf {v} _{n}\end{pmatrix}}&={\begin{pmatrix}\mathbf {v} _{1}&\cdots &\mathbf {v} _{n}\end{pmatrix}}{\begin{pmatrix}\lambda _{1}&0&\cdots &0\\0&\lambda _{2}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &\lambda _{n}\end{pmatrix}}\\&\Leftrightarrow &{\begin{pmatrix}A\mathbf {v} _{1}&\cdots &A\mathbf {v} _{n}\end{pmatrix}}&={\begin{pmatrix}\lambda _{1}\mathbf {v} _{1}&\cdots &\lambda _{n}\mathbf {v} _{n}\end{pmatrix}}\\&\Leftrightarrow &A\mathbf {v} _{1}&=\lambda _{1}\mathbf {v} _{1},\ldots ,A\mathbf {v} _{n}=\lambda _{n}\mathbf {v} _{n}.\end{aligned}}}$$ We have now proved that ${\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}}$ are eigenvectors. It remains to prove that they are linearly independent, which is true since they are linearly independent if and only if ${\displaystyle P}$ is invertible by the proposition about relationship between invertibility and linear independence. ${\displaystyle \Box }$

Remark.

• we can put the eigenvectors into ${\displaystyle P}$ as columns in arbitrary orders, as long as we put the eigenvalues into the corresponding column of ${\displaystyle D}$ e.g. we may put ${\displaystyle \mathbf {v} _{1}}$ into the 3rd column of ${\displaystyle P}$, but we need to put ${\displaystyle \lambda _{1}}$ into the 3rd column of ${\displaystyle D}$

• it follows that the expression for diagonalization is not unique, and there are actually infinitely many expressions

• we have $${\displaystyle A{\begin{pmatrix}\mathbf {v} _{1}&\cdots &\mathbf {v} _{n}\end{pmatrix}}={\begin{pmatrix}A\mathbf {v} _{1}&\cdots &A\mathbf {v} _{n}\end{pmatrix}},}$$ by definition of matrix multiplication. E.g., $${\displaystyle A{\begin{pmatrix}\mathbf {u} &\mathbf {w} \end{pmatrix}}:={\begin{pmatrix}1&2\\3&4\\\end{pmatrix}}{\begin{pmatrix}u_{1}&w_{1}\\u_{2}&w_{2}\end{pmatrix}}={\begin{pmatrix}u_{1}+2u_{2}&w_{1}+2w_{2}\\3u_{1}+4u_{2}&3w_{1}+4w_{2}\\\end{pmatrix}},A\mathbf {u} ={\begin{pmatrix}u_{1}+2u_{2}\\3u_{1}+4u_{2}\\\end{pmatrix}},A\mathbf {w} ={\begin{pmatrix}w_{1}+2w_{2}\\3w_{1}+4w_{2}\\\end{pmatrix}}}$$

Definition. (Characteristic polynomial) Let ${\displaystyle A}$ be an ${\displaystyle n\times n}$ matrix. The characteristic polynomial of ${\displaystyle A}$ in the variable ${\displaystyle t}$ is the polynomial ${\displaystyle \det(A-tI_{n})}$.

Remark.

• we can use arbitrary letter for the variable.
• equivalently, the characteristic polynomial of ${\displaystyle A}$ is the determinant of ${\displaystyle A}$ with its diagonal entries subtracted by ${\displaystyle t}$

Example. The characteristic polynomial of ${\displaystyle {\begin{pmatrix}1&2&4\\4&5&2\\0&0&9\\\end{pmatrix}}}$ is $${\displaystyle {\begin{vmatrix}1-t&2&4\\4&5-t&2\\0&0&9-t\\\end{vmatrix}}=(9-t){\begin{vmatrix}1-t&2\\4&5-t\\\end{vmatrix}}=(9-t)[(1-t)(5-t)-2(4)]=-x^{3}+15x^{2}-51x-27.}$$

Proposition. (Equivalent condition for eigenvalue) Let ${\displaystyle A}$ be an ${\displaystyle n\times n}$ matrix. Then, ${\displaystyle \lambda }$ is an eigenvalue of ${\displaystyle A}$ if and only if ${\displaystyle \det(A-\lambda I_{n})=0}$, i.e. it is a root of the characteristic polynomial of ${\displaystyle A}$.

Proof. $${\displaystyle {\begin{aligned}&&\lambda {\text{ is an }}&{\text{eigenvalue of }}A\\&\Leftrightarrow &A\mathbf {v} &=\lambda \mathbf {v} \qquad {\text{for some }}\mathbf {v} \neq \mathbf {0} \\&\Leftrightarrow &(A-\lambda I_{n})\mathbf {v} &=\mathbf {0} \qquad {\text{for some }}\mathbf {v} \neq \mathbf {0} \\&\Leftrightarrow &A-\lambda I_{n}&{\text{ is non-invertible}}\qquad {\text{by simplified invertible matrix theorem}}\\&\Leftrightarrow &\det(A-\lambda I_{n})&=0\end{aligned}}}$$

${\displaystyle \Box }$

Exercise.

Select all correct statement(s).

	A square matrix has ${\displaystyle n}$ different eigenvalues if its characteristic polynomial has ${\displaystyle n}$ roots.
	If ${\displaystyle A}$ with size ${\displaystyle 2\times 2}$ has two linearly independent eigenvectors ${\displaystyle (3,1)^{T},(0,2)^{T}}$ corresponding to the eigenvalues ${\displaystyle 4,5}$, then we can define an invertible matrix ${\displaystyle P={\begin{pmatrix}3&1\\0&2\\\end{pmatrix}}}$ and diagonal matrix ${\displaystyle D={\begin{pmatrix}4&0\\0&5\end{pmatrix}}}$ such that ${\displaystyle A=P^{-1}DP}$.
	If ${\displaystyle A}$ with size ${\displaystyle 2\times 2}$ has two linearly independent eigenvectors, then we can define an invertible matrix ${\displaystyle P}$ and a diagonal matrix ${\displaystyle D}$ such that ${\displaystyle A=P^{-1}DP}$.

Definition. (Eigenspace) Let ${\displaystyle A}$ be an ${\displaystyle n\times n}$ matrix. Suppose ${\displaystyle \lambda }$ is an eigenvalue of ${\displaystyle A}$. Then, the null space ${\displaystyle \operatorname {Null} (A-\lambda I_{n})}$, denoted by ${\displaystyle E_{\lambda }}$, is the eigenspace of ${\displaystyle A}$ corresponding to ${\displaystyle \lambda }$.

Remark.

• since the null space is a subspace of ${\displaystyle \mathbb {R} ^{n}}$, the eigensapce is also a subspace of ${\displaystyle \mathbb {R} ^{n}}$
• ${\displaystyle E_{\lambda }}$ consists of the zero vector (since it is subspace) and all eigenvectors corresponding to ${\displaystyle \lambda }$ since

$${\displaystyle A\mathbf {v} =\lambda \mathbf {v} \iff (A-\lambda I_{n})\mathbf {v} =\mathbf {0} \iff \mathbf {v} \in \operatorname {Null} (A-\lambda I_{n})=E_{\lambda }\quad (\mathbf {v} \neq \mathbf {0} {\text{ by definition}})}$$

Remark.

• it can be proved that eigenvectors of ${\displaystyle A}$ corresponding to distinct eigenvalues are linearly independent (its proof is skipped here)

• so the columns of ${\displaystyle P}$ are linearly independent, thus ${\displaystyle P}$ is invertible

• if ${\displaystyle A}$ has ${\displaystyle n}$ different eigenvalues, then ${\displaystyle A}$ is diagonalizable ^[1], since there are ${\displaystyle n}$ bases corresponding to the ${\displaystyle n}$ eigenvalues, which together contain ${\displaystyle n}$ vectors ${\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}}$.
• there are infinitely many possible bases for each eigenvalue, but we only need one of them

Example. (Diagonalization of ${\displaystyle 2\times 2}$ matrix) Recall the example in the motivation section, it is given that the matrix ${\displaystyle A={\begin{pmatrix}19&-8\\48&-21\\\end{pmatrix}}}$ is diagonalizable, and its expression in the form of ${\displaystyle PDP^{-1}}$ is also given. We will use the above procedure to derive the given expression.

First, $${\displaystyle {\begin{vmatrix}19-\lambda &-8\\48&-21-\lambda \end{vmatrix}}=0\iff (19-\lambda )(-21-\lambda )+8(48)=0\implies \lambda ^{2}+2\lambda -15=0\iff \lambda =3{\text{ or }}\lambda =-5.}$$ So, the eigenvalues of the matrix are ${\displaystyle \lambda =3}$ and ${\displaystyle \lambda =-5}$.

For the eigenvalue ${\displaystyle \lambda ={\color {green}3}}$, since ${\displaystyle (A-3I)\mathbf {x} =\mathbf {0} \implies {\begin{pmatrix}16&-8\\48&-24\\\end{pmatrix}}\mathbf {x} =\mathbf {0} }$, and it can be proved that its general solution is ${\displaystyle \mathbf {x} =(a,2a)^{T}}$, so a basis for ${\displaystyle E_{3}}$ is ${\displaystyle \{{\color {green}(1,2)^{T}}\}}$

For the eigenvalue ${\displaystyle \lambda ={\color {blue}-5}}$, since ${\displaystyle (A+5I)\mathbf {x} =\mathbf {0} \implies {\begin{pmatrix}24&-8\\48&-16\\\end{pmatrix}}\mathbf {x} =\mathbf {0} }$, and it can be proved that its general solution is ${\displaystyle \mathbf {x} =(b,3b)^{T}}$, so a basis for ${\displaystyle E_{-5}}$ is ${\displaystyle \{{\color {blue}(1,3)^{T}}\}}$

Then, we let ${\displaystyle P={\begin{pmatrix}{\color {green}1}&{\color {blue}1}\\{\color {green}2}&{\color {blue}3}\\\end{pmatrix}}}$ (since the two bases together contain two vectors), and ${\displaystyle D={\begin{pmatrix}{\color {green}3}&0\\0&{\color {blue}-5}\end{pmatrix}}.}$ Then, we can compute that ${\displaystyle P^{-1}={\begin{pmatrix}3&-1\\-2&1\\\end{pmatrix}}.}$ Therefore, we have $${\displaystyle A=PDP^{-1}={\begin{pmatrix}{\color {green}1}&{\color {blue}1}\\{\color {green}2}&{\color {blue}3}\\\end{pmatrix}}{\begin{pmatrix}{\color {green}3}&0\\0&{\color {blue}-5}\end{pmatrix}}{\begin{pmatrix}3&-1\\-2&1\\\end{pmatrix}},}$$ which is the same as the given form in the example in motivation section. In general, if we have ${\displaystyle A=PDP^{-1}}$, $${\displaystyle A^{n}=PD^{n}P^{-1},}$$ which is illustrated in the example in the motivation section. From the example in motivation section, $${\displaystyle A^{n}={\begin{pmatrix}3^{n+1}-2(-5)^{n}&(-5)^{n}-3^{n}\\6(3^{n})-6(-5)^{n}&3(-5)^{n}-2(3^{n})\\\end{pmatrix}}.}$$

Example. (Diagonalization of ${\displaystyle 3\times 3}$ matrix) Consider the matrix ${\displaystyle A={\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\\\end{pmatrix}}}$ (It is not ${\displaystyle I_{3}}$). We would like to find a formula for ${\displaystyle A^{n}}$. First, $${\displaystyle {\begin{vmatrix}-\lambda &0&1\\0&1-\lambda &0\\1&0&-\lambda \end{vmatrix}}=0\implies \lambda ^{2}(1-\lambda )-(1-\lambda )=0\implies (1-\lambda )(\lambda ^{2}-1)=0\implies \lambda =1{\text{(repeated) or }}\lambda =-1.}$$ So, the eigenvalues of the matrix are ${\displaystyle \lambda =1}$ and ${\displaystyle \lambda =-1}$.

For the eigenvalue ${\displaystyle \lambda ={\color {green}1}}$, since $${\displaystyle (A-I)\mathbf {x} =\mathbf {0} \implies {\begin{pmatrix}-1&0&1\\0&0&0\\1&0&-1\end{pmatrix}}\mathbf {x} =\mathbf {0} \implies \mathbf {x} =(b,a,b)^{T}=b(1,0,1)^{T}+a(0,1,0)^{T},}$$ (There are two independent unknowns, so the dimension of each basis for the eigenspace is ${\displaystyle 2}$, i.e. there should be two vectors in each basis) a basis for ${\displaystyle E_{1}}$ is ${\displaystyle \{{\color {green}(1,0,1)^{T}},{\color {green}(0,1,0)^{T}}\}}$.

For the eigenvalue ${\displaystyle \lambda ={\color {blue}-1}}$, since $${\displaystyle (A+I)\mathbf {x} =\mathbf {0} \implies {\begin{pmatrix}1&0&1\\0&2&0\\1&0&1\end{pmatrix}}\mathbf {x} =\mathbf {0} \implies \mathbf {x} =(c,0,-c)^{T},}$$ a basis for ${\displaystyle E_{-1}}$ is ${\displaystyle \{{\color {blue}(1,0,-1)^{T}}\}}$.

Then, we let ${\displaystyle P={\begin{pmatrix}{\color {green}1}&{\color {green}0}&{\color {blue}1}\\{\color {green}0}&{\color {green}1}&{\color {blue}0}\\{\color {green}1}&{\color {green}0}&{\color {blue}-1}\end{pmatrix}}}$, (since the two bases together contain three vectors) ${\displaystyle D={\begin{pmatrix}{\color {green}1}&0&0\\0&{\color {green}1}&0\\0&0&{\color {blue}-1}\end{pmatrix}}}$ (we have two eigenvectors corresponding to the eigenvalue ${\displaystyle \lambda =1}$, so this eigenvalue is repeated two times). Then, we can compute that ${\displaystyle P^{-1}={\begin{pmatrix}{\frac {1}{2}}&0&{\frac {1}{2}}\\0&1&0\\{\frac {1}{2}}&0&-{\frac {1}{2}}\end{pmatrix}}}$. It follows that $${\displaystyle A=PDP^{-1}={\begin{pmatrix}{\color {green}1}&{\color {green}0}&{\color {blue}1}\\{\color {green}0}&{\color {green}1}&{\color {blue}0}\\{\color {green}1}&{\color {green}0}&{\color {blue}-1}\end{pmatrix}}{\begin{pmatrix}{\color {green}1}&0&0\\0&{\color {green}1}&0\\0&0&{\color {blue}-1}\end{pmatrix}}{\begin{pmatrix}{\frac {1}{2}}&0&{\frac {1}{2}}\\0&1&0\\{\frac {1}{2}}&0&-{\frac {1}{2}}\end{pmatrix}},}$$ and $${\displaystyle A^{n}=PD^{n}P^{-1}={\begin{pmatrix}{\color {green}1}&{\color {green}0}&{\color {blue}1}\\{\color {green}0}&{\color {green}1}&{\color {blue}0}\\{\color {green}1}&{\color {green}0}&{\color {blue}-1}\end{pmatrix}}{\begin{pmatrix}{\color {green}1}^{n}&0&0\\0&{\color {green}1}^{n}&0\\0&0&({\color {blue}-1})^{n}\end{pmatrix}}{\begin{pmatrix}{\frac {1}{2}}&0&{\frac {1}{2}}\\0&1&0\\{\frac {1}{2}}&0&-{\frac {1}{2}}\end{pmatrix}}={\begin{pmatrix}{\frac {1+(-1)^{n}}{2}}&0&{\frac {1-(-1)^{n}}{2}}\\0&1&0\\{\frac {1-(-1)^{n}}{2}}&0&{\frac {1+(-1)^{n}}{2}}\\\end{pmatrix}}={\begin{cases}I_{3}&{\text{if }}n{\text{ is even}}\\A&{\text{if }}n{\text{ is odd}}\end{cases}}.}$$ This is an interesting result.

Example. (Complex eigenvalues) Let ${\displaystyle A={\begin{pmatrix}1&2\\-2&1\\\end{pmatrix}}.}$^[2] $${\displaystyle {\begin{vmatrix}1-\lambda &2\\-2&1-\lambda \end{vmatrix}}=0\implies (1-\lambda )^{2}+4=0\implies (1-\lambda )^{2}=-4\implies 1-\lambda =\pm 2i\implies \lambda =1\mp 2i.}$$ Since the eigenvalues are all complex (and thus there does not exist any corresponding real eigenvector), ${\displaystyle A}$ is not diagonalizable over real matrices. On the other hand, ${\displaystyle A}$ is diagonalizable over complex matrices, but diagonalization over complex matrices is not our main focus in this book, and we have not defined the operation of complex matrices. So, an expression of ${\displaystyle A}$ in the form of ${\displaystyle PDP^{-1}}$ is put in the following for reference: $${\displaystyle A={\begin{pmatrix}i&-i\\1&1\\\end{pmatrix}}{\begin{pmatrix}1-2i&0\\0&1+2i\end{pmatrix}}{\begin{pmatrix}i&-i\\1&1\end{pmatrix}}^{-1}.}$$

Example. (Non-diagonalizable matrix) Consider the matrix ${\displaystyle N={\begin{pmatrix}0&1\\0&0\\\end{pmatrix}}}$ (It is a nilpotent matrix, satisfying ${\displaystyle N^{2}=O}$).

First, since $${\displaystyle {\begin{vmatrix}-\lambda &1\\0&-\lambda \end{vmatrix}}=0\implies \lambda ^{2}=0\implies \lambda =0,}$$ the only eigenvalue is ${\displaystyle \lambda =0}$.

For eigenvalue ${\displaystyle \lambda =0}$, since $${\displaystyle (N+0I)\mathbf {x} =\mathbf {0} \implies N\mathbf {x} =\mathbf {0} \implies {\begin{pmatrix}0&1\\0&0\\\end{pmatrix}}\mathbf {x} =\mathbf {0} \implies \mathbf {x} =(a,0)^{T}}$$ So, a basis for ${\displaystyle E_{0}}$ is ${\displaystyle \{(1,0)^{T}\}}$. Since it only contains one vector, while the size of the matrix is ${\displaystyle 2\times 2}$, ${\displaystyle N}$ is not diagonalizable.

Exercise.

1

Compute ${\displaystyle {\begin{pmatrix}0&2\\3&0\\\end{pmatrix}}^{10}}$.

	${\displaystyle {\begin{pmatrix}1024&0\\0&59049\\\end{pmatrix}}}$.
	${\displaystyle {\begin{pmatrix}7776&0\\0&7776\end{pmatrix}}}$.
	${\displaystyle {\begin{pmatrix}7777&0\\0&7777\end{pmatrix}}}$.
	${\displaystyle {\begin{pmatrix}59049&0\\0&1024\\\end{pmatrix}}}$.

2

Select all correct statement(s).

	Over real matrices, eigenspace can be a zero space, i.e. only consists of zero vector.
	Eigenspace must consist of infinitely many eigenvectors.
	Each basis of the eigenspace consists of linearly independent vectors.
	Suppose through diagonalization, matrix ${\displaystyle A}$ can be expressed as ${\displaystyle A=PDP^{-1}}$ in which ${\displaystyle P}$ is an invertible matrix and ${\displaystyle D}$ is a diagonal matrix, then, ${\displaystyle P^{-1}A^{n}P}$ is a diagonal matrix for each positive integer ${\displaystyle n}$.

Example. (Fibonacci sequence) Consider the Fibonacci sequence ${\displaystyle F_{0},F_{1},\ldots }$, in which ${\displaystyle F_{0}=0}$, ${\displaystyle F_{1}=1}$ and ${\displaystyle F_{n+2}=F_{n+1}+F_{n}}$ for each nonngeative integer ${\displaystyle n}$. For each nonnegative integer ${\displaystyle n}$, this recurrence relation can be described as $${\displaystyle {\begin{pmatrix}F_{n+2}\\F_{n+1}\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\\\end{pmatrix}}{\begin{pmatrix}F_{n+1}\\F_{n}\end{pmatrix}}={\begin{pmatrix}F_{n+1}+F_{n}\\F_{n}\end{pmatrix}}.}$$

Let ${\displaystyle A={\begin{pmatrix}1&1\\1&0\\\end{pmatrix}}}$. Then, $${\displaystyle {\begin{pmatrix}F_{n+1}\\F_{n}\end{pmatrix}}=A{\begin{pmatrix}F_{n}\\F_{n-1}\end{pmatrix}}=A^{2}{\begin{pmatrix}F_{n-1}\\F_{n-2}\\\end{pmatrix}}=\cdots =A^{n}{\begin{pmatrix}F_{1}\\F_{0}\end{pmatrix}}=A^{n}{\begin{pmatrix}1\\0\\\end{pmatrix}}.}$$

To obtain the expression for ${\displaystyle F_{n}}$, it suffices to find the formula for ${\displaystyle A^{n}}$, and we may find it by diagonalization.

Since $${\displaystyle {\begin{vmatrix}1-\lambda &1\\1&-\lambda \end{vmatrix}}=0\implies (1-\lambda )(-\lambda )-1=0\implies \lambda ^{2}-\lambda -1=0\implies \lambda ={\frac {1+{\sqrt {5}}}{2}}{\text{ or }}{\frac {1-{\sqrt {5}}}{2}}.}$$ Let ${\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}}$ be the golden ratio, and ${\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}}$ be the conjugate of the golden ratio.

For eigenvalue ${\displaystyle \lambda ={\color {green}\varphi }}$, since for $${\displaystyle {\begin{pmatrix}1-\varphi &1\\1&-\varphi \end{pmatrix}}\mathbf {x} =\mathbf {0} ,}$$ we can transform the augmented matrix representing this SLE to RREF as follows: $${\displaystyle {\begin{pmatrix}1-\varphi &1&0\\1&-\varphi &0\\\end{pmatrix}}{\overset {\mathbf {r} _{1}\leftrightarrow \mathbf {r} _{2}}{\to }}{\begin{pmatrix}1&-\varphi &0\\1-\varphi &1&0\\\end{pmatrix}}{\overset {-(1-\varphi )\mathbf {r} _{1}+\mathbf {r} _{2}\to \mathbf {r} _{2}}{\to }}{\begin{pmatrix}1&-\varphi &0\\0&1+\varphi -\varphi ^{2}&0\\\end{pmatrix}}={\begin{pmatrix}1&-\varphi &0\\0&0&0\\\end{pmatrix}}.}$$ ^[3] So, the general solution is ${\displaystyle \mathbf {x} =(a\varphi ,a)^{T}}$, and thus a basis for ${\displaystyle E_{\varphi }}$ is ${\displaystyle \{{\color {green}(\varphi ,1)^{T}}\}}$.

For eigenvalue ${\displaystyle \lambda ={\color {blue}\psi }}$, since $${\displaystyle {\begin{pmatrix}1-\psi &1\\1&-\psi \end{pmatrix}}\mathbf {x} =\mathbf {0} ,}$$ and the RREF of the augmented matrix representing this SLE is $${\displaystyle {\begin{pmatrix}1&-\psi &0\\0&0&0\\\end{pmatrix}}}$$ by symmetry^[4]. So, the general solution is ${\displaystyle \mathbf {x} =(b\psi ,b)^{T}}$, and thus a basis for ${\displaystyle E_{\psi }}$ is ${\displaystyle \{{\color {blue}(\psi ,1)^{T}}\}}$.

Then, we let ${\displaystyle P={\begin{pmatrix}{\color {green}\varphi }&{\color {blue}\psi }\\{\color {green}1}&{\color {blue}1}\end{pmatrix}}}$, ${\displaystyle D={\begin{pmatrix}{\color {green}\varphi }&0\\0&{\color {blue}\psi }\end{pmatrix}}}$. We can compute that ${\displaystyle P^{-1}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}1&-\psi \\-1&\varphi \end{pmatrix}}}$ Then, ${\displaystyle A=PDP^{-1}}$, and thus $${\displaystyle A^{n}=PD^{n}P^{-1}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}{\color {green}\varphi }&{\color {blue}\psi }\\{\color {green}1}&{\color {blue}1}\end{pmatrix}}{\begin{pmatrix}{\color {green}\varphi }^{n}&0\\0&{\color {blue}\psi }^{n}\end{pmatrix}}{\begin{pmatrix}1&-\psi \\-1&\varphi \end{pmatrix}}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}\varphi ^{n+1}&\psi ^{n+1}\\\varphi ^{n}&\psi ^{n}\end{pmatrix}}{\begin{pmatrix}1&-\psi \\-1&\varphi \end{pmatrix}}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}\varphi ^{n+1}-\psi ^{n+1}&-\psi \varphi ^{n+1}+\varphi \psi ^{n+1}\\\varphi ^{n}-\psi ^{n}&-\psi \varphi ^{n}+\varphi \psi ^{n}\end{pmatrix}}}$$ Finally, we have $${\displaystyle {\begin{pmatrix}F_{n+1}\\F_{n}\end{pmatrix}}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}\varphi ^{n+1}-\psi ^{n+1}&-\psi \varphi ^{n+1}+\varphi \psi ^{n+1}\\\varphi ^{n}-\psi ^{n}&-\psi \varphi ^{n}+\varphi \psi ^{n}\end{pmatrix}}{\begin{pmatrix}1\\0\\\end{pmatrix}}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}\varphi ^{n+1}-\psi ^{n+1}\\\varphi ^{n}-\psi ^{n}\end{pmatrix}}.}$$ Thus, $${\displaystyle F_{n}={\frac {\varphi ^{n}-\psi ^{n}}{\varphi -\psi }}}$$ in which ${\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}}$ and ${\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}}$.

Exercise. Define a sequence ${\displaystyle a_{0},a_{1},\ldots }$ in which ${\displaystyle a_{0}=1}$ and ${\displaystyle a_{n+1}=2a_{n}}$ for each nonnegative integer ${\displaystyle n}$.

1

Which of the following is (are) correct formula(s) of ${\displaystyle a_{n}}$?

	${\displaystyle a_{n}=2^{n-1}}$.
	${\displaystyle a_{n}=8a_{n-2}}$.
	${\displaystyle a_{n}=32a_{n-5}}$.
	${\displaystyle a_{n}=128a_{n-6}}$.

2

Define another sequence ${\displaystyle b_{0},b_{1},\ldots }$ in which ${\displaystyle b_{0}=2}$ and ${\displaystyle b_{n+1}=b_{n}+a_{n}}$ for each nonnegative integer ${\displaystyle n}$. Which of the following is (are) correct formula(s) of ${\displaystyle b_{n}}$?

	${\displaystyle b_{n}=2^{n}+1}$.
	${\displaystyle b_{n}=3-2^{n}}$.
	${\displaystyle b_{n}=2^{n}}$.
	${\displaystyle b_{n}=2b_{n-1}}$.
	${\displaystyle b_{n}=b_{n-1}+2}$.

Example. (System of ODEs) Consider the system of ODEs $${\displaystyle {\begin{cases}{\frac {dx}{dt}}&=2x-3y\\{\frac {dy}{dt}}&=4x-5y\\\end{cases}}}$$ with initial conditions ${\displaystyle (x,y)=(1,2)}$ when ${\displaystyle t=0}$.

Using the dot notation for differentiation, the system can be rewritten as $${\displaystyle {\begin{pmatrix}{\dot {x}}\\{\dot {y}}\end{pmatrix}}=A{\begin{pmatrix}x\\y\end{pmatrix}}}$$ in which ${\displaystyle A={\begin{pmatrix}2&-3\\4&-5\\\end{pmatrix}}}$. Suppose we can write ${\displaystyle A=PDP^{-1}\iff D=P^{-1}AP}$ in which ${\displaystyle P}$ is an invertible matrix and ${\displaystyle D}$ is a diagonal matrix. Let ${\displaystyle P^{-1}={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}}$ in which ${\displaystyle a,b,c,d}$ are some real numbers. Also, let ${\displaystyle {\begin{pmatrix}u\\v\\\end{pmatrix}}=P^{-1}{\begin{pmatrix}x\\y\\\end{pmatrix}}={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}={\begin{pmatrix}ax+by\\cx+dy\\\end{pmatrix}}}$, which implies ${\displaystyle u=ax+by}$ and ${\displaystyle v=cx+dy}$, and ${\displaystyle {\begin{pmatrix}x\\y\\\end{pmatrix}}=P{\begin{pmatrix}u\\v\end{pmatrix}}}$ . It follows that ${\displaystyle {\dot {u}}=a{\dot {x}}+b{\dot {y}}}$ and ${\displaystyle {\dot {v}}=c{\dot {x}}+d{\dot {y}}}$. Thus, $${\displaystyle {\begin{pmatrix}{\dot {u}}\\{\dot {v}}\end{pmatrix}}={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}{\begin{pmatrix}{\dot {x}}\\{\dot {y}}\end{pmatrix}}=P^{-1}{\begin{pmatrix}{\dot {x}}\\{\dot {y}}\end{pmatrix}}=P^{-1}A{\begin{pmatrix}x\\y\\\end{pmatrix}}=P^{-1}AP{\begin{pmatrix}u\\v\\\end{pmatrix}}=D{\begin{pmatrix}u\\v\end{pmatrix}}.}$$ Let ${\displaystyle D={\begin{pmatrix}d_{1}&0\\0&d_{2}\end{pmatrix}}}$, then the system can be simplified to $${\displaystyle {\begin{cases}{\frac {du}{dt}}&=d_{1}u\\{\frac {dv}{dt}}&=d_{2}v\end{cases}}\implies {\begin{cases}{\frac {1}{u}}\,du&=d_{1}\,dt\\{\frac {1}{v}}\,dv&=d_{2}\,dt\end{cases}}\implies {\begin{cases}\int {\frac {1}{u}}\,du&=\int d_{1}\,dt\\\int {\frac {1}{v}}\,dv&=\int d_{2}\,dt\end{cases}}\implies {\begin{cases}\ln |u|&=d_{1}t+C_{1}\\\ln |v|&=d_{2}t+C_{2}\end{cases}}\implies {\begin{cases}u&=\pm e^{d_{1}t+C_{1}}\\v&=\pm e^{d_{2}t+C_{2}}\end{cases}}\implies {\begin{cases}u&=C_{3}e^{d_{1}t}\\v&=C_{4}e^{d_{2}t}\end{cases}}}$$ in which ${\displaystyle C_{1},C_{2}}$ are arbitrary constants, and ${\displaystyle C_{3}=\pm e^{C_{1}},C_{4}=\pm e^{C_{2}}}$.