Linear Algebra

Mathematics

Author

Rafiq Islam

Published

January 24, 2021

Some Linear Algebra Proofs

1. Let $n$ be a positive integer. Show that every matrix $A \in M_{n \times n}(\mathbb{R})$ can be written as the sum of two non-singular matrices.

Proof: To prove this, we will use two known properties of matrices.

$det(A)=$ Product of the eigenvalues of $A$
If $\lambda$ is an eigenvalue of $A$ then $\lambda+n$ is an eigenvalue of $A+nI$ matrix.

Since, $A\in M_{n\times n}(\mathbb{R}),$ let $\lambda_i$ for $1\le i\le n$ be the eigenvalues of $A$. The matrix $A+(n+1)I$ has eigenvalues $\lambda_{i+n+1}$ for $1\le i\le n$.

Let,
\[\begin{align*} n&=max\{|\lambda_i| : 1\le i \le n\}\\ \implies& -n\le \lambda_i \le n \text{ for all }1\le i \le n\\ \implies& -n+n+1\le \lambda_i+n+1 \le n+n+1\text{ for all }1\le i \le n\\ \implies& 1\le \lambda_i+n+1 \le 2n+1\text{ for all }1\le i \le n \end{align*}\] Thus, $\lambda_i+n+1\ge 1$ that is $\lambda_i+n+1 \ne 0$ and $0$ is not an eigenvalue of $A$.

Now, from property (1), we have,
$det(A)=\prod_{i=1}^{n}\lambda_i$
and
\[\begin{align*} det(A+(n+1)I)&=\prod_{i=1}^{n}(\lambda_i+n+1)\ne 0\\ \end{align*}\] $\implies A\text{ or }A+(n+1)I$ both are non-singular.

$-(n+1)I$ is of course non-singular.

Then
$A=(A+(n+1)I)+(-(n+1)I)$

2. Let $\alpha \in \mathcal{L}(V)$ and $\dim V=n< \infty$

Suppose that $\alpha$ has two distinct eigenvalues $\lambda$ and $\mu$. Prove that if $\dim E_{\lambda}=n-1$ then $\alpha$ is diagonalizable.

Proof: Since $\mu$ and $\lambda$ are two distinct eigenvalues associated with $\alpha$, so $V=E_{\lambda}\bigoplus E_{\mu}$ and $\dim E_{\lambda}(\alpha)+\dim E_{\mu}(\alpha)=n$.

Here, $\dim E_{\mu}(\alpha)\ge 1$ and $\dim E_{\lambda}(\alpha)=n-1$. So,

$\dim E_{\lambda}(\alpha)+\dim E_{\mu}(\alpha)=n-1+1=n$

3. Let $\alpha\in \mathcal{L}(V)$ and $0\ne v\in V$ where $\dim V=n< \infty$.

Prove that there is a unique monic polynomial $p(t)$ of the smallest degree such that $p(\alpha)(v)=0$

Proof: Since $V$ is finite-dimensional so there exists smallest $k$ such that $\{v,\alpha(v),\cdots,\alpha^{k-1}(v)\}$ is linearly independent but $\{v,\alpha(v),\cdots,\alpha^{k-1}(v),\alpha^k(v)\}$ is linearly dependent. So there exists $c_0,c_1,\cdots,c_k\in \mathbb{F}$ not all zero such that

$c_0v+c_1\alpha(v)+c_2\alpha^2(v)+\cdots+c_{k-1}\alpha^{k-1}(v)+c_k\alpha^k(v)=0$

Without loss of generality, let’s assume that $c_k\ne 0$. Then

$a_0v+a_1\alpha(v)+a_2\alpha^2(v)+\cdots+a_{k-1}\alpha^{k-1}(v)+\alpha^k(v)=0$

where, $a_i=\frac{c_i}{c_k}$ for $1\le i \le k$.

Thus, $p(t)=a_0+a_1t+a_2t^2+\cdots+a_{k-1}t^{k-1}+t^k$, a unique monic polynomial such that $p(\alpha)(v)=0$

(ii) Prove that $p(t)$ from (i) divides the minimal polynomial of $\alpha$

Proof: By polynomial division we have,

$m(t)=p(t)h(t)+r(t)$ where, $m(t)$ is the minimal polynomial.

Then, $m(\alpha)(v)=p(\alpha)h(\alpha)(v)+r(\alpha)(v)$

$\implies 0=0+r(\alpha)(v)$

$\implies r(\alpha)(v)=0$

4. Let $A,B\in M_{n\times n}(\mathbb{F})$ such that there exists an invertible matrix $S\in M_{n\times n}(\mathbb{F})$ such that $SAS^{-1}$ and $SBS^{-1}$ are upper triangular matrices. Show that every eigenvalue of $AB-BA$ is zero

Proof: To prove the above statement, it is enough to show that $spec(AB-BA)=\{0\}$

We know that if $C$ and $D$ are upper triangular matrices then $spec(CD-DC)=\{0\}$. Now let’s assume that $C=SAS^{-1}$ and $D=SBS^{-1}$. Then,

$CD-DC=SAS^{-1}SBS^{-1}-SBS^{-1}SAS^{-1}$

$\implies CD-DC=SABS^{-1}-SBAS^{-1}$

$\implies CD-DC=S(AB-BA)S^{-1}$

Hence $CD-DC$ and $AB-BA$ are similar matrices. So they have the same eigenvalues, that is $spec(AB-BA)=\{0\}$.

5. Caley-Hamilton Theorem

Theorem: Let $p(t)$ be the characteristic polynomial of a matrix $A$. Then $p(A)=0$

Before we start proving the theorem, we need to discuss some basics.

For Linear Operator: If $\alpha\in \mathcal{L}(V)$ and $A=\Phi_{BB}(\alpha)$ is a representation matrix of $\alpha$ with respect to the basis $B$, then $p(A)$ is the representation matrix of $p(\alpha)$. Thus we also have $p(\alpha)=0$ if $p$ is the characteristic polynomial of $\alpha$

Adjoint Matrix Method: If we have a matrix $A=[a_{ij}]\in M_{n\times n}(\mathbb{F})$ then we define the $\textit{adjoint}$ of $A$ to be the matrix $adj(A)=[b_{ij}]\in M_{n\times n}(\mathbb{F})$, where $b_{ij}=(-1)^{i+j}|A_{ji}|$ for all $1 \le i,j \le n$

And, $A_{ji}$ is the matrix obtained by deleting the i-th row and j-th column.

Example: Let $A=\left(\begin{array}{ccc}1 &4 &7\\2 &5 &8\\3 &6 &9\end{array}\right)$

$b_{11}=(-1)^{1+1}|A_{11}|=det\left(\begin{array}{cc}5 &8\\6 &9\end{array}\right)=-3$

$b_{12}=(-1)^{1+2}|A_{21}|=-det\left(\begin{array}{cc}4 &7\\6 &9\end{array}\right)=6$

$b_{13}=(-1)^{1+3}|A_{21}|=det\left(\begin{array}{cc}4 &7\\5 &8\end{array}\right)=-3$

$b_{21}=(-1)^{2+1}|A_{12}|=-det\left(\begin{array}{cc}2 &8\\3 &9\end{array}\right)=6$

$b_{22}=(-1)^{2+2}|A_{22}|=det\left(\begin{array}{cc}1 &7\\3 &9\end{array}\right)=-12$

$b_{23}=(-1)^{2+3}|A_{32}|=-det\left(\begin{array}{cc}1 &7\\2 &8\end{array}\right)=6$

$b_{31}=(-1)^{3+1}|A_{13}|=det\left(\begin{array}{cc}2 &5\\3 &6\end{array}\right)=-3$

$b_{32}=(-1)^{3+2}|A_{23}|=-det\left(\begin{array}{cc}1 &4\\3 &6\end{array}\right)=6$

$b_{33}=(-1)^{3+3}|A_{33}|=-det\left(\begin{array}{cc}1 &4\\2 &5\end{array}\right)=-3$

Thus,

$adj(A)=\left(\begin{array}{ccc}-3 &6 &-3\\6 &-12 &6\\-3 &6 &-3\end{array}\right)$.

The important formula that we are going to use is that,

$AA^{-1}=I \implies A\frac{adj(A)}{det(A)}=I \implies A.adj(A)=det(A).I$ (*)

Proof: Let $A\in M_{n\times n}(\mathbb{F})$ have the minimal polynomial $p(t)=t^n+\sum_{i=0}^{n-1}a_it^i$.

Now let, $adj(tI-A)=[g_{ij}(t)]=\left(\begin{array}{cccc}g_{11}(t) &g_{12}(t) &\cdots &g_{1n}(t)\\g_{21}(t) &g_{22}(t) &\cdots &g_{2n}(t)\\\vdots &\vdots &\ddots &\vdots\\g_{n1}(t) &g_{n2}(t) &\cdots &g_{nn}(t)\end{array}\right)$ be the adjoint matrix of $(tI-A)$.

Since each $g_{ij}(t)$ is a polynomial of degree at most $n-1$, we can write this as, $adj(tI-A)=\sum_{i=1}^{n}B_it^{n-i}$ where $B_i\in M_{n\times n}(\mathbb{F})$.

Then by (*) we have,

\[\begin{align*} p(t)I&=det(tI-A).I=(tI-A)adj(tI-A)=(tI-A)\sum_{i=1}^{n}B_it^{n-i}\\ \implies& (a_0+a_1t+a_2t^2+\cdots+a_{n-1}t^{n-1}+t^n)I=(tI-A)B_1t^{n-1}+\cdots+(tI-A)B_{n-1}t+(tI-A)B_n\\ \implies& (a_0+a_1t+a_2t^2+\cdots+a_{n-1}t^{n-1}+t^n)I=B_1t^n-AB_1t^{n-1}+B_2t^{n-1}-AB_2t^{n-2}+\cdots+B_{n-1}t^2-AB_n \end{align*}\]

By comparing the coefficients, we get

$B_1=I$

$B_2-AB_1=a_{n-1}I$

$B_3-AB_2=a_{n-2}I$

$\vdots$

$B_n-AB_{n-1}=a_1I$

$-AB_n=a_0I$

Now multiply $A^{n+1-j}$ to the $j-th$ equation, and then sum up both sides we get,

$A^{n+1-1}B_1=IA^{n+1-1}\hspace{2.3in} \implies A^nB_1=A^n$

$A^{n+1-2}(B_2-AB_1)=a_{n-1}A^{n+1-2}\hspace{1in} \implies A^{n-1}B_2-AB_1=a_{n-1}A^{n-1}$

$A^{n+1-3}(B_3-AB_2)=a_{n-2}A^{n+1-3}\hspace{1in} \implies A^{n-2}B_3-A^{n-1}B_2=a_{n-1}A^{n-2}$

$\vdots\hspace{5in} \vdots$

$A^{n+1-n}(B_n-AB_{n-1})=a_1A^{n+1-n}\hspace{1in} \implies AB_n-A^2B_{n-1}=a_1A$

$-AB_n=a_0I\hspace{3.2in}\implies -AB_n=a_0I$

If we add both sides then we obtain, $p(A)=0$

6. Prove that the spectral radius of the $\textit{Markov}$ matrix is less than or equal to 1

We need to prove that if $A$ is a Markov matrix then $\rho(A)\le 1$. Now, what is a Markov matrix?

Markov Matrix: A matrix $A=[a_{i,j}]_{n\times n}$ is called a Markov matrix if $a_{i,j}\ge 0$ for all $1\le i,j \le n$ and $\sum_{j=1}^{n} a_i=1$, that is the sum of the elements in any row is equal to 1.

Example: If we have a matrix like this, $A=\left(\begin{array}{ccc}0.2 &0.4 &0.4\\0.1 &0.4 &0.5\\0.9 &0.1 &0\end{array}\right)$ then $A$ is a Markov matrix.

Proof: Let $\lambda \in spec(A)$ and $x=\left(\begin{array}{c}x_1\\x_2\\\vdots\\x_n\end{array}\right)\ne 0$ be a column vector. Then we define, $x_h=max\{|x_i|: 1\le i \le n\}$ & $>0$. Here we are assuming $x_h >0$ because $x\ne 0$, as a result at least one of the coordinate of $x$ must be greater than $0$.

Now,

$Ax=\lambda x=\left(\begin{array}{c}\lambda x_1\\ \lambda x_2 \\ \vdots \\ \lambda x_n\end{array}\right)$

$\implies \lambda x_h =\sum_{j=1}^{n} a_{hj}x_j$

$\implies |\lambda x_h|=|\lambda |.|x_h|=|\sum_{j=1}^{n} a_{hj}x_j|\le \sum_{j=1}^{n} |a_{hj}| |x_j|$

$\implies |\lambda |.|x_h| \le (\sum_{j=1}^{n} |a_{hj}|) |x_h|=1. |x_h|$

$\implies |\lambda| \le 1$

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BibTeX citation:

@online{islam2021,
  author = {Islam, Rafiq},
  date = {2021-01-24},
  url = {https://mrislambd.github.io/posts/someproofs/},
  langid = {en}
}

For attribution, please cite this work as:

Islam, Rafiq. 2021. January 24, 2021. https://mrislambd.github.io/posts/someproofs/.

--- date: "2021-01-24" author: Rafiq Islam categories: [Linear Algebra, Mathematics] citation: true search: true lightbox: true image: lin1.jpg listing: contents: "/../../posts" max-items: 3 type: grid categories: false date-format: full fields: [image, date, title, author, reading-time] --- # Some Linear Algebra Proofs ## 1. Let $n$ be a positive integer. Show that every matrix $A \in M_{n \times n}(\mathbb{R})$ can be written as the sum of two non-singular matrices. **Proof**: To prove this, we will use two known properties of matrices. 1. $det(A)=$ Product of the eigenvalues of $A$ 2. If $\lambda$ is an eigenvalue of $A$ then $\lambda+n$ is an eigenvalue of $A+nI$ matrix. Since, $A\in M_{n\times n}(\mathbb{R}),$ let $\lambda_i$ for $1\le i\le n$ be the eigenvalues of $A$. The matrix $A+(n+1)I$ has eigenvalues $\lambda_{i+n+1}$ for $1\le i\le n$. Let, \begin{align*} n&=max\{|\lambda_i| : 1\le i \le n\}\\ \implies& -n\le \lambda_i \le n \text{ for all }1\le i \le n\\ \implies& -n+n+1\le \lambda_i+n+1 \le n+n+1\text{ for all }1\le i \le n\\ \implies& 1\le \lambda_i+n+1 \le 2n+1\text{ for all }1\le i \le n \end{align*} Thus, $\lambda_i+n+1\ge 1$ that is $\lambda_i+n+1 \ne 0$ and $0$ is not an eigenvalue of $A$. Now, from property (1), we have, $det(A)=\prod_{i=1}^{n}\lambda_i$ and \begin{align*} det(A+(n+1)I)&=\prod_{i=1}^{n}(\lambda_i+n+1)\ne 0\\ \end{align*} $\implies A\text{ or }A+(n+1)I$ both are non-singular. $-(n+1)I$ is of course non-singular. Then $A=(A+(n+1)I)+(-(n+1)I)$ ## 2. Let $\alpha \in \mathcal{L}(V)$ and $\dim V=n< \infty$ Suppose that $\alpha$ has two distinct eigenvalues $\lambda$ and $\mu$. Prove that if $\dim E_{\lambda}=n-1$ then $\alpha$ is diagonalizable.Proof: Since $\mu$ and $\lambda$ are two distinct eigenvalues associated with $\alpha$, so $V=E_{\lambda}\bigoplus E_{\mu}$ and $\dim E_{\lambda}(\alpha)+\dim E_{\mu}(\alpha)=n$.Here, $\dim E_{\mu}(\alpha)\ge 1$ and $\dim E_{\lambda}(\alpha)=n-1$. So,$\dim E_{\lambda}(\alpha)+\dim E_{\mu}(\alpha)=n-1+1=n$ ## 3. Let $\alpha\in \mathcal{L}(V)$ and $0\ne v\in V$ where $\dim V=n< \infty$. (i) Prove that there is a unique monic polynomial $p(t)$ of the smallest degree such that $p(\alpha)(v)=0$Proof: Since $V$ is finite-dimensional so there exists smallest $k$ such that $\{v,\alpha(v),\cdots,\alpha^{k-1}(v)\}$ is linearly independent but $\{v,\alpha(v),\cdots,\alpha^{k-1}(v),\alpha^k(v)\}$ is linearly dependent. So there exists $c_0,c_1,\cdots,c_k\in \mathbb{F}$ not all zero such that$c_0v+c_1\alpha(v)+c_2\alpha^2(v)+\cdots+c_{k-1}\alpha^{k-1}(v)+c_k\alpha^k(v)=0$Without loss of generality, let's assume that $c_k\ne 0$. Then$a_0v+a_1\alpha(v)+a_2\alpha^2(v)+\cdots+a_{k-1}\alpha^{k-1}(v)+\alpha^k(v)=0$where, $a_i=\frac{c_i}{c_k}$ for $1\le i \le k$. Thus, $p(t)=a_0+a_1t+a_2t^2+\cdots+a_{k-1}t^{k-1}+t^k$, a unique monic polynomial such that $p(\alpha)(v)=0$ (ii) Prove that $p(t)$ from (i) divides the minimal polynomial of $\alpha$Proof: By polynomial division we have,$m(t)=p(t)h(t)+r(t)$ where, $m(t)$ is the minimal polynomial.Then, $m(\alpha)(v)=p(\alpha)h(\alpha)(v)+r(\alpha)(v)$$\implies 0=0+r(\alpha)(v)$$\implies r(\alpha)(v)=0$ ## 4. Let $A,B\in M_{n\times n}(\mathbb{F})$ such that there exists an invertible matrix $S\in M_{n\times n}(\mathbb{F})$ such that $SAS^{-1}$ and $SBS^{-1}$ are upper triangular matrices. Show that every eigenvalue of $AB-BA$ is zero Proof: To prove the above statement, it is enough to show that $spec(AB-BA)=\{0\}$We know that if $C$ and $D$ are upper triangular matrices then $spec(CD-DC)=\{0\}$. Now let's assume that $C=SAS^{-1}$ and $D=SBS^{-1}$. Then,$CD-DC=SAS^{-1}SBS^{-1}-SBS^{-1}SAS^{-1}$$\implies CD-DC=SABS^{-1}-SBAS^{-1}$$\implies CD-DC=S(AB-BA)S^{-1}$Hence $CD-DC$ and $AB-BA$ are similar matrices. So they have the same eigenvalues, that is $spec(AB-BA)=\{0\}$. ## 5. Caley-Hamilton Theorem Theorem: Let $p(t)$ be the characteristic polynomial of a matrix $A$. Then $p(A)=0$Before we start proving the theorem, we need to discuss some basics.For Linear Operator: If $\alpha\in \mathcal{L}(V)$ and $A=\Phi_{BB}(\alpha)$ is a representation matrix of $\alpha$ with respect to the basis $B$, then $p(A)$ is the representation matrix of $p(\alpha)$. Thus we also have $p(\alpha)=0$ if $p$ is the characteristic polynomial of $\alpha$Adjoint Matrix Method: If we have a matrix $A=[a_{ij}]\in M_{n\times n}(\mathbb{F})$ then we define the $\textit{adjoint}$ of $A$ to be the matrix $adj(A)=[b_{ij}]\in M_{n\times n}(\mathbb{F})$, where $b_{ij}=(-1)^{i+j}|A_{ji}|$ for all $1 \le i,j \le n$And, $A_{ji}$ is the matrix obtained by deleting the i-th row and j-th column.Example: Let $A=\left(\begin{array}{ccc}1 &4 &7\\2 &5 &8\\3 &6 &9\end{array}\right)$$b_{11}=(-1)^{1+1}|A_{11}|=det\left(\begin{array}{cc}5 &8\\6 &9\end{array}\right)=-3$$b_{12}=(-1)^{1+2}|A_{21}|=-det\left(\begin{array}{cc}4 &7\\6 &9\end{array}\right)=6$$b_{13}=(-1)^{1+3}|A_{21}|=det\left(\begin{array}{cc}4 &7\\5 &8\end{array}\right)=-3$$b_{21}=(-1)^{2+1}|A_{12}|=-det\left(\begin{array}{cc}2 &8\\3 &9\end{array}\right)=6$$b_{22}=(-1)^{2+2}|A_{22}|=det\left(\begin{array}{cc}1 &7\\3 &9\end{array}\right)=-12$$b_{23}=(-1)^{2+3}|A_{32}|=-det\left(\begin{array}{cc}1 &7\\2 &8\end{array}\right)=6$$b_{31}=(-1)^{3+1}|A_{13}|=det\left(\begin{array}{cc}2 &5\\3 &6\end{array}\right)=-3$$b_{32}=(-1)^{3+2}|A_{23}|=-det\left(\begin{array}{cc}1 &4\\3 &6\end{array}\right)=6$$b_{33}=(-1)^{3+3}|A_{33}|=-det\left(\begin{array}{cc}1 &4\\2 &5\end{array}\right)=-3$Thus, $adj(A)=\left(\begin{array}{ccc}-3 &6 &-3\\6 &-12 &6\\-3 &6 &-3\end{array}\right)$.The important formula that we are going to use is that, $AA^{-1}=I \implies A\frac{adj(A)}{det(A)}=I \implies A.adj(A)=det(A).I$ (*) Proof: Let $A\in M_{n\times n}(\mathbb{F})$ have the minimal polynomial $p(t)=t^n+\sum_{i=0}^{n-1}a_it^i$. Now let, $adj(tI-A)=[g_{ij}(t)]=\left(\begin{array}{cccc}g_{11}(t) &g_{12}(t) &\cdots &g_{1n}(t)\\g_{21}(t) &g_{22}(t) &\cdots &g_{2n}(t)\\\vdots &\vdots &\ddots &\vdots\\g_{n1}(t) &g_{n2}(t) &\cdots &g_{nn}(t)\end{array}\right)$ be the adjoint matrix of $(tI-A)$. Since each $g_{ij}(t)$ is a polynomial of degree at most $n-1$, we can write this as, $adj(tI-A)=\sum_{i=1}^{n}B_it^{n-i}$ where $B_i\in M_{n\times n}(\mathbb{F})$. Then by (*) we have, \begin{align*} p(t)I&=det(tI-A).I=(tI-A)adj(tI-A)=(tI-A)\sum_{i=1}^{n}B_it^{n-i}\\ \implies& (a_0+a_1t+a_2t^2+\cdots+a_{n-1}t^{n-1}+t^n)I=(tI-A)B_1t^{n-1}+\cdots+(tI-A)B_{n-1}t+(tI-A)B_n\\ \implies& (a_0+a_1t+a_2t^2+\cdots+a_{n-1}t^{n-1}+t^n)I=B_1t^n-AB_1t^{n-1}+B_2t^{n-1}-AB_2t^{n-2}+\cdots+B_{n-1}t^2-AB_n \end{align*} By comparing the coefficients, we get$B_1=I$$B_2-AB_1=a_{n-1}I$$B_3-AB_2=a_{n-2}I$$\vdots$$B_n-AB_{n-1}=a_1I$$-AB_n=a_0I$Now multiply $A^{n+1-j}$ to the $j-th$ equation, and then sum up both sides we get,$A^{n+1-1}B_1=IA^{n+1-1}\hspace{2.3in} \implies A^nB_1=A^n$$A^{n+1-2}(B_2-AB_1)=a_{n-1}A^{n+1-2}\hspace{1in} \implies A^{n-1}B_2-AB_1=a_{n-1}A^{n-1}$$A^{n+1-3}(B_3-AB_2)=a_{n-2}A^{n+1-3}\hspace{1in} \implies A^{n-2}B_3-A^{n-1}B_2=a_{n-1}A^{n-2}$$\vdots\hspace{5in} \vdots$$A^{n+1-n}(B_n-AB_{n-1})=a_1A^{n+1-n}\hspace{1in} \implies AB_n-A^2B_{n-1}=a_1A$$-AB_n=a_0I\hspace{3.2in}\implies -AB_n=a_0I$If we add both sides then we obtain, $p(A)=0$ ## 6. Prove that the spectral radius of the $\textit{Markov}$ matrix is less than or equal to 1 We need to prove that if $A$ is a Markov matrix then $\rho(A)\le 1$. Now, what is a Markov matrix?Markov Matrix: A matrix $A=[a_{i,j}]_{n\times n}$ is called a Markov matrix if $a_{i,j}\ge 0$ for all $1\le i,j \le n$ and $\sum_{j=1}^{n} a_i=1$, that is the sum of the elements in any row is equal to 1.Example: If we have a matrix like this, $A=\left(\begin{array}{ccc}0.2 &0.4 &0.4\\0.1 &0.4 &0.5\\0.9 &0.1 &0\end{array}\right)$ then $A$ is a Markov matrix. Proof: Let $\lambda \in spec(A)$ and $x=\left(\begin{array}{c}x_1\\x_2\\\vdots\\x_n\end{array}\right)\ne 0$ be a column vector. Then we define, $x_h=max\{|x_i|: 1\le i \le n\}$ \& $>0$. Here we are assuming $x_h >0$ because $x\ne 0$, as a result at least one of the coordinate of $x$ must be greater than $0$.Now,$Ax=\lambda x=\left(\begin{array}{c}\lambda x_1\\ \lambda x_2 \\ \vdots \\ \lambda x_n\end{array}\right)$$\implies \lambda x_h =\sum_{j=1}^{n} a_{hj}x_j$$\implies |\lambda x_h|=|\lambda |.|x_h|=|\sum_{j=1}^{n} a_{hj}x_j|\le \sum_{j=1}^{n} |a_{hj}| |x_j|$$\implies |\lambda |.|x_h| \le (\sum_{j=1}^{n} |a_{hj}|) |x_h|=1. |x_h|$$\implies |\lambda| \le 1$ **Share on** <div id="fb-root"></div> <script async defer crossorigin="anonymous" src="https://connect.facebook.net/en_US/sdk.js#xfbml=1&version=v20.0"></script> <div class="share-buttons"> <div class="fb-share-button" data-href="https://mrislambd.github.io/posts/someproofs/" data-layout="button_count" data-size="small"><a target="_blank" href="https://www.facebook.com/sharer/sharer.php?u=https%3A%2F%2Fmrislambd.github.io%2Fposts%2Fsomeproofs%2F&src=sdkpreparse" class="fb-xfbml-parse-ignore">Share</a></div> <script src="https://platform.linkedin.com/in.js" type="text/javascript">lang: en_US</script> <script type="IN/Share" data-url="https://mrislambd.github.io/posts/someproofs/"></script> <a href="https://twitter.com/share?ref_src=twsrc%5Etfw" class="twitter-share-button" data-url="https://mrislambd.github.io/posts/someproofs/" data-show-count="true">Tweet</a> <script async src="https://platform.twitter.com/widgets.js" charset="utf-8"></script> </div> <div class="fb-comments" data-href="https://mrislambd.github.io/posts/someproofs/" data-width="" data-numposts="5"></div> **You may also like**

Some Linear Algebra Proofs

1. Let \(n\) be a positive integer. Show that every matrix \(A \in M_{n \times n}(\mathbb{R})\) can be written as the sum of two non-singular matrices.

2. Let \(\alpha \in \mathcal{L}(V)\) and \(\dim V=n< \infty\)

3. Let \(\alpha\in \mathcal{L}(V)\) and \(0\ne v\in V\) where \(\dim V=n< \infty\).

4. Let \(A,B\in M_{n\times n}(\mathbb{F})\) such that there exists an invertible matrix \(S\in M_{n\times n}(\mathbb{F})\) such that \(SAS^{-1}\) and \(SBS^{-1}\) are upper triangular matrices. Show that every eigenvalue of \(AB-BA\) is zero

5. Caley-Hamilton Theorem

6. Prove that the spectral radius of the \(\textit{Markov}\) matrix is less than or equal to 1

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