How to prove the maximum eigenvalue of $A^2$ is greater than or equal to $2$ and greater than $4$ if $A$ is positive definite?

Question

Given $\alpha \in \mathbb{R}$ and matrix

\begin{equation*}A=\begin{pmatrix} \alpha& 2 \\ 2 & \alpha\end{pmatrix} \end{equation*}

try the following

b)If $\lambda$ is the largest eigenvalue of $A^2$ shows that $\lambda \geq 2\\$

c)If A is positive definite and $\lambda$ is its largest eigenvalue shows that $\lambda>4$

solution b)

$D=A^2$

\begin{equation*}D-I\lambda=\begin{pmatrix} \alpha^2+4-\lambda& 4\lambda \\ 4\lambda & \alpha^2+4-\lambda\end{pmatrix} \end{equation*}

then \begin{equation*}\det(D)=\lambda^2-\lambda(2\alpha^2 +8)+(\alpha^4-8\alpha^2+16)\end{equation*}

Equating the equation to zero and solving for the quadratic we have that \begin{equation*} \lambda_1=\alpha^2+4+4\alpha \end{equation*} \begin{equation*} \lambda_2=\alpha^2+4-4\alpha \end{equation*}

that means that \begin{equation*} \lambda_{\text{max}}=\max\left \{ \lambda_1,\lambda_2 \right \} \end{equation*}

I think it should be assumed if $\alpha\geq0$ then \begin{equation*} \lambda_{max}= \lambda_1 \end{equation*} then \begin{equation*} \lambda_{max}= \alpha^2+4+4\alpha \end{equation*} then \begin{equation*} \lambda_{max}= (\alpha+2)^2 \end{equation*}

But I don't know what to do after this to prove that the expression is greater than or equal to 2

c)I calculate $\lambda$ and it gives me lambda $\lambda_1=\alpha+2$ and $\lambda_2=\alpha-2$ for it to be positively defined these two must be positive arriving at $\alpha\geq 2$ and $\alpha\geq -2$ but I don't know how it arrives at $\lambda\geq 4$

Answer 1

if $\alpha\geq0$ then $\lambda_{max}= (\alpha+2)^2$

But I don't know what to do after this to prove that the expression is greater than or equal to 2

$$(0+2)^2 \ge 4$$

c)I calculate $\lambda$ and it gives me lambda $\lambda_1=\alpha+2$ and $\lambda_2=\alpha-2$ for it to be positively defined these two must be positive arriving at $\alpha> 2$ and $\alpha> -2$ but I don't know how it arrives at $\lambda\geq 4$

$$\alpha> 2\cap\alpha> -2\Longrightarrow \alpha>2$$ Therefore,

$$\max(\lambda_1,\lambda_2)=\lambda_1=\alpha+2>2+2>4$$

Answer 2

Given $\alpha \in \mathbb{R}$ and matrix

\begin{equation*}A=\begin{pmatrix} \alpha& 2 \\ 2 & \alpha\end{pmatrix} \end{equation*}

1. If each row sum is equal to a fixed scalar then the fixed scalar would be an eigen value $($hint: $x=e_1+e_2$ and $Ax=?)$.

2. $\textrm{trace}(A)$ is the sum of eigen values counting with multiplicity.

3. $\lambda\in\textrm{spec}(A) \implies \lambda^2\in\textrm{spec}(A^2) $

Using 1), 2), 3) we can quickly compute the eigenvalues of $A^2$ : $$\lambda_1=(\alpha-2) ^2, \lambda_2=(\alpha+2) ^2$$

b) Suppose $\max\{\lambda_1, \lambda_2\}<2$

$\begin{align}&(\alpha+2) ^2+(\alpha-2)^2<4\\&2(\alpha^2+4)<4\\&\alpha^2<-2\end{align}$

Which is absurd as $\alpha\in\Bbb{R}, \alpha^2\ge 0$

c) Given $A$ is positive definite.Then $\alpha-2>0, \alpha+2>0$

Implies $(\alpha>2 $ and $\alpha>-2)$ implies $\alpha>2$

Implies $\max\{\lambda_1, \lambda_2\}=\alpha+2>4$