# 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?

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 $$\lambda\geq 4$$

• (2) If $\max\{\lambda_1,\lambda_2\}<2$, then both $\lambda_1$ and $\lambda_2$ are less than $2$. $\implies 8+2\alpha^2=\lambda_1+\lambda_2<2+2=4$, which is absurd. May 11, 2023 at 2:16

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

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,

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