To prove that trace of $A^2+A+I$ has minimum value of $3$ where $A$ is a 4 × 4 real symmetric matrix.

Question

Let $A$ be an 4×4 symmetric real matrix. Then show that the trace of $A^2+A+I$ cannot be less than $3$.

I know that for a real symmetric matrix, trace is equal to sum of its eigenvalues that are present as the diagonal entries. Also, the sum of matrices $A^2+A+I$ is also a symmetric one but this logic is not helping me to prove the result.

Answer 1

Hint: real symmetric matrices have real eigenvalues, and: $$ \lambda^2+\lambda+1=\Big(\lambda+\frac{1}{2}\Big)^2+\frac{3}{4}\geq\frac{3}{4} $$

Answer 2

As $A$ is real symmetric matrix, it has all real eigenvalues that we denote $(\lambda_i)_{i=1,..,4}$.

Suppose $P$ is the matrix of the eigenvectors. As $A$ is symmetric, we can write $$A=P\Lambda P^{-1} \tag{1}$$ Where $\Lambda$ is the diagonal matrix with the diagonal $(\lambda_i)_{i=1,..,4}$.

From $(1)$, it easy to prove that the trace of $A^2$ is $$\text{trace}(A^2) = \text{trace}(P\Lambda^2P^{-1}) =\text{trace}(\Lambda^2) = \sum_{i=1}^4 \lambda_i^2$$

Then, $$\begin{align} \text{trace}(A^2+A+I)&= \sum_{i=1}^4 \lambda_i^2+\sum_{i=1}^4 \lambda_i+4\\ &=\sum_{i=1}^4\left( \lambda_i^2+ \lambda_i + \frac{1}{4} \right) +3\\ &=\sum_{i=1}^4\left( \lambda_i+\frac{1}{2} \right)^2 +3 \ge 3\\ \end{align}$$ Q.E.D

Answer 3

Since $\operatorname{tr}\mathbf I=4$ for a $4\times 4$ matrix, this is equivalent to showing that $\operatorname{tr}(\mathbf A+\mathbf A^2)\geq-1$.

As you know, the trace is related to the sum of the eigenvalues. Let's say we know the eigenvalues of $\mathbf A$ - call them $\{\lambda _i\}_{i\in\{1,\dots ,4\}}$. We know that for the eigenvectors $\{\boldsymbol v_i\}_{i\in\{1,\dots ,4\}}$ that $$\mathbf A\boldsymbol v_i=\lambda_i\boldsymbol v_i$$ So, $$\mathbf A\mathbf A\boldsymbol v_i=\mathbf A\lambda_i\boldsymbol v_i \\ \mathbf A^2\boldsymbol v_i=\lambda_i(\mathbf A\boldsymbol v_i) \\ \mathbf A^2\boldsymbol v_i={\lambda_i}^2\boldsymbol v_i$$

Therefore, the eigenvalues of $\mathbf A^2$ are $\{{\lambda_i}^2\}_{i\in\{1,\dots,4\}}$. Hence $$\operatorname{tr}(\mathbf A+\mathbf A^2)=\sum_{i=1}^4 \lambda_i+{\lambda_i}^2$$ But, for any $\lambda\in\mathbb R$, we have $\lambda+\lambda^2\geq -1/4$, and hence $$\sum_{i=1}^4\lambda_i+{\lambda_i}^2\geq \sum_{i=1}^4 \frac{-1}{4}=-1$$

So $\operatorname{tr}(\mathbf A+\mathbf A^2)\geq-1$, and the desired result follows.