TITLE: Faster method for solving the inequality $\sqrt{λ_1^2+ λ_2^2+ λ_3^2}$ $\leq$ $\sqrt{1949}$ for a 3x3 matrix
QUESTION [5 upvotes]: λ1, λ2, λ3 are the eigen values of the matrix \begin{bmatrix}26&-2&2\\2&21&4\\4&2&28\end{bmatrix}
Show that $\sqrt{λ_1^2+ λ_2^2+ λ_3^2}$ $\leq$ $\sqrt{1949}$

Here's my approach at solving this problem -
Performing Single Value Decomposition of $A$, we get
$$A^TA = \begin{bmatrix}696&-2&172\\-2&449&136\\172&136&804\end{bmatrix}$$
$$\Sigma^T\Sigma = \begin{bmatrix}389.78&0&0\\0&604.83&0\\0&0&954.38\end{bmatrix}$$
Therefore,
$$\Sigma = \begin{bmatrix}\sqrt{389.78}&0&0\\0&\sqrt{604.83}&0\\0&0&\sqrt{954.38}\end{bmatrix}$$
Now,
$$\sum_{i=1}^n|\lambda_i| \leq \sum_{i=1}^n|\sigma_{ii}|$$
Squaring on both sides,
$$\sum_{i=1}^n|\lambda_i|^2 \leq \sum_{i=1}^n|\sigma_{ii}|^2$$
We get
$$\lambda_1^2+\lambda_3^2+\lambda_3^2 \leq 1948.99\approx1949$$
Taking square root, we get
$$\sqrt{\lambda_1^2+\lambda_3^2+\lambda_3^2} \leq \sqrt{1949}$$
Is the step where I square the summation inequality correct? I have never come across taking squares of summation so I have my doubts whether it's correct or not. Also, is there any faster way of solving this problem. Computing $A^TA$ and then finding it's eigenvalues takes too much time.
EDIT :
$$(\sum_{i=1}^n|\lambda_i|)^2 \leq (\sum_{i=1}^n|\sigma_{ii}|)^2$$
$$\sum_{i=1}^n|\lambda_i|^2 + 2\sum_{j=1}^{n-1}\sum_{i=j+1}^{n}|\lambda_i||\lambda_j| \leq tr(A^TA)$$
Since $2\sum_{j=1}^{n-1}\sum_{i=j+1}^{n}|\lambda_i||\lambda_j| > 0$,
$$|\lambda_1|^2+|\lambda_3|^2+|\lambda_3|^2 \leq 1949$$
$$\sqrt{|\lambda_1|^2+|\lambda_3|^2+|\lambda_3|^2} \leq \sqrt{1949}$$

REPLY [4 votes]: Since $\lambda_1,\lambda_2,\lambda_3$ are the eigenvalues of $A$, we have that $\lambda_1^2,\lambda_2^2,\lambda_3^2$ are the eigenvalues of $A^2$. So once you compute $$A^2 = \begin{bmatrix}680 & -90 & 100 \\ 110 & 445 & 200 \\ 220 & 90 & 800 \end{bmatrix},$$ it is easy to get that $$\sqrt{\lambda_1^2+\lambda_2^2+\lambda_3^2} = \sqrt{\text{tr}(A^2)} = \sqrt{680+445+800} = \sqrt{1925} < \sqrt{1949}.$$

Alternatively, you can first compute $A^TA$ like you did, and then apply the inequality $$\sqrt{|\lambda_1|^2+|\lambda_2|^2+|\lambda_3|^2} \le \sqrt{\text{tr}(A^TA)} = \sqrt{696+449+804} = \sqrt{1949}.$$ Then, you just need to check that $\lambda_1,\lambda_2,\lambda_3$ are real to make sure $\sqrt{|\lambda_1|^2+|\lambda_2|^2+|\lambda_3|^2} = \sqrt{\lambda_1^2+\lambda_2^2+\lambda_3^2}$.