Faster method for solving the inequality $\sqrt{λ_1^2+ λ_2^2+ λ_3^2}$ $\leq$ $\sqrt{1949}$ for a 3x3 matrix

Question

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 $2\sum_{j=1}^{n-1}\sum_{i=j+1}^{n}|\lambda_i||\lambda_j| > 0$,

$$|\lambda_1|^2+|\lambda_3|^2+|\lambda_3|^2 \leq 1949$$

$$\sqrt{|\lambda_1|^2+|\lambda_3|^2+|\lambda_3|^2} \leq \sqrt{1949}$$

Answer 1

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 $$\sqrt{\lambda_1^2+\lambda_2^2+\lambda_3^2} = \sqrt{\text{tr}(A^2)} = \sqrt{680+445+800} = \sqrt{1925} < \sqrt{1949}.$$

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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 $\sqrt{|\lambda_1|^2+|\lambda_2|^2+|\lambda_3|^2} = \sqrt{\lambda_1^2+\lambda_2^2+\lambda_3^2}$.