Proving an inequality in Eigenvalues of a matrix

If $\lambda_1$, $\lambda_2$, $\lambda_3$ are the eigenvalues of the matrix : $$\begin{pmatrix} 26 & -2 & 2 \\ 2 & 21 & 4 \\ 4 & 2 & 28 \\ \end{pmatrix}$$ Show that : $$\sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2} \le \sqrt{1949}$$

I found the characterstic equation as $$\lambda^3 - 75\lambda^2 + 1850\lambda - 15000 = 0$$ This gave $$\lambda_1 + \lambda_2 + \lambda_3 = 75$$ and $$\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_1\lambda_3 = 1850$$ Also $${(\lambda_1 + \lambda_2 + \lambda_3)}^2 = \lambda_1^2 + \lambda_2^2 + \lambda_3^2 + 2{(\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_1\lambda_3)}$$ From this I found out $$\sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2} = \sqrt{1925}$$ But am not satisfied with the way I have approached the problem. Does anyone know of an easier way of doing it?

• I would say this is well done and easily done except the part of finding out characteristic equation :) – user87543 Aug 1 '13 at 13:36

Hint: If your matrix is $A$ with eigenvalues $\lambda_1,\ldots,\lambda_n$, then $$\operatorname{tr} A^2 = \sum_{i=1}^n \lambda_i^2$$

• But the issue here is that we have to prove the inequality $$\sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2} \le \sqrt{1949}$$ The way you are suggesting will give me the value of the $$\lambda$$s directly which I can put in to find the equality and not the inequality. – Sheetal Sarin Aug 1 '13 at 13:44
• @SheetalSarin: In your case, $A$ is given, so you can compute the left-hand side easily. – ccorn Aug 1 '13 at 13:46
• @SheetalSarin: See my other answer for an explanation of the $1949$. – ccorn Aug 1 '13 at 14:32

OK, another one, to give a hint where the $1949$ comes from: If your Matrix is $A=((a_{ij}))_{i,j=1,\ldots,n}$ with real entries, then the Frobenius norm of $A$ is defined as $$\|A\|_{\mathrm{F}} = \sqrt{\operatorname{tr}(A^{\mathsf{T}} A)} = \sqrt{\sum_{i,j=1,\ldots,n} a_{ij}^2}$$ Note that $$\sum_{i,j=1,\ldots,n} a_{ij}^2 - \sum_{i,j=1,\ldots,n} a_{ij}\,a_{ji} = \sum_{i,j=1,\ldots,n;i<j} (a_{ij} - a_{ji})^2\geq 0$$ In other words, $$\operatorname{tr}(A^2) \leq \operatorname{tr}(A^{\mathsf{T}} A)$$ which in your case means $$\sum_{i=1}^n\lambda_i^2 \leq 1949$$ So this is where the $1949$ comes from, it's the square of the Frobenius norm of $A$.

Edit: typo corrections

The question seems to me just ask a bound and not an exact value that's we can found by a little calculus. However there's a method which localize the eigenvalues of a squared matrix and called Gershgorin circle theorem. Let's apply this method in our example: we obtain the three discs: $$D(26,4)\quad;\quad D(21,4)\quad;\quad D(28,6)$$ and then by triangle inequality we find $$\sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2} \le \sqrt{30^2+25^2+34^2}=\sqrt{2681}$$

• It's $D(21,6)$ and thus $27^2$ – ccorn Aug 1 '13 at 14:36
• There's two Discs $D(21,6)$ and $D(21,4)$ corresponding to the row and the column and we can take the smaller. – user63181 Aug 1 '13 at 14:41
• Ah, thanks for the clarification! – ccorn Aug 1 '13 at 14:50