Prove that a given $3\times 3$ symmetric matrix is diagonalizable without using the spectral theorem.

Question

I have to prove that following matrix is diagonalizable over $\mathbb R$. $$ \begin{bmatrix} 1 & 2 & -5 \\ 2 & 4 & -1 \\ -5 & -1 & 3 \\ \end{bmatrix} $$ I found characteristic equation for this matrix as $ x^3-8x^2-11x+81=0$ Now if I prove that this equation has distinct roots then it follows that matrix is diagonalizable. For that I can consider its derivative which will be quadratic equation whose roots can be found easily. Then I can show that these 2 roots are not roots of characteristic equation.Then it has distinct roots. But, this is long calculation. Also, after that I need to show its 3 roots are real. So, Is there any other method to solve this problem? Edit : It is Symmetric Matrix and it will be always diagonalizable. But, can we solve it without using this fact ? I mean if such difficult characteristic equation came for some other matrix which is not symmetric then how can I proceed?

Answer 1

You can prove it with elementary tools:

Let $p(x)=x^3-8x^2-11x+81$. Then $p'(x)=3x^2-16x-11$ has a positive and a negative root since the constant term and the leading coefficient have opposite signs.

Let $\alpha <0<\beta$ these roots. The polynomial function $p(x)$ has a local maximum $M=p(\alpha)$ and a local minimum $m=p(\beta)$. All we have to prove is that $p(\alpha) >0\:$ and $\:p(\beta) <0$.

Perform the Euclidean division of $p(x)$ by $p'(x)$: $$p(x)=\frac{3x-8}9\,p'(x)+\frac{641-194\,x}9,$$ so that $p(\alpha)=\frac19(641-194\,\alpha)$, $\:p(\beta)=\frac19(641-194\,\beta)$.

• Since $\alpha <0$, we obviously have $p(\alpha) >0$.
• As to $p(\beta)$, observe that $\beta >4$ since $p'(4) <0$, therefore $$p(\beta)=\frac19(641-194\,\beta)<\frac19(641-194\cdot 4)<0.$$

Answer 2

An easy way of seeing there are $3$ distinct zeros is to use the intermediate value theorem and a bit of guessing and checking.

First off, note that at $x=-1000$, $p(x) < 0$.

Then, at $x = 0$, $p(x) = 81 > 0$. Thus there is a real zero between $-100$ and $0$.

Next, at $x = 5$, $p(x) = -49 < 0$. Thus there is a real zero between $0$ and $5$.

Finally, at $x=1000$, $p(x) > 0$. Thus we've found our third distinct real root between $5$ and $1000$!

Answer 3

If (and only if) the cubic discriminant is greater than zero, the cubic has three distinct real roots.

Answer 4

A cubic with a positive leading coefficient goes to negative infinity as x goes to negative infinity, and positive infinity as x goes to positive infinity. A local maximum and minimum occur at the roots of the derivative. If the derivative doesn't have distinct real roots, neither does the cubic equation. If the derivative has distinct real roots, the values of the cubic at these roots can both be positive, both be negative, one or the other can be zero, or the first (least x) can be positive and the second negative. The cubic has three distinct roots in the last case.

Usually values near the roots can be tested instead of the exact roots, since all that needs to be shown is that the cubic goes positive then negative. The roots of the derivative just tell you where to look.

Answer 5

Here is a different route: use singular value decomposition. Similar to the fact that the spectral theorem admits a topological proof, the existence of SVD can be directly proved via a topological argument (see e.g. theorem 2.2 of Paweł Wójcik, A simple proof of the polar decomposition theorem). Therefore the use of SVD here does not result in a circular argument.

Let $A$ be a real symmetric matrix. Then $A+cI$ represents a positive quadratic form when $c>0$ is sufficiently large. Therefore, by Sylvester's law of inertia, $A+cI=P^TP$ for some matrix $P$. Let $P=USV^T$ be a SVD. Then $$ A=P^TP-cI=VS^2V^T-cI=V(S^2-cI)V^T $$ is an orthogonal diagonalisation.

(Remark. We are essentially proving the spectral theorem without using eigenvalues. This works only for symmetric/Hermitian matrices, of course.)