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## Proof

Prove that $M$ is symmetric if and only if $A T M A$ is symmetric.

We can prove this statement by induction, as follows.

It is trivial for 1 by 1 matrices, and so we assume it true for all $n$ by $n$ matrices, and suppose $M$ is $n + 1$ by $n + 1$ .

If $M$ has any real eigenvalue $z$ and an eigenvector corresponding to it, we can find an eigenvector for it and use that eigenvector, suitably normalized, as the first basis vector in the new basis, with all the other basis vectors arbitrarily chosen to be normal to it.

Now consider what the matrix $A T M A$ will look like when we have done this.

The first column of $A$ will be this eigenvector, $M$ will multiply this vector by $z$ and $A T$ will take it back to the vector whose first component is $z$ and the rest are 0. Thus the first column of $A T M A$ will have only its first entry non-zero, and by symmetry the same thing is true of its first row.

We now invoke the induction hypothesis to deduce that the rest of the matrix after removing the first row and column can be diagonalized, (there are $n$ other eigenvectors for it all having 0 first components) which handles the whole thing.

So all that needs proving is that $M$ has an eigenvalue and eigenvector.

Now if $M$ has a real eigenvalue, so that the equation $det ⁡ ( M − x I ) = 0$ has a real root, $x = r$ , then there must be a non trivial vector $v ⟶$ such that $( M − r I ) v ⟶ = 0 ⟶$ , and we will have an eigenvector.

We need only prove then that this characteristic equation has a real root for any symmetric matrix $M$ .

We know that $det ⁡ ( M − x I )$ is a polynomial in $x$ of degree $n$ , which goes to both plus and minus infinity for $x$ very large positive or negative when $n$ is odd.

Since all polynomials are continuous, it must pass through 0 and have a real root somewhere for $n$ odd.

When $n$ is even, this polynomial approaches positive infinity for large $x$ of either sign.

We need only show that it is non-positive somewhere, to show that there is a real root to the characteristic equation, again by continuity of the determinant polynomial.

We can find a non-positive value for $det ⁡ ( M − x I )$ most conveniently by using the induction hypothesis to find a basis of eigenvectors of the first $n$ rows and columns of $M$ (which form an $n$ by $n$ symmetric matrix and by the induction hypothesis it has a real basis of eigenvectors.)

If we use this basis on the entire matrix $M$ then the only non-zero off diagonal elements will occur in the last row and last column.

With $M$ in this form our conclusion can be deduced from the following observations.

When $M$ has the form just described, the only terms that contribute to $det ⁡ ( M − x I )$ are either the product of all of $M − x I$ 's diagonal elements, or the product of $n − 1$ of its first $n$ diagonal elements multiplied by a non-positive number. (minus the square of the off diagonal element $M j n − 1$ where $j$ is the missing diagonal element)

Exercise: prove this last claim to your satisfaction.

If we let $x$ be the smallest of the first $n − 1$ diagonal elements of $M$ , then the only non-vanishing terms in $det ⁡ ( M − x I )$ will be negative, so that the characteristic polynomial evaluated at that argument cannot be positive, and the characteristic equation must therefore have a real root by continuity.

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