# Showing that real symmetric matrices are diagonalizable

• JD_PM
In summary: So if A is diagonalisable, then the minimal polynomial of A splits into distinct linear factors, and therefore so does the characteristic polynomial, and therefore so does the minimal polynomial of A- \lambda I, and therefore so does the characteristic polynomial of A- \lambda I, and therefore A- \lambda I is diagonalisable.

#### JD_PM

Summary:: Let ##A \in \Bbb R^{n \times n}## be a symmetric matrix and let ##\lambda \in \Bbb R## be an eigenvalue of ##A##. Prove that the geometric multiplicity ##g(\lambda)## of ##A## equals its algebraic multiplicity ##a(\lambda)##.

Let ##A \in \Bbb R^{n \times n}## be a symmetric matrix and let ##\lambda \in \Bbb R## be an eigenvalue of ##A##. Prove that the geometric multiplicity ##g(\lambda)## of ##A## equals its algebraic multiplicity ##a(\lambda)##.

We know that if ##A## is diagonalizable then ##g(\lambda)=a(\lambda)##. So all we have to show is that ##A## is diagonalizable.

I found a proof by contradiction: Assuming ##A## is not diagonalizable we have

$$(A- \lambda_i I)^2 v=0, \ (A- \lambda_i I) v \neq 0$$

Where ##\lambda_i## is some repeated eigenvalue. Then

$$0=v^{\dagger}(A-\lambda_i I)^2v=v^{\dagger}(A-\lambda_i I)(A-\lambda_i I) \neq 0$$

Which is a contradiction (where ##\dagger## stands for conjugate transpose).

But I do not really understand it; what is meant by 'generalized eigenvalues of order ##2## or higher'?

We could go over an alternative proof of course, the aim is understand how to show that real symmetric matrices are diagonalizable.

Thank you!

Well, there is a quite elegant analysis argument for that using compactness of a certain group of matrices (and other machinery), but I'll treat it as a linear algebra problem.

Let $E\neq 0$ be a real Euclidean space. It is well known that a transformation $\varphi$ on $E$ is symmetric if and only if $E$ admits an orthonormal basis of eigenvectors of $\varphi$.

Now let $A$ be a symmetric real matrix and fix an orthonormal basis $e$ on $E$ (take any basis, give it the Gram-Schmidt treatment and normalise it, say). The corresponding transformation $\varphi$ is then symmetric. Let $f$ be an orthonormal basis of eigenvectors of $\varphi$.

The matrix of $\varphi$ with respect to $f$, call it $B$, is diagonal.

Let $T$ be the transition matrix from $e$ to $f$. Then $B=T^{-1}AT$, thus $A$ is similar to a diagonal matrix i.e is diagonalisable. Transition matrix between orthonormed bases is orthogonal, so can replace $T^{-1}=T^t$.

Remark: in the complex case, this is not true. However, Hermitian matrices are diagonalisable.

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JD_PM
You have two separate questions, I feel. So I'll briefly address the second one. Yes, diagonalisability is equivalent to $g(\lambda) = a(\lambda)$. This is because equality of multiplicities is equivalent to the space admitting a basis of eigenvectors. Also, similar matrices have the same characteristic polynomial.

## 1. What is a real symmetric matrix?

A real symmetric matrix is a square matrix where the elements are real numbers and the matrix is equal to its own transpose.

## 2. What does it mean for a matrix to be diagonalizable?

A matrix is diagonalizable if it can be transformed into a diagonal matrix through a similarity transformation.

## 3. Why is it important to show that real symmetric matrices are diagonalizable?

It is important because diagonal matrices are much easier to work with mathematically and can provide valuable insights into the properties of the original matrix.

## 4. What is the process for showing that a real symmetric matrix is diagonalizable?

The process involves finding the eigenvalues and corresponding eigenvectors of the matrix, constructing a matrix with the eigenvectors as columns, and then using this matrix to diagonalize the original matrix through a similarity transformation.

## 5. Are all real symmetric matrices diagonalizable?

Yes, all real symmetric matrices are diagonalizable.