TeachingLinear Algebra – Symmetric matrices and quadratic forms
Definition: A matrix is called symmetric if \(A^T=A\).
Remark: So a symmetric matrix can only be square.
Symmetric matrices have nice properties. The most important property of a symmetric matrix is that it is always diagonaizable. We will show this in a few steps. We start with:
Theorem: If \(A\) is a symmetric matrix, then the eigenvectors of \(A\) corresponding to different eigenvalues are orthogonal.
Proof: Suppose that \(A\) is symmetric (so: \(A^T=A\)), \(A\mathbf{v}_1=\lambda_1\mathbf{v}_1\) and \(A\mathbf{v}_2=\lambda_2\mathbf{v}_2\) with \(\lambda_1\neq\lambda_2\). Then we have:
\[\lambda_1(\mathbf{v}_1\cdot\mathbf{v}_2)=(\lambda_1\mathbf{v}_1)\cdot\mathbf{v}_2=(\lambda_1\mathbf{v}_1)^T\mathbf{v}_2 =(A\mathbf{v}_1)^T\mathbf{v}_2=\mathbf{v}_1^TA^T\mathbf{v}_2=\mathbf{v}_1^TA\mathbf{v}_2=\mathbf{v}_1^T(\lambda_2\mathbf{v}_2) =\mathbf{v}_1\cdots(\lambda_2\mathbf{v}_2)=\lambda_2(\mathbf{v}_1\cdot\mathbf{v}_2).\]Hence: \((\lambda_1-\lambda_2)(\mathbf{v}_1\cdot\mathbf{v}_2)=0\). However \(\lambda_1\neq\lambda_2\), so: \(\mathbf{v}_1\cdot\mathbf{v}_2=0\). Hence the vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) are orthogonal.
Definition: A matrix is called orthogonally diagonalizable if theres exist an orthogonal matrix \(P\) and a diagonal matrix \(D\) such that \(A=PDP^{-1}=PDP^T\).
An orthogonal matrix is a square matrix with orthonormal columns. So for an orthogonal matrix \(P\) we have that \(P^TP=I\) and that \(P\) is square. Hence: \(P\) is invertible and \(P^{-1}=P^T\).
Theorem: A square matrix is orthogonally diagonalizable if and only if \(A\) is symmetric.
So this theorem says that every symmetric matrix is not only diagonizable, but even orthogonally diagonalizable. Moreover, every matrix that is orthogonally diagonalizable is a symmetric matrix. The latter property is easy to see:
Proof: If \(A\) is orthogonally diagonalizable, then we have: \(A=PDP^T\) for some orthogonal matrix \(P\) and diagonal matrix \(D\). This implies that:
\[A^T=(PDP^T)^T=(P^T)^TD^TP^T=PD^TP^T=PDP^T=A.\]Hence: \(A^T=A\). Or equivalently: \(A\) is symmetric.
The proof of the converse is more difficult. In the exercises 23 and 24 of §5.5 in Lay it is shown that a symmetric matrix has only real eigenvalues. We have seen that eigenvectors of symmetric matrix corresponding to different eigenvalues are orthogonal. So these eigenvectors can also be chosen to be orthonormal. If an eigenvalue of a symmetric matrix has a geometric multiplicity larger than \(1\), then using the Gram-Schmidt process we can easily construct an orthonormal basis of the corresponding eigenspace. However, it is not so easy to show that for each eigenvalue the algebraic multiplicity is equal to the geometric multiplicity. That part of the proof is skipped here.
Last modified on May 2, 2021
