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Symmetry is Preserved Under Congruence

If $A$ is a symmetric matrix and $M$ is any matrix of compatible dimensions, the product $M^\top A M$ is also symmetric. This operation — transforming $A$ by sandwiching with $M$ and $M^\top$ — is called a congruence transformation.

The proof is quick: $(M^\top A M)^\top = M^\top A^\top (M^\top)^\top = M^\top A M$ , where we used $A^\top = A$ (symmetry of $A$ ) and $(M^\top)^\top = M$ . So the result equals itself, hence it is symmetric.

This fact matters because when we change coordinates (Section 8.14), the matrix of the quadratic form changes by exactly a congruence transformation. Knowing that symmetry is preserved means the diagonalized form is also symmetric — a diagonal matrix with eigenvalues on the diagonal.

Formal View

Lemma 8.2 — Congruence Preserves Symmetry

A

is a symmetric

n \times n

matrix and

M

is any

n \times k

matrix, then

M^\top A M

is symmetric.

Proof: $(M^\top A M)^\top = M^\top A^\top M = M^\top A M$ since $A^\top = A$ .

Why This Matters

Congruence transformations arise in changes of coordinates, and knowing symmetry is preserved ensures the transformed quadratic retains all the structural properties we care about.

Principal component analysis: the covariance matrix $\Sigma$ transforms to $U^\top \Sigma U = \Lambda$ under a rotation $U$ — still symmetric and diagonal.
Finite element methods: transforming stiffness matrices to reduced coordinates preserves their symmetry and positive-definiteness.
Normal equations: $A^\top A$ is always symmetric, a consequence of congruence (with $M = A$ , $A^\top A$ is $M^\top I M$ ).

Learning Resources

Matrix transpose properties

Khan Academy

Properties of the transpose used in proofs about symmetric matrices.

8 min

Symmetric matrices and spectral theorem

MIT OpenCourseWare

Gilbert Strang on symmetric matrices and their eigenvalue structure.

45 min

Quiz

Question 1

If $A$ is symmetric and $M$ is any matrix, then $M^\top A M$ is always symmetric.

Question 2

What is the key property of $A$ used to show $(M^\top A M)^\top = M^\top A M$ ?

Common Mistakes

Thinking $MAM^\top$ (wrong order) also preserves symmetry — it does if $A$ is symmetric, but the correct congruence form is $M^\top A M$ .
Assuming $M$ must also be symmetric — there is no requirement on $M$ .