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Given a real symmetric matrix $A$, is it true that any $A$-invariant subspace is formed by the linear span of some subset of eigenvectors of $A$?

user2684957
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    This is correct. Take the restriction of the linear operator A to an invariant subspace and take advantage of the fact that a symmetric linear operator over a field of real numbers has a basis of eigenvectors. – kabenyuk Sep 07 '21 at 16:20

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An explicit way to do this would be via using projections and commutativity.

let $P$ be an orthogonal projection with image = to this $A-$invariant subspace $W$. Then for $w\in W$: $P(Aw) = Aw = APw$ and for $v \in W^\perp$, you see $W^\perp$ is A-invariant because $w^T(Av)=(w^TA)v = (w')^Tv = 0$, so $(Av) \in W^\perp\implies \mathbf 0 = APv =P(Av)$.

Thus $AP = PA$. They commute and each are diagonalizable (indeed both are real symmetric), hence they are simultaneously diagonalizable. (Thus $\text{im }P$ has a basis given by eigenvectors of $A$ and so does $\ker P$.)

user8675309
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