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Let's say we have a symmetric linear transform, $T$ over some space over $\Bbb R^n$ represented as a matrix $M$. Now, we find some sub-space, $W$ over $\Bbb R^k$ with $k<n$ such that applying $T$ on any vector, $u$ in $W$ results in a vector also in $W$. We now define a linear transform $T^*$ that is just $T$ when applied over the $W$ sub-space. Is it true that we can find a basis for $T^*$ such that the associated $k \times k$ matrix is also symmetric? In other words, $T^*$ is also a symmetric linear transform?

The reason I'm asking this is to complete a proof that symmetric matrices are diagonalizable: Help verifying proof that symmetric matrices are diagonalizable.

Rohit Pandey
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First to get a small technical gripe out the way, it doesn't really make sense to talk about a linear transformation being symmetric without additional structure (as opposed to talking about the matrix representing $T$ w.r.t a specific basis). This is because being symmetric isn't invariant under changes of basis, so by the spectral theorem we have that

There exists a basis $B$ such that $T$ is symmetric $\iff$ $T$ is diagonalisable

The situation changes if you impose an inner product on the space, as then you can talk about orthonormal vectors, and being symmetric is preserved under orthogonal changes of basis

So with this information the question becomes:

Is a diagonalisable matrix still diagonalisable when restricted to an invariant subspace

Of which the answer is yes, as it then becomes a repeat of this question

Red5551
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  • Thanks. I should have mentioned that the reason I want to prove this is to complete a proof for the very fact that symmetric matrices are diagonalizable (https://math.stackexchange.com/questions/4357152/help-verifying-proof-that-symmetric-matrices-are-diagonalizable). So invoking diagonalization becomes circular reasoning. Of course, no way you could have known this. – Rohit Pandey Jan 23 '22 at 00:31