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For finite sets $A$ and $B$, it is clear that $A \subseteq B$ and $|A| \geq |B|$ implies $A = B$. While an obvious fact, it can sometimes be a nice shortcut in proofs.

Analogously, if $V$ and $W$ are finite-dimensional vector spaces such that $V \subseteq W$ and $dim\ V \geq dim\ W$ then $V = W$. This is an especially useful tool when $V$ is defined parametrically and $W$ is defined implicitly. Then you can easily prove $V \subseteq W$ by plugging the parametric expression for $V$ into the equations for $W$. Counting the dimensions can take more work, but you sometimes get lucky.

For a while I've been wondering how this extends to algebraic varieties.

Here's an attempted application to proving the spectral theorem. Fix the dimension $n$; all matrices will be $n \times n$. The theorem says that for every symmetric matrix $S$ there exists an orthogonal matrix $Q$ and a diagonal matrix $D$ such that

$$Q^T\ D\ Q = S.$$

Let $A$ and $B$ respectively denote the matrices of the form on the left-hand and right-hand side. $A$ is defined parametrically by a function $f$ from $D$ and $Q$, and $B$ is defined implicitly by the symmetry condition. We want to prove $A = B$. It is clear that $A \subseteq B$:

$$(Q^T\ D\ Q)^T = Q^T\ D^T\ (Q^T)^T = Q^T\ D\ Q,$$

so $Q^T\ D\ Q$ is symmetric.

It's easy to see that $f$'s

The domain of $f$ has dimension $d dim\ D + q$ dim\ Q$where$d dim\ D = n$comes from$D$and$q dim\ Q = (n-1) + \cdots + 1$comes from$Q$. And , while$S$'s space has dimension$n + (n-1) + \cdots + 1$, so the dimensions seem to matchon the face of it. But what about$f$'s degree of injectivity? It isn't perfectly injective: if$D$and$D'$equal the identity matrix then$Q^T\ D\ Q = Q'^T\ D'\ Q'$for any independent combination of$Q$and$Q'$. My question is thus twofold: What is the right generalization of the theorem for vector spaces to algebraic varieties? and Can the attempted proof of the spectral theorem be salvaged with a genericity argument? I'm happy with the extant proofs of the spectral theorem, so this is more curiosity than anything else. 3 edited title # Proving equality of varieties by containmentand dimension counting 2 edited body; deleted 2 characters in body For finite sets$A$and$B$, it is clear that$A \subseteq B$and$|A| \geq |B|$implies$A = B$. While an obvious fact, it can sometimes be a useful nice shortcut in proofs. Analogously, if$V$and$W$are finite-dimensional vector spaces such that$V \subseteq W$and$dim\ V \geq dim\ W$then$V = W$. This is an especially useful tool when$V$is defined parametrically and$W$is defined implicitly. Then you can easily prove$V \subseteq W$by plugging the parametric expression for$V$into the equations for$W$. Counting the dimensions can take more work, but you sometimes get lucky. For a while I've been wondering how this extends to algebraic varieties. Here's an attempted application to proving the spectral theorem. Fix the dimension$n$; all matrices will be$n \times n$. The theorem says that for every symmetric matrix$S$there exists an orthogonal matrix$Q$and a diagonal matrix$D$such that $$Q^T\ D\ Q = S.$$ Let$A$and$B$respectively denote the matrices of the form on the left-hand and right-hand side.$A$is defined parametrically by a function$f$from$D$and$Q$, and$B$is defined implicitly by the symmetry condition. We want to prove$A = B$. It is clear that$A \subseteq B$: $$(Q^T\ D\ Q)^T = Q^T\ D^T\ (Q^T)^T = Q^T\ D\ Q,$$ so$Q^T\ D\ Q$is symmetric. It's easy to see that$f$'s domain has dimension$d + q$where$d = n$comes from$D$and$q = (n-1) + \cdots + 1$comes from$Q$. But And$S$'s space has dimension$n + (n-1) + \cdots + 1$, so dimensions match on the face of it. But what about$f$'s degree of injectivity? It isn't perfectly injective: if$D$and$D'$equal the identity matrix then$Q^T\ D\ Q = Q'^T\ D'\ Q'$for any independent combination of$Q$and$Q'\$.

My question is thus twofold:

What is the right generalization of the theorem for vector spaces to algebraic varieties?

and

Can the attempted proof of the spectral theorem be salvaged with a genericity argument?

I'm happy with the extant proofs of the spectral theorem, so this is more curiosity than anything else.

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