I was proving the spectral theorem for normal operators on finite-dimensional complex vector spaces today during a test, when I arrived at the point in which

If $T\in\operatorname{End}(V)$ is normal, then if $W$ is a subspace of $V$ is $T$-invariant "surely" the restriction $T|_W\colon W\to W$ must be normal too.

But the professor said he could find a counterexample. Now, the claim above was true in that case, but I'm still curious about that example. Practically, he said that it can happen that the restriction of the adjoint $T^*$ may not coincide with the adjoint of $T$ in $W$, in other words $$ (T|_W)^*\ne (T^*)|_W. $$

When is it true (in finite-dimensional vector spaces)?


2 Answers 2


It is always true for finite dimensional vector spaces. If $T \colon V \rightarrow V$ is a normal operator and $W$ is $T$-invariant, then $W$ is also $T^*$-invariant and hence $T|_W$ is normal (and $(T|_W)^* = T^*|_W$). You can find a proof in Axler's book "Linear Algebra Done Right", Chapter 7.

  • 1
    $\begingroup$ Very helpful. Update : in the third edition this in Chapter 9. $\endgroup$
    – Niels
    Commented Jan 11, 2021 at 14:13

For arbitrary operators, it could be that $W$ is a $T$-invariant subspace, but not a $T^*$ invariant subspace. For example, take $$ T = \pmatrix{1&1\\0&2} $$ Note that $W = \operatorname{span}[(1,0)]$ is $T$-invariant, but not $T^*$ invariant.

On the other hand, if $W$ is both $T$-invariant and $T^*$-invariant, we will necessarily have $(T|_W)^* = (T^*)|_W$. It's easy to see that this should be the case by noting that we can define the adjoint by the property $$ \langle T|_W x, y \rangle = \langle x,(T|_W)^*y \rangle \quad \forall x,y \in W $$

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    $\begingroup$ I see. In my case $W$ was indeed $T$- and $T^*$-invariant. We could have $\langle T(x),y\rangle=\langle T|_W(x),y\rangle=\langle x,(T|_W)^*(y)\rangle=\langle x,T^*(y)\rangle$ if $x,y\in W$. Is that it? Does $T^*(y)=(T|_W)^*(y)$ for all $y\in W$ imply $(T^*)|_W=(T|_W)^*$? $\endgroup$
    – yellon
    Commented Jun 25, 2015 at 8:57
  • $\begingroup$ @DavideF looks good to me $\endgroup$ Commented Jun 25, 2015 at 12:59
  • $\begingroup$ Oh and yes, two functions (linear maps) are the same if they take all the same values $\endgroup$ Commented Jun 25, 2015 at 23:27

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