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# Condition for a linear subspace E of R^n to be A-invariant

The exercise is as follows:

a) Let E be a linear subspace of R^n. Show that E is a closed subspace of R^n.

b) Let A be a real n x n matrix. Show that a linear subspace E of R^n is A-invariant if and only if E is e^{tA} -invariant for all t in R, where e^{tA} is the exponential matrix associated to A.

#### Solution Preview

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a) Let E be a linear subspace of R^n. Show that E is a closed subspace of R^n.

There are different ways to do this, one is as follows. A linear space E admits a basis of orthonormal vectors v_1,...,v_m,
here the dimension of E is m<= n.

Now to show that E is closed we show that given a sequence {x_i}_{i} with x_i in E for all i, and x in R^n such
that x_i converges to x as i tends to infinity, then x is in E.

First step: Show that the sequence is bounded. This follows since by convergence there exists i_0 such that for all
i>= i_0, ||x_i-x||<1/2, say. Then the triangular inequality gives ||x_i||<1/2+||x|| for all i>= i_0. Therefore

||x_i||<= 1/2+||x||+max_{1<= j<= i_0}||x_j|| for all i>= 1,

showing that the sequence is bounded.

Second step: Now use the basis of E to write each x_i in E as a linear combination of the v_i's

x_i=c_{1,i}v_1+...+c_{m,i}v_m.

Because the v_i's are orthonormal we have that ||x_i||=(c_{1,i}^2+...+c_{m,i}^2)^{1/2} and ...

#### Solution Summary

It is proved that a closed subset of n-dimensional Euclidean space is closed. Then, given a matrix A, it is proved that a linear subset of R^n is A-invariant, if and only if the linear subset is e^{tA}- invariant for all t.

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