The subspace spanned by the eigenvectors of a matrix, or a linear transformation, can be expressed as a direct sum of eigenspaces.
- ;
- .
Another way of expressing this is to say that every vector in can be written uniquely as a sum of i) a vector in and ii) a vector in . If is a direct sum of and , we write .
A similar description applies more generally to an -fold direct sum: is a direct sum of subspaces , if
- Every can be written as for ;
- .
- Proof
- Assume that . Suppose also that where . If the representation is unique for all then the sum is a direct sum by definition. On the other hand, suppose for some . We can assume without loss of generality that . Then implying , so that the sum is not a direct sum.
When is a direct sum of the subspaces we indicate this by writing either , or .
Show that if is a linearly independent set of vectors in , a linearly independent set
of vectors in , and with , then is linearly independent.
Define to be the supspace of spanned by the eigenvectors of (this may or may not be all of ). As we have seen, the number of distinct possible eigenvalues of is at most when is an matrix. In particular, it is finite.
- Proof
- For the statement is trivially true. So suppose that , and with . Then
implying . As , this implies .
Inductively we can assume that the span of the union of the eigenspaces is the direct sum of these subspaces. We wish to show that . So let . Then where . Multiplying the above equality on the left by gives Subtracting gives As the eigenvalues are distinct, the coefficient is non-zero for each . Then Induction allows us to assume which together with the last equation implies for all . As we conclude finally that . Since was taken to be an arbitrary element of this shows the intersection must be zero, completing the proof.
More generally, let be a linear self-map on (as before). Without appeal to bases, we write for the linear span of the eigenvectors of in . Then by exactly the same argument one has
In the case , this yields a direct sum decomposition of into eigenspaces of , which is a very useful thing to know (in the cases it occurs). We will investigate this property in more detail next.