My book writes the following statement:

For a square matrix $A$ of order $n$ to be diagonalizable, the sum of the dimensions of the eigenspaces must be equal to $n$. One way this can happen is when $A$ has $n$ distinct eigenvalues.

I'm a little confused on the last sentence. Are they trying to say that if $A$ has $n$ eigenvalues, then it must be the case that each eigenspace is $1$-dimensional? Couldn't it be possible for one or more of the eigenspaces to be $2$-dimensional or three-dimensional, etc?

  • $\begingroup$ The geometric multiplicity is always at most the algebraic multiplicity. If there are $n$ distinct eigenvalues, each has algebraic multiplicity of 1. $\endgroup$ – symplectomorphic Mar 2 '17 at 6:12
  • $\begingroup$ Put it another way, eigenvectors for different eigenvalues are linearly independent. So, the sum of dimensions of eigenspaces of a matrix must be bounded above by the dimension of the vector space. $\endgroup$ – user1551 Mar 2 '17 at 10:47

The intersection of the eigenspaces corresponding to different eigenvalues is always $\{0\}$. Indeed if $x \in E_m$, the eigenspace of $m$ and $x \in E_n$ then we have $Ax = mx$ and $Ax = nx$ so $(m-n)x = 0$ so, since $m \neq n$, $x$ must be $0$. Now let $d_i$ be the dimension of the eigenspace of eigenvalue $m_i$ with $i \in \{1 \ldots n\}$ then the only way $d_1+d_2 + \ldots +d_n = n$ is when all the $d_i = 1$.

  • $\begingroup$ It is not enough, really, that the intersection of the eigenspaces be zero: you need them to be in direct sum for this to work. $\endgroup$ – Mariano Suárez-Álvarez Mar 3 '17 at 17:05
  • $\begingroup$ I was only responding to the OP''s concern that the dimensions could be $>1$. $\endgroup$ – Marc Bogaerts Mar 3 '17 at 17:20
  • $\begingroup$ Well, the dimensions could all be equal to 2 and the intersection zero. For example, the intersection of the three coordinate planes in $R^3$ is zero. $\endgroup$ – Mariano Suárez-Álvarez Mar 3 '17 at 17:41
  • $\begingroup$ I thought to have proven that for any two eigenspaces their intersection was zero.? $\endgroup$ – Marc Bogaerts Mar 3 '17 at 17:45
  • $\begingroup$ In $\mathbb R^4$ consider for each $\lambda\in\mathbb R$ the subspace $V_\lambda=\langle e_1+\lambda e_2,e_3+\lambda e_4\rangle$. You can easily check that all of these subspaces have dimension $2$ and that their pairwise intersections are all trivial. $\endgroup$ – Mariano Suárez-Álvarez Mar 3 '17 at 23:29

Well if it has n distinct eigenvalues then yes, each eigenspace must have dimension one. This is because each one has at least dimension one, there is n of them and sum of dimensions is n, if your matrix is of order n it means that the linear transformation it determines goes from and to vector spaces of dimension n.

If you have 2 equal eigenvalues then no, you may have a eigenspace with dimension greater than one.

  • $\begingroup$ But even if A is nxn, couldn't you have more than n eigenvectors? Like let's say A is 3x3, couldn't you have two two-dimensional eigenspaces? It's not like I'm talking about having a four-dimensional eigenspace, just two two-dimensional ones $\endgroup$ – dagny Mar 2 '17 at 6:14
  • $\begingroup$ You must remember that each pair of eigenspaces is "disjoint" so if your space is of dimension 3 you cannot have 2 spaces of dimensions 2 without them intersecting in other point than zero and then not being disjoint. To see why they are disjoint remember that the eigenvalues determine a partition of the vector space. $\endgroup$ – Jonathaniui Mar 2 '17 at 6:21
  • $\begingroup$ Thank you, can you expand on why the eigenvalues determine a partition of the vector space? I'm afraid I don't know what a partition of a vector space is. $\endgroup$ – dagny Mar 2 '17 at 6:48
  • $\begingroup$ Let $a_1, \dots , a_n $ be the eigenvalues, the eigenspace associated to $a_i $ is $Ker (f-a_i I)$, now we have to see that 2 eigenspaces only intersect in zero, lets suppose $v \in Ker (f-a_i I)\cap Ker (f-a_j I) $ then $(f-a_i I)(v)=f (v)-a_i v)=0=(f- a_j)(v)=f (v)-a_j v $ ths implies that $v=0$ because qe are supposing $a_i $ is distinct from $a_j $ then we have that different eigenvalues give different eigenspaces and that 2 eigenspaces from different eigenvectors have intersection zero. A partition means that it divides the vector space in disjoint pieces. Hope it helps. $\endgroup$ – Jonathaniui Mar 2 '17 at 13:50
  • $\begingroup$ @Jonathaniui, it is not enough to show that the pairwise intersections are trivial to be ble to conclude. See the comments on Marc´s answer. $\endgroup$ – Mariano Suárez-Álvarez Mar 3 '17 at 23:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.