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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?

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  • $\begingroup$ 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
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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 $d_i = 1$.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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