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Hey everyone here's the problem:

Let V be a vector space with dim(V)=n

For a particular linear transformation,f, we are given that there are two distinct eigenvalues, λ1 and λ2, with corresponding eigenspaces, E(λ1) and E(λ2).

dim(E(λ1))=m, dim(E(λ2))=n-m

I'm just struggling to figure out why the Jordan normal form of this linear transformation is diagonal and what it looks like. Would anyone be able to give me a reason that the Jordan normal form of this linear transformation is strictly diagonal?

Thanks!

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The Jordan form is diagonal if the matrix has a basis of eigenvectors. If you have a basis $\{ x_1,x_2,\cdots x_m \}$ of $E(\lambda_1)$ and a basis $\{ y_1,y_2,\cdots,y_{n-m}\}$ of $E(\lambda_2)$, then the union of these two bases is a basis of the full space because, if $$ (\alpha_1 y_1 + \alpha_2 y_2 + \cdots+\alpha_mx_m)+(\alpha_{m+1}y_1+\alpha_{m+2}y_2+\cdots+\alpha_ny_{n-m}) = 0, $$ then you can apply $A-\lambda_2 I$ to the above in order to conclude $$ (\lambda_1-\lambda_2)(\alpha_1 y_1 + \alpha_2 y_2 + \cdots+\alpha_mx_m) = 0. $$ Because $\lambda_1-\lambda_2 \ne 0$ then $\alpha_1 y_1+\alpha_2 y_2+\cdots+\alpha_m y_m=0$, which means $\alpha_1=\alpha_2=\cdots=\alpha_m = 0$ because $\{ y_j \}$ is a linearly independent set of vectors. The same argument gives $\alpha_{m+1}=\alpha_{m+2}=\cdots=\alpha_{n}=0$. Once you have a basis of eigenvectors for $A$, then $A$ is similar to a diagonal matrix. So the Jordan form must also be diagonal.

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  • $\begingroup$ This is really helpful! So once I justify that the two eigenspaces allow me to form a basis of eigenvectors for the whole of the vector space V then I can say that the Jordan normal form is diagonal? Is that kind of like an A=(P^-1)DP type thing? $\endgroup$ – user263626 Aug 22 '15 at 4:06
  • $\begingroup$ @Brock : Yes. Jordan form is invariant under similarity transformations. And a diagonal form is a Jordan form. $\endgroup$ – Disintegrating By Parts Aug 22 '15 at 11:28

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