If $A$ has two eigenvalues $\lambda _1, \lambda_2$ and $\dim (E_{\lambda_1})=n-1$, then $A$ is diagonalizable

Question

Suppose that $A \in M_{n\times n}(\Bbb F)$ has two distinct eigenvalues $\lambda_{1}$ and $\lambda_{2}$ and that $\dim (E_{\lambda_1})=n-1$ show that $A$ is diagnolizable.

score 4 · Answer 1 · answered Nov 18 '16 at 21:19

Being diagonalisable means that the sum of the different eigenspaces is equal to the whole space. It is a fundamental fact that this sum is always a direct sum, so its dimension is the sum of the dimensions of the eigenspaces; all that remains to show is that this sum attains the value$~n$. Now $\dim(E_{\lambda_1})=n-1$ is given, and $\dim(E_{\lambda_2})\geq1$ since otherwise $\lambda_2$ would not be an eigenvalue. This proves that $\dim(E_{\lambda_1})+\dim(E_{\lambda_2})\geq n$, which can only be equality, so one can also conclude $\dim(E_{\lambda_2})=1$.

I believe this is the easier way to think about it. It really is as easy as $n = n-1 + 1$. — NoseKnowsAll, Nov 19 '16 at 08:44

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

3

Hint: Since $\dim(E_{\lambda _1})=n-1$, there exist $v_1, \ldots , v_{n-1}$ linearly independent eigenvectors of $\lambda _1$. Let $v_n$ be an eigenvector of $\lambda _2$. Now consider the $n\times n$ matrix $P$ whose $i^{\text{th}}$column is $v_i$. The invertibility of $P$ follows from this.

Can you take it from here?

edited Apr 13 '17 at 12:20

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answered Dec 07 '13 at 14:00

Git Gud

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ya thanx ... :) – Amit Dec 08 '13 at 15:30
@Amit Glad to help. – Git Gud Dec 08 '13 at 15:42
Not sure how forming the matrix $P$ really helps; either you know (or argue) that the eigenspaces for $\lambda_1,\lambda_2$ form a direct sum, and that suffices, or you don't, in which case $P$ will not help you any further either. – Marc van Leeuwen Nov 18 '16 at 09:58
@MarcvanLeeuwen Something must be escaping me. At the very least I provided a matrix $P$ such that $AP=P\text{diag}(\lambda_1, \lambda_2)$. All that is left is to prove that $P$ is invertible... OK, I think I get you now, but I still think the matrix $P$ is useful, after all, providing $P$ is proving the required element for the definition of diagonalizability. Now I think I should explain what the eigenvectors are linearly independent for different eigenvalues. I'll do it later. Did I get your point correctly? – Git Gud Nov 18 '16 at 20:32
Well I guess the main confusion comes from the definition of being diagonalisable that is being assumed. For me it is either the existence of a basis of eigenvectors, or better the fact that the sum of the eigenspaces is the whole space, that gives the diagonalisable condition. But if instead you take the definition to require similarity to a diagonal matrix, then there is use to having a change-of-basis matrix $P$ to apply. – Marc van Leeuwen Nov 18 '16 at 21:26

score 0 · Answer 3 · answered May 07 '20 at 06:36

$\dim (E_{\lambda_1})=n-1$ indicates there are $(n-1)$ linearly independent eigenvectors, forming the set $\beta$, corresponding to $\lambda_1$.

$\lambda_2$ is an eigenvalue for $A$, implying there are at least $1$ eigenvectors, $v$, for $A$ corresponding to $\lambda_2$.

Then $\beta\cup\{v\}$ forms the set consisting of $n$ linearly independent vectors in $F^n$, and it can be the basis of $F^n$.

Because $F^n$ has the basis consisting of $A$'s eigenvectors, A is diagonalizable.

If $A$ has two eigenvalues $\lambda _1, \lambda_2$ and $\dim (E_{\lambda_1})=n-1$, then $A$ is diagonalizable

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