$$A=\left(\begin{array}{ccccc}
1 &1 &\cdots &1 &1 \\
1 &0 &\cdots & 0 & 1\\
\vdots & & \ddots & & \vdots\\
1 & 0 & \cdots & 0 &1 \\
1 &1 &\cdots &1 &1
\end{array}\right)\in M_{n}(\mathbb{R})$$
It has 1's around it and 0's everywhere else, and it is of size $n$.
I need to decide whether this matrix is Diagonalizable,what's it's eigenvalues, eigenvectors and it's eigenspace (${v|Av=גv}$)
Ok, So first I want to make one thing clear to myself and get your approval for that, If a matrix have n different eigenvalues it means that it's diagonalizable for sure, but it's not "iff" right? I mean, in this case I can't decide that it's not Diagonalizable since I can clearly see that it doesn't have $n$ different eigenvalues, Is it true?
I notice that 1 and 0 are the only eigenvalues of this matrix since the Characteristic polynomial is $|XI-A|$ and $f_{A}(0)=|0I-A|=0$ and I would like to find it's eigenspace- Is it defined by $\dim \mathrm{Ker}(0I-A)$? and if so, It should be $n-1$, So does it actually say that it has $n-1$ independent eigenvectors for this eigenvector?
Another eigenvector is 1, in a similar way I get that $\dim \mathrm{Ker}(1I-A)$ is 1 and finally I get that it has base of $n$ independent eigenvectors and it is diagonalizable?
Thank you very much guys.
Best Answer
$A$ is diagonalizable because $A$ is symmetric. The eigenvalues and eigenvectors can indeed be found out using your method.