# Is A diagonalizable?

$$A=\left(\begin{array}{ccccc} 1 &1 &\cdots &1 &1 \\ 1 &0 &\cdots & 0 & 1\\ \vdots & & \ddots & & \vdots\\ 1 & 0 & \cdots & 0 &1 \\ 1 &1 &\cdots &1 &1 \end{array}\right)\in M_{n}(\mathbb{R})$$

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.

• The matrix A is diagonalizable if and only if the sum of the dimensions of its eigenspaces is n. You seem to say that 0 and 1 are the only eigenvalues so you have to check that multiplicity of 0 plus multiplicity of 1 equals n. Now the eigenspace of 0 is defined by the equations x(1)+...+x(n)=x(1)+x(n)=0 hence it has dimension n-2. It remains to determine whether the eigenspace of 1 has dimension 2.
– Did
Jul 20, 2011 at 12:49
• Nir: while it doesn't really help for this question, a quick sanity check shows that it can't be true that the matrix is diagonalizable and has only $0$ and $1$ as eigenvalues. For suppose $A = TDT^{-1}$ with $D$ diagonal and only ones and zeros on the diagonal. Then $D^2 = D$ gives $A^2 = TDT^{-1}TDT^{-1} = TD^2T^{-1} = TDT^{-1} = A$. Considering the top left entry of $A^2$ you see that it is $n$, so $A^2 = A$ can't be true.
– t.b.
Jul 20, 2011 at 20:54

$A$ is diagonalizable because $A$ is symmetric. The eigenvalues and eigenvectors can indeed be found out using your method.

• We can always say that a matrix is diagonalizable if it's symmetric?
– user6163
Jul 20, 2011 at 12:59
• Yes, normality suffices. Jul 20, 2011 at 13:00
• en.wikipedia.org/wiki/…
– Did
Jul 20, 2011 at 13:00
• Thank you very much Didier and Sunni.
– user6163
Jul 20, 2011 at 13:07

I notice that 1 and 0 are the only eigenvalues

This is wrong, try with $n=2$ for example, 2 is quite clearly an eigenvalue of $M$. Besides the matrix has rank 2, therefore its null space has dimension $n-2$ and 0 has multiplicity $n-2$ as an eigenvalue, not $n-1$. Sunni has answered for the diagonalizibility.

The other two eigenvalues have sum 2 (trace) and product 4-2n. This can be done either "directly" or by noticing that this product is the coefficient of $x^{n-2}$ in the characteristic polynomial, and computing it via the sum of permutations formula (the only permutations contributing being identity and transpositions).