Diagonalise without finding eigenvalues

Question

I am asked to find the Jordan normal form (in this case, diagonalise) the $n\times n$ matrix $M$ defined:

$$M_{ij}=1+\delta_{ij}\,x$$ I am then asked to deduce the minimal polynomial, eigenvalues and their multiplicities. I believe the answer to the first part is $M'_{ij}=\delta_{ij} (x+\delta_{in}n)$, but I cannot see how to arrive at this result without using the material I am meant to deduce from the first part. Can anyone help?

Answer 1

Denote by $E$ the $(n\times n)$-matrix having all entries $=1$. $E$ has rank $1$; therefore the vector space $K:=\{x\in{\mathbb R}^n \>|\> Ex=0\}$ has dimension $n-1$. Choose a basis $(f_2,\ldots,f_n)$ of $K$, and put $f_1:=(1,1,\ldots,1)\notin K$. Then $(f_1,f_2,\ldots, f_n)$ is a basis of ${\mathbb R}^n$. The matrix $\bigl[E_{ik}\bigr]$ of $E$ with respect to this basis has $E_{11}=n$ and $E_{ik}=0$ otherwise. It follows that the matrix of $M=E+x I$ with respect to this basis is given by $$\bigl[M\bigr]={\rm diag}(n+x,x,\ldots, x)\ ,$$ and this is obviouly the Jordan normal form of $M$.

Answer 2

To diagonalise the given matrix $M$, we need to know its eigenvalues. So we start with finding the eigenvalues, thus solving $M\mathbf{v}=\lambda\mathbf{v}$.

Rewriting: $$O_n\mathbf{v}+xI_n\mathbf{v}=\lambda\mathbf{v}$$ Here $O_n$ is an $n\times n$-matrix with all entries $1$ and $I_n$ is the $n \times n$ unit matrix. Rearranging gives: $$O_n\mathbf{v}=(\lambda-x)\mathbf{v}$$

But $O_n\mathbf{v}$ is just a $n$-dimensional vector with all entries the same: the sum of all elements from $\mathbf{v}$. So if $\mathbf{v}$ is an eigenvector, it must have all entries equal. Because eigenspaces are linear, we are allowed to assume that $\mathbf{v}=(1,1,\dots,1)^t$, if the sum of elements from $\mathbf{v}$ is nonzero. The equation simplifies to

$$n\mathbf{v}=(\lambda-x)\mathbf{v}$$

This gives an eigenvalue of $\lambda=x+n$.

If the sum of elements from $\mathbf{v}$ is zero, then $O_n\mathbf{v}=\mathbf{0}$, which yields an eigenvalue $\lambda=x$.

Inserting $\lambda=x$ into the matrix $M-\lambda I_n$ gives $O_n\mathbf{v}=\mathbf{0}$, but because all $n$ equalities are the same, this is one equality in $n$ unknowns, which gives a multiplicity of $n-1$.

For $\lambda=x+n$ we know that $\mathbf{v}=(1,1,\dots,1)^t$, so this eigenvalue has multiplicity $1$.

Therefore the minimal polynomial is $p(\lambda)=(\lambda-x)^{n-1}(\lambda-x-n)$.

The last step to diagonalize the given matrix, is to construct an eigenvector basis. We already know one eigenvector $(1,1,\dots,1)^t$.

Now we have to calculate the other $n-1$ eigenvectors. They are $(1,-1,0,\dots,0)$, $(1,0,-1,0,\dots,0)$, $(1,0,0,-1,0,\dots,0)$ etc., and finally $(1,0,\dots,0,-1)$. These are $n-1$ vectors, linear independent. We can insert these $n$ eigenvectors into a mapping matrix, and have to invert this matrix. This matrix is invertible, because all columns are linearly independent.

While writing this answer, I saw that Algebraic Pavel was quicker, but I hope this answer is helpful. Because the mapping matrix is usually the matrix with the eigenvectors, I don't know a method without finding eigenvalues. Eigenvalues are very important to diagonalize a matrix.