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Take a different real matrix,Its only eigenvalue is $2$, with algebraic multiplicity two. However, the polynomial $q(t)=t-2$, which vanishes at that eigenvalue, gives $q(B)=N\ne 0$. Merely checking values at eigenvalues cannot prove a matrix identity. The full characteristic polynomial does work:The matrix is a Jordan block, and its minimal polynomial is also $(t-2{)}^2$.

The binomial theorem gives, for positive integers $n$,Terms containing $N^2$ or a higher matrix power vanish. After removing the common growth factor $2^n$, the remaining linear map is the shear mapping $I+(n/2)N$. It moves $(x,y)$ to $(x+ny/2,y)$. The single eigenvalue controls the exponential scale, while the nilpotent matrix adds a factor growing linearly in $n$.

In terms of polynomial interpolation, reduction modulo $(t-2{)}^2$ retains both a value and a derivative:Indeed, polynomial division gives $f(t)=f(2)+f^{\prime }(2)(t-2)+(t-2{)}^{2}h(t)$. This is a simple Hermite interpolation condition. For $f(t)=t^n$, it gives exactly the displayed matrix power. Repeated eigenvalues retain information about how a polynomial changes, not just where it vanishes.

Consider the diagonal matrix $D=2I_2$ and a Jordan blockBoth have characteristic polynomial $(t-2{)}^2$, trace $4$, determinant $4$, and eigenvalue $2$ with algebraic multiplicity two. But $\ker (D-2I)$ is the whole plane, whereas $(J-2I)(a,b{)}^T=(b,0{)}^T$ vanishes only when $b=0$. Their eigenspaces have different dimensions, so eigenvectors under similarity rules out matrix similarity. There is an even shorter proof: $S^{-1}(2I)S=2I$ for every invertible matrix $S$.

The missing information is how the linear operator acts beyond its eigenvectors. Here $D-2I=0$, while $J-2I\ne 0$ but $(J-2I{)}^2=0$. Their minimal polynomials are respectively $t-2$ and $(t-2{)}^2$. Over a field where the characteristic polynomial splits, the full sizes of the Jordan blocks, encoded by the Jordan normal form, determine the matrix similarity class.

An eigenvalue is simple when it has algebraic multiplicity one.

A finite-dimensional linear operator has simple spectrum when each eigenvalue has algebraic multiplicity one. For a Hermitian matrix, this means its eigenvalues are all distinct, so eigenvectors chosen for different eigenvalues are automatically orthogonal.

For $u=(1,\dots ,1{)}^T$, $J_n=u{u}^T$ maps $v$ to $({\sum }_i{v}_i)u$. Therefore $u$ is an eigenvector with eigenvalue $n$, while the zero-sum vector subspace is its kernel. For $n\ge 2$ a basis of the kernel is $e_1-e_n,\dots ,e_{n-1}-e_n$. These vectors and $u$ form an eigenbasis, proving diagonalization and characteristic polynomial $t^{n-1}(t-n)$. Both algebraic multiplicity and geometric multiplicity are one at $n$ and $n-1$ at zero. At $n=1$ the sole eigenvalue is 1; the zero-sum space has dimension zero and contributes no additional root.