How to Compute the Semisimple Part of a Matrix

The semisimple part of a matrix is the part that accounts for its eigenvalues without its Jordan-block nilpotent terms. In the usual Jordan (or Jordan–Chevalley) decomposition, a matrix is written as $A=S+N$, where $S$ is diagonalizable, $N$ is nilpotent, and $SN=NS$.

The most direct method is to put $A$ in Jordan form, replace each Jordan block by its eigenvalue times the identity, and change back to the original basis. If you do not want to construct a Jordan basis, a polynomial in $A$ gives the same result.

Compute it from Jordan form

Assume first that $A$ is a complex square matrix. Choose an invertible matrix $P$ such that

$$P^{-1}AP=J.$$

Each Jordan block for an eigenvalue $\lambda$ has the form $J_k(\lambda)=\lambda I_k+U$, where $U$ has ones on the superdiagonal and is nilpotent. The semisimple part of this block is $\lambda I_k$; its nilpotent part is $U$. Thus, if $J$ has blocks for eigenvalues $\lambda_1,\ldots,\lambda_r$, replace each block $J_k(\lambda_i)$ by $\lambda_i I_k$ to get a block-diagonal matrix $D$. Then

$$S=PDP^{-1},\qquad N=A-S.$$

This construction gives $A=S+N$, with $S$ diagonalizable, $N$ nilpotent, and $SN=NS$. These properties characterize the decomposition uniquely over the complex numbers.

An equivalent description uses generalized eigenspaces. The space splits as a direct sum of generalized eigenspaces $V_\lambda=\ker(A-\lambda I)^m$ for sufficiently large $m$. On each $V_\lambda$, the semisimple part acts as multiplication by $\lambda$, while the nilpotent part acts as $A-\lambda I$.

Worked example

Consider

$$A=\begin{pmatrix}2&1&0\\0&2&0\\0&0&-1\end{pmatrix}.$$

This matrix is already in Jordan form. Its first two coordinates form a size-two Jordan block for eigenvalue $2$, and its last coordinate is a size-one block for eigenvalue $-1$. Replace the size-two block by $2I_2$ and keep the last block as $-1$:

$$S=\begin{pmatrix}2&0&0\\0&2&0\\0&0&-1\end{pmatrix}.$$

Subtract to find the nilpotent part:

$$N=A-S=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}.$$

Here $N^2=0$, and $S$ commutes with $N$. Therefore $A=S+N$ is the required decomposition. Notice that the off-diagonal entry in the Jordan block belongs to $N$, not $S$.

Compute it as a polynomial in A

The semisimple part can also be found without explicitly computing a Jordan basis. Let the minimal polynomial factor as

$$m_A(t)=\prod_{i=1}^r(t-\lambda_i)^{e_i}.$$

Find a polynomial $p(t)$ satisfying

$$p(t)\equiv\lambda_i\pmod{(t-\lambda_i)^{e_i}}\quad\text{for each }i.$$

Then $S=p(A)$. The congruences require more than just $p(\lambda_i)=\lambda_i$: they also force the higher-order terms to vanish on each Jordan block. The Chinese remainder theorem guarantees a solution, unique modulo the minimal polynomial. After computing $S$, take $N=A-S$.

For the example, $m_A(t)=(t-2)^2(t+1)$. A polynomial satisfying the required congruences is

$$p(t)=\frac{-t^2+4t+2}{3}.$$

Indeed, $p(t)-2$ is divisible by $(t-2)^2$, and $p(-1)=-1$. Evaluating it at the matrix gives

$$S=p(A)=\frac{-A^2+4A+2I}{3} =\begin{pmatrix}2&0&0\\0&2&0\\0&0&-1\end{pmatrix}.$$

This polynomial method is useful when you know the minimal polynomial or can compute polynomial expressions in $A$ more easily than a Jordan basis.

Real matrices and the meaning of diagonalizable

For a real matrix, it is common to define the semisimple part as the part diagonalizable over the complex numbers. The Jordan-form method can then be carried out over $\mathbb{C}$; for a real matrix, the resulting $S$ and $N$ are real matrices. If instead “semisimple” is being used to mean diagonalizable over $\mathbb{R}$, that is a stricter condition and may not apply when there are nonreal eigenvalues. Check which convention your course or text uses.

Common mistakes

• Keeping the Jordan off-diagonal entries in $S$. Those entries make a Jordan block fail to be diagonalizable; they belong to $N$.
• Using only eigenvalues to build the polynomial. The conditions $p(\lambda_i)=\lambda_i$ alone may not remove nilpotent terms. Use the full congruences modulo $(t-\lambda_i)^{e_i}$.
• Assuming $S$ is an orthogonal projection. It generally is not. It acts as a scalar on each generalized eigenspace, and need not be symmetric or idempotent.
• Checking only $A=S+N$. Also verify that $S$ is diagonalizable, $N$ is nilpotent, and $SN=NS$; the sum alone does not establish the Jordan decomposition.