Problem: Suppose $A \in \mathbb{R}^{n \times n}$ has eigenvalues $\{\lambda_1,....,\lambda_n\} \subset \mathbb{C}$ (not necessarily distinct) where $Re\{\lambda_i\} \le Re\{\lambda_{i+1}\}$ for $i \in \{1,...,n-1\}$. Show that for any $k \in \{1,...,n-1\}$ there exists an invertible matrix $T \in \mathbb{R}^{n \times n}$ such that:

$TAT^{-1}=\begin{bmatrix} A_1 & 0 \\ 0 & A_2 \end{bmatrix}$, where the matrix $A_1 \in \mathbb{R}^{k \times k}$ has eigenvalues $\{\lambda_1,....,\lambda_k\} \subset \mathbb{C}$ and $A_2 \in \mathbb{R}^{(n-k) \times (n-k)}$ has eigenvalues $\{\lambda_{k+1},....,\lambda_n\} \subset \mathbb{C}$.

Attempted solution: We know that $A$ is similar to a real Jordan canonical form that has this diagonal block-like structure. Moreover, we can identify the eigenvalues associated with each block by considering the characteristic polynomial, which can be found by taking advantage of the Jordan form structure when evaluating the determinant.

However, the exact sizes of the blocks depend upon the algebraic multiplicity and geometric multiplicity of the eigenvalues. Therefore it seems difficult to show the sizes of the block diagonals are such that $A_1 \in \mathbb{R}^{k \times k}$ and $A_2 \in \mathbb{R}^{(n-k) \times (n-k)}$. Furthermore, in most textbooks, it is shown that $A$ is similar to some Jordan form, ignoring the order of eigenvalues. Of course, once we transform onto a Jordan form we could permute the rows to recapture the order the eigenvalues should appear in but then we will lose the block-like structure.

According to this question, it does seem that the order in which the Jordan blocks appear does not matter since they're all similar:

Does the Jordan form of the matrix depend on where the Jordan blocks are placed?

Question: Is there an alternative way to prove this (possibly without using Jordan canonical forms)? This would allow us to avoid reproducing the same/similar argument used in showing the existence Jordan canonical forms, worrying about the order in which the eigenvalues for certain blocks appear and how the algebraic and geometric multiplicity of each eigenvalue affects the block sizes.

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    $\begingroup$ You could use the "primary decomposition theorem" or "rational canonical form" instead of Jordan form $\endgroup$ Commented Feb 18, 2022 at 17:13
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    $\begingroup$ Also, I don't understand why you believe that permuting the rows and columns to reorder the eigenvalues could lead to "losing the block-like structure" $\endgroup$ Commented Feb 18, 2022 at 17:14
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    $\begingroup$ Would you be willing to use the Cayley Hamilton theorem? $\endgroup$ Commented Feb 18, 2022 at 17:20
  • $\begingroup$ I meant loosing the block structure because $\begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 &2 \\ 0 & 0 & 2\end{bmatrix}$ permutes to $\begin{bmatrix} 0 & 1 & 2 \\ 2 & 0 & 0\\ 0 & 0 & 2\end{bmatrix}$ which no longer has the nice block diagonal structure. Maybe you could permute before doing the Jordan norm? Yes I would be willing to use the Cayley Hamilton theorem. $\endgroup$ Commented Feb 18, 2022 at 17:29
  • $\begingroup$ Maybe you can also permute the columns to get the block structure back? It would be nice if all the heavy lifting in the proof was done elsewhere and referenced to get the result. $\endgroup$ Commented Feb 18, 2022 at 17:36

1 Answer 1


Let $S_1 = \{\lambda_1,\dots,\lambda_k\},S_2 = \{\lambda_{k+1},\dots,\lambda_m\}$ be disjoint sets comprising the (distinct!) eigenvalues of $A$ such that $\lambda \in S_1 \implies \bar \lambda \in S_1$. Let $p(x) = (x-\lambda_1)\dots(x-\lambda_k)$ and $q(x) = (x-\lambda_{k+1})\dots(x-\lambda_m)$. Because of the way that $S_1,S_2$ were defined, both $p,q$ are polynomials with real coefficients.

As a consequence of the Cayley-Hamilton theorem, it holds that $[p(A)]^n[q(A)]^n = 0$. Note that $U_1 = \ker(p(A)),U_2 = \ker(q(A))$ are invariant subspaces of $\Bbb R^n$.

Claim: $\Bbb R^n = U_1 \oplus U_2$.

Proof of claim: Note that because $p,q$ are relatively prime, there exist polynomials $f,g$ such that $f(x)p(x) + g(x)q(x) = 1$, from which it follows that $f(A)p(A) + g(A)q(A) = I$.

To see that $U_1,U_2$ are disjoint (i.e. have intersection $\{0\}$), note that if $v \in U_1 \cap U_2$, it follows that $$ v = Iv = [f(A)p(A) + g(A)q(A)]v = f(A)[p(A)v] + g(A)[q(A)v] = 0. $$ To see that $U_1 + U_2 = \Bbb R^n$, note that any $v$ can be decomposed into $$ v = q(A)g(A)v + p(A)f(A)v. $$ Because $p(A)q(A) = q(A)p(A) = 0$, it is easy to see that $q(A)g(A)v \in U_1$ and $p(A)f(A)v \in U_2$. $\square$

Now, if $v_1,\dots,v_d$ is a basis of $U_1$ and $v_{d+1},\dots,v_n$ is a basis of $U_2$, it follows that the matrix of $A$ relative to the basis $\{v_1,\dots,v_n\}$ has the desired block-diagonal form.

  • $\begingroup$ Nice answer! So you're saying you need $S_1$ to be such that it contains the pair $(\lambda,\bar{\lambda})$? For instance the result does not hold if $\lambda_k=5 +2i$ and $\lambda_{k+1}= 5 -2i$? $\endgroup$ Commented Feb 18, 2022 at 18:14
  • $\begingroup$ @MorganJones That's right. Otherwise, we have no guarantee that there is a suitable matrix $T$ with real entries. $\endgroup$ Commented Feb 18, 2022 at 18:16
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    $\begingroup$ @MorganJones Also, note the further point that $S_1,S_2$ must be disjoint. For example, the matrix $$ A = \pmatrix{0&1&0\\0&0&1\\0&0&0} $$ with eigenvalues $0,0,0$ cannot be brought into a block-diagonal form. $\endgroup$ Commented Feb 18, 2022 at 18:18
  • $\begingroup$ Does the part $[p(A)]^n [q(A)]^n=0$ follow because the characteristic polynomial of $A$, $P_A$, has roots $\lambda_1,...,\lambda_n$. So by the fundamental theorem of algebra $P_A(\lambda)=\Prod_{i=1}^n (\lambda-\lambda_i)= p(\lambda)q(\lambda)$. Then by the Cayley Hamilton theorem $0=P_A(A)=p(A)q(A)$. $\endgroup$ Commented Feb 18, 2022 at 18:20
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    $\begingroup$ @MorganJones Your argument is not quite correct: it holds that $P_A(\lambda)$ divides $p(\lambda)q(\lambda)$, but not that it is equal. $\endgroup$ Commented Feb 18, 2022 at 18:29

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