I am looking for a condition on $b$ so that the matrix $B$ below can be diagonalized:

$$B=\begin{pmatrix}2b+1 &-2b-2 &b-2\\ 2b-1&-2b&b-2\\ -6&6&-1\end{pmatrix}.$$

I know that if $B$ has $3$ linearly independent eigenvectors, $B$ is diagonalizable, but I want to avoid going all the way to get eigenvalues and eigenvectors of $B$.

Do we have any other easier ways?

  • 2
    $\begingroup$ there is a way to do it based on a Jordan Chevalley type decomposition. You have $B= R + bN,$ where $R,N$ are constant integer matrices, $N$ is nilpotent, we really do have $RN=NR.$ The reasonable approach is to find $P$ such that $J = P^{-1}RP$ is its Jordan form, then find $P^{-1}NP$ which is still nilpotent, finally put variable $b$ back and calculate $P^{-1}(R+bN)P$ and see when, if ever, it is diagonal. That's what I did. $\endgroup$
    – Will Jagy
    Aug 19, 2020 at 4:44
  • $\begingroup$ Do you count diagonalizability over $\mathbb C$ or just $\mathbb R$? $\endgroup$
    – Erick Wong
    Aug 19, 2020 at 4:46
  • $\begingroup$ @erickwong It does not specify which, only saying b is a real number. $\endgroup$ Aug 19, 2020 at 5:00
  • $\begingroup$ @WillJagy are the elements of N the coefficients of $b$? $\endgroup$ Aug 19, 2020 at 5:23
  • 2
    $\begingroup$ @Will If you've already made the observation that $B = R + bN$, where $RN = NR$ and $N$ is nilpotent, then we already know that the eigenvalues of $B$ are the eigenvalues of $R$, so it suffices to consider the eigenspace of $B$ associated with the repeated eigenvalue $\lambda = -1$ of $R$. In other words, it suffices to consider the rank of $B + I$. $\endgroup$ Aug 19, 2020 at 8:22

3 Answers 3


This may not be what you want, but the characteristic polynomial of $B$ is $$x^3-3x-2=(x-2)(x+1)^2,$$ independent of $b$, so a necessary and sufficient condition for $B$ to be diagonalizable is when the minimal polynomial of $B$ is $(x-2)(x+1)$. Using this, it can be checked directly that the condition for $B$ to be diagonalizable is $$b=2.$$


Observe that the first two rows of $B$ are nearly identical and the first two columns of $B$ are almost the negative of each other. This motivates one to inspect the following matrix that is similar to $B$: $$ C=\pmatrix{1&-1&0\\ 0&1&0\\ 0&0&1} \pmatrix{2b+1&-2b-2&b-2\\ 2b-1&-2b&b-2\\ -6&6&-1} \pmatrix{1&1&0\\ 0&1&0\\ 0&0&1} =\left(\begin{array}{c|cc}2&0&0\\ \hline2b-1&-1&b-2\\ -6&0&-1\end{array}\right). $$ The only repeated eigenvalue of $C$ is the eigenvalue $-1$ of multiplicity $2$. Hence $B$ and $C$ are diagonalisable if and only if $\operatorname{nullity}(C+I)=2$, i.e. if and only if $b=2$.


Defintion. Rank of real matrix $\textrm{A}$ (symb. $\textrm{rank}\left(\textbf{A}\right)$) is the number of lineary independent columns (or rows) of $\textrm{A}$.

Example. If $$ A=\left(\begin{array}{cc} 1\\ 1\\ 1\\ 1 \end{array}\begin{array}{cc} 1\\ 2\\ 3\\ 1 \end{array}\begin{array}{cc} 2\\ 1\\ 0\\ 0 \end{array} \begin{array}{cc} 0\\ 1\\ 2\\ 0 \end{array} \begin{array}{cc} -1\\ 1\\ 3\\ 4 \end{array} \right). $$ The columns are $A_1=(1,1,1,1)$, $A_2=(1,2,3,1)$, $A_3=(2,1,0,0)$, $A_4=(0,1,2,0)$, $A_5=(-1,1,3,4)$. With elementary operations (Gauss operations) this can be writen as $$ \begin{array}{cc} A_1\\ A_2\\ A_3\\ A_4\\ A_5 \end{array} \left|\begin{array}{cc} 1\\ 1\\ 2\\ 0\\ -1 \end{array}\begin{array}{cc} 1\\ 2\\ 1\\ 1\\ 1 \end{array}\begin{array}{cc} 1\\ 3\\ 0\\ 2\\ 3 \end{array}\begin{array}{cc} 1\\ 1\\ 0\\ 0\\ 4 \end{array}\right|\equiv\left|\begin{array}{cc} 1\\ 0\\ 0\\ 0\\ 0 \end{array}\begin{array}{cc} 1\\ 1\\ 0\\ 0\\ 0 \end{array}\begin{array}{cc} 1\\ 2\\ 0\\ 0\\ 0 \end{array}\begin{array}{cc} 1\\ 0\\ 1\\ 0\\ 0 \end{array}\right| $$ Hence $\textrm{rank}(\textbf{A})=3$

Theorem.(see rank-nullity theorem) A square $n\times n$ matrix $\textbf{A}$ with real elements is diagonalizable in $\textbf{R}$ if and only if all the roots of the characteristic polynomial $X_{\textbf{A}}(t)=\textrm{det}(\textbf{A}-t\cdot \textbf{I}_n)$ are real numbers $t_1,t_2,\ldots,t_{r}$ (eigenvalues) and the multiplicity $m_1,m_2,\ldots,m_r$ of each eigenvalue equals the dimension of the of the eigenspace (the space $E(\lambda_j)$ of proof below) i.e. iff $\textrm{rank}(A-t_j\textbf{I}_n)=n-m_j$, $j=1,2,\ldots,r$.

Proof. Let $\lambda_j$ be eigenvalue of $A$ with algebraic multiplicity $m_j$. We define the space $$ E(\lambda_j)=\left\{X\in\textbf{R}^n:\textbf{A}X^T=\lambda_j X^T\right\}. $$ It is known that if $V$ is any $\textbf{R}-$vector space and $f$ a linear map of $V$, then $f$ is diagonalizable if and only if all roots of the characteristic polynomial of $f$ are in $\textbf{R}$ and the multiplicity $m_j$ of every eigenvalue $\lambda_j$ is equal to the dimension of $E(\lambda_j)$. Obviously we can attach to $f$ its matrix $\textbf{A}$ and the oposite. Hence $\textbf{A}$ is diagonalizable iff $\textrm{dim}(E(\lambda_j))=m_j$, $\forall j$. But from rank-nullity theorem for the matrix $\textbf{M}_j=\textbf{A}-\lambda_j\textbf{I}_n$, we have $\textrm{rank}(\textbf{M}_j)+\textrm{Nullity}(\textbf{M}_j)=n$. Hence we get equivalently the condition $$ \textrm{dim}(E(\lambda_j))=\textrm{Nullity}(\textbf{M}_j)=n-\textrm{rank}(\textbf{A}-\lambda_j\textbf{I}_n)=m_j $$
and the proof follows.

Example. In your question you have $X_{\textbf{B}}(t)=-(t-2)(t+1)^2$. Hence $t_1=2$ with multiplicity $m_1=1$ and $t_2=-1$ with multiplicity $m_2=2$. But as someone can see $\textrm{rank}(\textbf{A}-2\textbf{I}_n)=2$ and if $b\neq 2$, then $\textrm{rank}(\textbf{A}-(-1)\textbf{I}_n)=2\neq 3-2$. Hence $\textbf{B}$ is not diagonazible for any $b\neq 2$. However for $b=2$, we have $$ \textrm{rank}(\textbf{A}+\textbf{I}_n)=\textbf{rank} \left(\begin{array}{cc} 6\\ 3\\ -6 \end{array}\begin{array}{cc} -6\\ -3\\ 6 \end{array}\begin{array}{cc} 0\\ 0\\ 0 \end{array}\right)=1 $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .