# What is a condition on my parameter for this matrix to be diagonalizable?

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?

• 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. Aug 19, 2020 at 4:44
• Do you count diagonalizability over $\mathbb C$ or just $\mathbb R$? Aug 19, 2020 at 4:46
• @erickwong It does not specify which, only saying b is a real number. Aug 19, 2020 at 5:00
• @WillJagy are the elements of N the coefficients of $b$? Aug 19, 2020 at 5:23
• @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 $B + I$. Aug 19, 2020 at 8:22

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 $$\textrm{rank}(\textbf{A}+\textbf{I}_n)=\textbf{rank} \left(\begin{array}{cc} 6\\ 3\\ -6 \end{array}\begin{array}{cc} -6\\ -3\\ 6 \end{array}\begin{array}{cc} 0\\ 0\\ 0 \end{array}\right)=1$$