I know that a symmetric matrix with positive eigenvalues (i.e. a positive-definite matrix) is diagonalizable. But what if the matrix is not symmetric? Can we still diagonalize it? i.e. if $A$ is a matrix with all positive eigenvalues, can we write it as $A=P^{-1}DP$ where $D$ is a diagonal matrix with eigenvalues on the diagonal?

  • 4
    $\begingroup$ Nope. Example: $\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$. $\endgroup$
    – amsmath
    Commented Oct 10, 2017 at 15:31
  • 2
    $\begingroup$ It is true, however, that every symmetric matrix is diagonalizable. $\endgroup$ Commented Oct 10, 2017 at 15:34
  • 2
    $\begingroup$ And if the eigenvalues are distinct, the matrix can always be diagonalized as well. $\endgroup$
    – Peter
    Commented Oct 10, 2017 at 15:35
  • 2
    $\begingroup$ Yes, $(I + AB)$ is similar to $I + B^{1/2} A B^{1/2}$, which is self adjoint, so diagonalizable. So $I + A B$ is also diagonlizable. $\endgroup$
    – orangeskid
    Commented Oct 10, 2017 at 16:03
  • 2
    $\begingroup$ @Mah Note that adding $I$ to a matrix has absolutely no effect on diagonalizabilty, so your particular case boils down to diagonalizing $AB$ and has nothing to do with positivity of the eigenvalues. $\endgroup$
    – Erick Wong
    Commented Oct 10, 2017 at 20:55

1 Answer 1


The answer is no. As pointed out by amsmath in his comment, the matrix

$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \tag 1$

cannot be diagonalized, and a wider class of examples is provided by size $n$ square matrices of the form

$B = \lambda I + T, \tag 2$

where $I$ is the $n \times n$ identity matrix, and $T \ne 0$ is an $n \times n$ strictly upper triangular matrix, that is, $T$ takes the form

$T = [t_{ij}] \ne 0, \tag 3$


$t_{ij} = 0, \; i \le j. \tag 4$

We note that

$T^n = 0, \tag 5$

and this implies that the only eigenvalue of $B$ is $\lambda$, since if

$B \vec v = \mu \vec v \tag 6$

for some $\vec v \ne 0$, by (2) we would have

$(\mu - \lambda) \vec v= (B - \lambda I)\vec v = T\vec v, \tag 7$

and this in turn yields

$(\mu - \lambda)^n \vec v = (B - \lambda)^n \vec v = T^n \vec v = 0; \tag 8$

since $\vec v \ne 0$ we conclude that

$(\mu - \lambda)^n = 0, \tag 9$


$\mu = \lambda. \tag{10}$

Since $\lambda$ is the only eigenvalue of $B$, if some matrix $P$ diagonalized $B$ we would have

$\Lambda = PBP^{-1}, \tag{11}$

$\Lambda$ being the diagonal matrix

$\Lambda = \lambda I; \tag {12}$

inserting $B$ as in (2) into (11) yields

$\Lambda = P(\lambda I + T)P^{-1} = P(\lambda I)P^{-1} + PTP^{-1}$ $= \lambda PIP^{-1} + PTP^{-1} = \lambda I + PTP^{-1} = \Lambda + PTP^{-1}, \tag {14}$


$PTP^{-1} = 0, \tag{15}$

so that in fact

$T = 0. \tag{16}$

We have proved that if a matrix of the form (2) is diagonalizable then $T = 0$; thus, by contraposition, $T \ne 0$ implies $B$ cannot be diagonalied; in this way a great many instances of non-diagonalizable matrices may be constructed. Note that we do not require $\lambda > 0$.

Of course, if the eigenvalues of any matrix are all distinct, then it may be diagonalized, as is well-known.

Finally, it appears upon scrutiny of the above argument that we may relax the condition that $T$ is strictly upper triangular to the more general hypothesis that $T$ is nilpotent; that is

$T^m = 0 \tag{17}$

for some positive integer $m$. The essential details of the proof are left undisturbed by this assumption.


You must log in to answer this question.

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