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I have a symmetric matrix where all non-diagonal elements are positive and identical, and all diagonal elements are identical as well. For example, the $3 \times 3$ version of this matrix has the following form: $$ \left( \begin{array}{ccc} 2a+b & a & a \\ a & 2a+b & a \\ a & a & 2a+b \end{array} \right) $$

Note that $a>0\ , b>0$. For such a simple form, is there an easy way of determining that the above matrix is positive definite in the general $n \times n$ case? I'd like to show that the matrix is still positive definite when the dimension is higher. Thank you.

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    $\begingroup$ The criterion of Sylvester is well-adapted to your case. $\endgroup$
    – Giuseppe Negro
    May 24, 2018 at 21:37
  • $\begingroup$ @GiuseppeNegro Beat me to it. $\endgroup$
    – Kendall
    May 24, 2018 at 21:53

3 Answers 3

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Yes. Your matrix can be written as $$(a+b)I + a ee^T$$ where $I$ is the identity matrix and $e$ is the vector of ones. This is a sum of a symmetric positive definite (SPD) matrix and a symmetric positive semidefinite matrix. Hence it is SPD.

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  • $\begingroup$ Thank you! This is very helpful $\endgroup$
    – SelinH
    May 24, 2018 at 21:57
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    $\begingroup$ It would be cool to actually show that $ee^T$ is positive semidefinite: $x^T e e^T x = (e^Tx)^T(e^Tx)$; since $e^Tx$ is a scalar $(e^T x)^T = e^T x$ whence $x^T e e^T x = (e^T x)^2 \ge 0$. Endorsed! Cheers! $\endgroup$
    – Robert Lewis
    May 24, 2018 at 23:18
  • $\begingroup$ @SelinH You are very welcome. $\endgroup$
    – Carl Christian
    May 25, 2018 at 9:37
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    $\begingroup$ @RobertLewis Thank your for your kind words. I often find it difficult to select the right level of detail. $\endgroup$
    – Carl Christian
    May 25, 2018 at 9:43
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In addition to Carl's answer (+1) you can use the Gershgorin circle theorem which is one I'll often try when I see diagonally dominant matrices like this one.

For each row, the sum of the absolute values of the off-diagonal entries is $2a$. The theorem then says that every eigenvalue is in a closed disc of radius $2a$ centered at the diagonal elements $2a + b$, so for any eigenvalue $\lambda$ we have $\lambda \geq b > 0$.

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$$\begin{bmatrix} 2a+b & a & a\\ a & 2a+b & a\\ a & a & 2a+b\end{bmatrix} = (a+b) \, \mathrm I_3 + a \, 1_3 1_3^\top = (a+b) \, \mathrm I_3 + 3a \left(\frac{\, 1_3}{\sqrt{3}}\right) \left(\frac{\, 1_3}{\sqrt{3}}\right)^\top$$

Hence, the eigenvalues are $4a + b$ and $a+b$, with multiplicities $1$ and $2$, respectively. Since $a, b > 0$, both eigenvalues are positive, i.e., the given symmetric matrix is positive definite.

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