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Given, $$A = \begin{bmatrix} 1 &1 & 1\\ 1&1 & 1\\ 1& 1& 1 \end{bmatrix}.$$

I want to see if the matrix $A$ positive (negative) (semi-) definite.

Using Method 1:

Define the quadratic form as $Q(x)=x'Ax$.

Let $x \in \mathbb{R}^{3}$, with $x \neq 0$.

So, $Q(x)=x'Ax = \begin{bmatrix} x_{1} &x_{2} &x_{3} \end{bmatrix} \begin{bmatrix} 1 &1 & 1\\ 1&1 & 1\\ 1& 1& 1 \end{bmatrix} \begin{bmatrix} x_{1}\\x_{2} \\x_{3} \end{bmatrix}$.

After multiplying out the matrices I am left with $$Q(x) = x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2(x_{1}x_{2} + x_{1}x_{3}+x_{2}x_{3}).$$

So by Method 1, I get that $A$ is positive definite.

Using Method 2:

I calculate all principle minors $A$ and if they're all positive, then the matrix is positive definite (learned recently from @hermes).

So $|A_{1}| =1> 0$, $|A_{2}| = 0$, and $|A_{3}| = |A| = 0$.

So $A$ is positive semi-definite.

Which method am I making a mistake?

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The matrix is actually positive semi-definite. All your calculations are correct; only the conclusion in Method 1 is incorrect.

Note that for Method 1, you got $$Q(x)=x_1^2+x_2^2+x_3^2+2(x_1x_2+x_1x_3+x_2x_3)=(x_1+x_2+x_3)^2\ge0$$ with equality whenever $x_1+x_2+x_3=0$.

For a solid example, you can take $x=(-1,0,1)$ to get $Q(x)=0$.

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First is wrong. $Q(x)$ doesn't mean $A$ is positive definite.

Moreover $rank(A)=1$. Since $A$ is not of full rank, it can not be positive definite.

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What about a Method 3)? Your matrix has rank $1$ and its eigenvalues are $3,0,0$. Since it is a symmetric matrix, it is positive semi-definite.

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  • $\begingroup$ Didn't know about the method but good point. Could you please elaborate on Method 3 a bit. Like a step-by-step procedure on how to would implement it for this problem. $\endgroup$ – OGC Sep 3 '15 at 16:39
  • $\begingroup$ @OGC: well, that is a rank-$1$ matrix, hence two eigenvalues are zero. $(1,1,1)^T$ is an eigenvector associated with the eigenvalue $\lambda=3$. By the spectral theorem, your matrix is conjugated (through an orthogonal transformation) with a diagonal matrix having $3,0,0$ as diagonal elements, hence your matrix is positive semi-definite since the last matrix is. $\endgroup$ – Jack D'Aurizio Sep 3 '15 at 16:49

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