Show that a 2x2 matrix A is symmetric positive definite if and only if A is symmetric, trace(A) > 0 and det(A) > 0

Question

I need to show two parts of the implication are true. First: if $A$ is $2\times 2$ and is symmetric positive definite then $trace(A)>0$ and $\det(A)>0$.

Second: if $trace(A)>0$ and $\det(A)>0$ then $A$ is symmetric positive definite.

For the first part I was thinking: $A$ is symmetric and positive definite then $A$ has its eigenvalues positive. If $A$ is $2\times 2$ then characteristic polynomial of $A$ is $x^2−x.trace(A)+\det(A)=0$ If we compute the discriminant we get $(tr(A))^2 ≥ 4.\det(A)$ Now $tr(A)$ is squared so it is positive. How do I know that $\det(A)$ is also positive?

After copper.hat's response we argue that the eigenvalues of A are all positive because $A$ is spsd and the $\det(A)$ is the product of its eigenvalues. Thus $\det(A)$ is strictly positive.

Now, how do I verify the second part?

Thanks.

Answer 1

The classification of symmetric $2\times 2$ real matrices (or bilinear symmetric $2$-forms, or quadratic $2$-forms) through trace and determinant can be obtained in different ways, depending on the machinery one accepts. From more to less:

1) Spectral theorem. Then one knows the classification is done through eigenvalues. For instance positive semidefinite means two positive eigenvalues $\lambda>0,\mu>0$, which is equivalent to $\lambda\cdot\mu>0,\,\lambda+\mu>0$, that is determinant and trace both positive. Honestly, I think that for $2\times 2$ matrices this is too heavy.

2) Canonical forms. Any symmetric $2\times 2$ real matrix $A$ is equivalent to one of the following five canonical forms $$ \begin{pmatrix}1&0\\0&1\end{pmatrix},\, \begin{pmatrix}-1&0\\0&-1\end{pmatrix},\, \begin{pmatrix}1&0\\0&0\end{pmatrix},\, \begin{pmatrix}-1&0\\0&0\end{pmatrix},\, \begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$ The matrix $A$ shares with its canonical form the sign of the determinant (including being $0$). Thus we see that $\det>0$ immediately gives $A$ definite, and it remains to distinguish whether $A$ is positive or negative. In any case, the two entries in the diagonal of $A$ have the same sign, hence the sign of their sum, which is the trace of $A$. Thus $\det(A)>0$, tr$(A)>0$ means positive definite.

3) Nothing. In other words just from the definition. Let $A= \begin{pmatrix}a&b\\b&c\end{pmatrix}$.

Then the corresponding quadratic form is $q(x,y)=ax^2+2bxy+cy^2$, and we have to study the sign variations of this function for $(x,y)\ne(0,0)$. For instance $f(x,0)=ax^2>0$ if and only if $a>0$. Then if $y\ne0$ we can write: $$ \frac{1}{y^2}q(x,y)=at^2+2bt+c=P(t),\quad t=\frac{x}{y}, $$ and we discuss the signs of $P(t)$. For $t$ big enough, $P(t)>0$, since $a>0$. Then $P(t)>0$ for all $t$ means the polinomial has no zero, that is, its discriminant is negative, which gives $$ 0>\varDelta=b^2-ac=-\det(A). $$ And we get the condition $\det(A)>0$. Thinking this over one realizes this characterizes being positive semidefinite (that is, is a back and forth argument). And trace? Since $0<\det=ac-b^2$, and $a>0$, necessarily $c>0$ and trace$=a+c>0$.

Answer 2

What isn't quite proved here is 2nd part in number 3 of Jesus RS' answer, I think?

Anyhoo:

Proof:

1. Real symmetric is orthogonally diagonalisable. (I think we use just diagonalisable._ Let $A=QDQ^T$.

2. We want to show $x^TAx=v^TDv > 0$ if $v=Q^Tx \ne 0$ (iff $x \ne0$ since $Q^T$ is invertible (since $Q^T$ is orthogonal. again, I think just diagonalisable is fine))'.

3. For $D=diag[d_1,d_2]$, we have $v^TDv=d_1v_1^2+d_2v_2^2$.

4. We have $0 < tr(A)=tr(D) = d_1 + d_2$ and $0 < \det(A)=\det(D) = d_1 d_2$, i.e. $d_1,d_2>0$.

5. By (3) and (4), we have $v^TDv > 0$ if $v \ne 0$. QED

Answer 3

Well, this question is old but it seems to me that nobody answered the original question even if it has been seen 34 thousand times. Above answers use methods different from the question. However, it is quite simple to complete @AmirHosein reasoning without any quadratic form nor linear algebra theory :

Denote $d = \det(A), t = trace(A), \Delta = t^2-4d$. The equation roots are $\lambda_1 = (t+\sqrt{\Delta})/2, \lambda_2 = (t-\sqrt{\Delta})/2$.

We want $\lambda_1>0$ and $\lambda_2>0$, which is equivalent to $\lambda_1 \lambda_2>0$ and $\lambda_1 +\lambda_2>0$.

But $\lambda_1 \lambda_2 = d$ therefore $d>0$.

And $\lambda_1 + \lambda_2 = t$ therefore $t>0$.

We proceeded by equivalence therefore:

$\lambda_1$ and $\lambda_2>0$ is equivalent to $d>0$ and $t>0$.