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

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.

• The determinant is the product of the eigenvalues. Mar 3 '15 at 18:17
• Look at the possible five canonical forms of $2\times2$ symmetric matrices. The sign of the determinant is an invariant, hence if $\det\ne0$ must be definite or indefinite. Indefinite have $\det<0$. Thus if $\det>0$ is definite and both entries in the diagonal have the same sign, which is the sign of the trace. Mar 3 '15 at 18:29
• @Jesus RS What do you mean by canonical forms?
– AVP
Mar 4 '15 at 18:11
• I am interested in proving the second part of the implication. Meaning, if $A$ is symmetric and $trace(A)>0$ and $det(A)>0$ then $A$ is symmetric positive definite.
– AVP
Mar 4 '15 at 18:18
• The canonical forms are $\begin{pmatrix}1&0\\0&1\end{pmatrix}$, $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$, $\begin{pmatrix}-1&0\\0&0\end{pmatrix}$, $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Any symmetric $2\times 2$ matrix is congruent-similar to one of these. Mar 4 '15 at 20:07

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$=a+c>0$.

• For part 3): in other words, geometrically, fix $y$ at any non-zero value, taking a cross-section of the 2D quadratic form surface, look at the resulting 1D quadratic as $x$ varies, find its minimum, and make sure that it's zero or greater. (A more visual way of saying the same thing about the discriminant.) Sep 30 '16 at 2:13
• What isn't quite proved here is 2nd part in number 3 of Jesus RS' answer, I think? Added an answer. Nov 13 '20 at 15:37

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

• I don’t really understand what’s missing... and the point in this part was to use nothing. I show first that PD implies det>0 & tr>0 (just discussing a second degree equation). For the converse: det>0 implies ac>0, plus tr>0 gives a>0, and then read backwards the discussion of P>0. Nov 13 '20 at 18:22
• @JesusRS You mean something like all the steps in your part 3 are reversible? Nov 14 '20 at 9:52
• That’s it (I believe...) Nov 15 '20 at 11:53