Question

If$\mathbf{A}$is positive definite, then all the entries of Amust be positive or zero.

Short answer

The given statement is FALSE.

Step-by-step solution

Step 1: Definition of positive definite matrix

A matrix is said to be positive definite matrix, when it is symmetric matrix and all its eigenvalues are positive.

Step 2: To Find TRUE or FALSE

Let’s verify the given statement with the help of an example.

Consider a matrix $\mathrm{A}=\left[\begin{array}{cc}1& -1\\ -1& 3\end{array}\right]$

Find the symmetric of the matrix to prove that the given matrix is positive definite.

Because the symmetric of the matrix is said to be a positive definite matrix.

Now,

$$\begin{gathered} {\mathrm{A}}^{\mathrm{T}}=\left[\begin{array}{cc}1& -1\\ -1& 3\end{array}\right] \\ =\mathrm{A} \end{gathered}$$

Here, ${\mathrm{A}}^{\mathrm{T}}=\mathrm{A}$which means it’s symmetric. So, the given matrix is positive definite.

Also, the principle sub-matrices of $\mathrm{A}$have positive determinants.

Hence, $\mathrm{A}$is positive definite matrix because it is symmetric.

However,$\mathrm{A}$ have negative entries in it.

Final Answer

In view of this example, we can confirm that all the entries of a positive definite matrix are positive is FALSE.

So, the given statement is FALSE.