TIFR CSE 2018 | Part A | Question: 12

Question

An $n \times n$ matrix $M$ with real entries is said to be positive definite if for every non-zero $n$-dimensional vector $x$ with real entries, we have $x^{T}Mx>0.$ Let  $A$ and $B$ be symmetric, positive definite matrices of size $n\times n$ with real entries.

Consider the following matrices, where $I$ denotes the $n\times n$ identity matrix:

1. $A+B$
2. $ABA$
3. $A^{2}+I$

Which of the above matrices must be positive definite?

• A. Only $(2)$
• B. Only $(3)$
• C. Only $(1)$ and $(3)$
• D. None of the above matrices are positive definite
• E. All of the above matrices are positive definite

Comments

@Arjun Sir, Please format the question.

Answer 1

Note

- The sum of two positive definite matrices is positive definite. $(x^T(A+B)x = x^TAx + x^TBx > 0)$.

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$i)$ $A+B$ is positive definite from above property.

$ii)$ $x^T(ABA)x = (x^TA)B(Ax) = (Ax)^TB(Ax) > 0$

$iii)$ Since $A$ is symmetric $A^2$ is also symmetric. And $A^2 + I$ is also positive definite since $I$ is positive definite. 

Thus, all of the above matrices are positive definite. 

Comments

In case $(iii)$ $A^2$ is symmetric doesn't imply that $A^2$ is positive definite.

Instead, $x^T(A^2+I)x = (xA)^T(xA) + x^Tx$

But here $(xA)$ is nx1 matrix and hence any $x^Tx$ is equivalent to sum of squares which is always > 0

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case(iii), another easy way to prove $A^{2}$ is positive definite:

as $A$ is positive definite, all Eigen values of $A$ are positive (property of positive definite matrices).

And we know Eigen values of  $A^{2}$ are squares of Eigen values of $A$ (everyone knows, I guess).

$\therefore$ $A^{2}$ is also positive definite.

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Thanks @smn98 ! The positive definite is new to me, wasn't aware of the property !!