Show that the following function PdSMat returns a positive...

Show that the following function PdSMat returns a positive definite symmetric matrix of size $n \times n$ ``` function A=PdSMat(n) A=SymmetricMat(n); // defined in Exercise 2.2 [P,D]=eig(A);D=abs(D); D=D+norm(D)*eye(size(D)); A=P*D*inv(P); ``` 1. For different values of $n$, compute the determinant of $\mathrm{A}=\mathrm{PdSMat}(\mathrm{n})$. What do you observe? Justify. 2. Fix $n=10$. For $k$ varying from 1 to $n$, define a matrix $A_k$ of size $k \times k$ by $\mathrm{Ak}=\mathrm{A}(1: \mathrm{k}, 1: \mathrm{k})$. Check that the determinants of all the matrices $A_k$ are positive. Prove this result. 3. Are the eigenvalues of $A_k$ eigenvalues of $A$ ?

Show that the following function PdSMat returns a positive definite symmetric matrix of size $n \times n$
```
function A=PdSMat(n)
A=SymmetricMat(n); // defined in Exercise 2.2
[P,D]=eig(A);D=abs(D);
D=D+norm(D)*eye(size(D));
A=P*D*inv(P);
```
1. For different values of $n$, compute the determinant of $\mathrm{A}=\mathrm{PdSMat}(\mathrm{n})$. What do you observe? Justify.
2. Fix $n=10$. For $k$ varying from 1 to $n$, define a matrix $A_k$ of size $k \times k$ by $\mathrm{Ak}=\mathrm{A}(1: \mathrm{k}, 1: \mathrm{k})$. Check that the determinants of all the matrices $A_k$ are positive. Prove this result.
3. Are the eigenvalues of $A_k$ eigenvalues of $A$ ?

Transcript

00:01 We're asked to prove theorem 714.

00:05 Recall what this theorem says.

00:07 It says that a matrix a, two by two real matrix, with entries a, b, c, d is positive definite.

00:18 If and only if, diagonal elements a and d are positive.

00:34 And the determinant of a, which is ad minus, well, this is really the same as these two entries.

00:52 It's a symmetric.

00:53 Yeah, because he's like, actually what is going on here? the theorem 714 only applies to, sorry, symmetric real two by two matrices.

01:05 So a is a matrix of the form ab, b, b.

01:08 And therefore it's positive definite if and only if a and d are positive and the determined ad minus b squared is also positive.

01:19 To prove this statement, i'll begin by defining you to be the transpose of x, y.

01:35 Then it follows that f of u, this is u.

01:50 F of u this is u transpose times a times u.

01:56 This is x y row vector times a times the column vector xy which multiplying things out this is a -x squared plus 2b times x times y plus b squared.

02:18 Plus d is a -squared x -squared plus 2b x -y plus d times y -square, not an a -squared.

02:35 It's a times x squared plus 2b x -y plus d times y squared.

02:42 We're going to suppose that our function f of u is positive for every non -zero use.

02:52 Well, then this implies that f of 1, the vector 1 -0, well, this is, if you plug this in, a, is positive.

03:07 Also, f of the row vector 01, which is d, is also positive...

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