[10]7- Let F be a field. A matrix A ∈ Mnxn(F) is called...

[10]7- Let F be a field. A matrix $A \in M_{nxn}(F)$ is called skew-symmetric if $A^t = -A$ (where $A^t$ is the transpose of A). (a) Prove that if $A \in M_{nxn}(\mathbb{R})$ is skew-symmetric and n is odd, then $det(A) = 0$. (b) Does the assertion of (a) remain valid if $\mathbb{R}$ is replaced by an arbitrary field F? Prove or give a counterexample. (Hint: think characteristic)

Transcript

00:01 So i looked it up on the internet and i think what you're calling characteristic values of a, i would call the eigenvalues of a.

00:08 And let's let matrix a be a die or excuse me, upward triangle and matrix.

00:19 Then it has the form lambda 1 ,0, 0, dot, dot, 0.

00:26 And then you have a 1 -2 here, a 1 -3 here.

00:32 Then lambda 2 then everything else below it is 0 then a 2 3 here then lambda 3 everything below it is 0 and continue that pattern i'll wait until you get to a 1n a 2n a 3n and you have a lambda n down here so this is what your this is what your matrix looked like i'm going to abbreviate it sometimes i'm going to abbreviate by making it lambda 1, lambda 2, lambda n.

01:12 And i draw a squiggle to mean all the a values i have in the upper right corner.

01:18 I draw a 0 here to mean that the bottom below the diagonal is all 0.

01:22 This is what i kind of used to abbreviate an upper triangle matrix.

01:27 But anyway, the characteristic values solve the problem, this problem.

01:34 There's some vector is equal to lambda.

01:39 Times the vector.

01:43 And this can be a rearranged to be a .v minus lambda v equals zero, which you can write as av minus lambda i times v equal to zero.

02:05 So to solve for the characteristic values, you want to find lambda such that av, v, minus lambda i has a non -trivial null space and here's the thing having a non -trivial null space is equivalent to having a determinant of zero so really all you have to do is you have to take the determinant of a minus lambda i and set it equal to zero and then solve for lambda and what that looks like is it looks like the determinant of lambda 1 minus lambda since lambda i is just a diagonal matrix with lambdas on the diagonal and zeros everywhere else and lambda 2 minus lambda all the way to lambda n minus lambda and there's stuff up here and then there's zeros below everything and said equals zero now hopefully you know how to take the determinant of a matrix so what you can do is you can take what's called a minor...

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