eigen value

[deep arrow]

How to solve this one? Any short hack

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[unborn cloud]

Well here 0 is an eigenvalue (the rank of the matrice is 2) and if L is a non zero eigenvalue you can use the fact that the trace of the matrice is the sum of the eigenvalues

[deep arrow]

So sum of eigen values is 20

L1+l2=20
So l1 is 0 so l2=20-l1

[outer galleon]

why's there got to be a "short hack" ?

apply the definition of eigenvalue and check..

[deep arrow]

I meant we need to save time in exam hall soooo

[outer galleon]

then learn your definitions well so you save time by not wallowing in doubt soooo..

[deep arrow]

Eigen value says if we do |A-lambda I| then it will be 0 determinant

[outer galleon]

k is an eigenvalue of A if there exists a nonzero vector x such that Ax = kx

[deep arrow]

Yes

The problem is that i am facing difficulties with definition and their applications

Tell me how will you apply this Ax=kx

[outer galleon]

for starters just consider whether this problem is well posed

[deep arrow]

??

[outer galleon]

what's the trace of A and what do you know about the trace of similar matrices?

1. show that 0 is an eigenvalue of A
2. trace A is the sum of eigenvalues (why?)
3. there exists nonzero eigenvalue k
4. how do we get k + m = 20

[deep arrow]

It comes when we solve the determinant equation

|A-lambda I|=0

So it is like cubic equation and roots of it sum and multiplication will be Determinant

[outer galleon]

well, you wanted a shorter way

[deep arrow]

Here trace is 20 so eigen values sum will be equal to it

Let's check determinant is 0 so one eigen value is 0

[outer galleon]

correct

[deep arrow]

How did you check for 0 eigen value?

Same process or any different one?

[outer galleon]

determinant = 0 iff 0 is an eigenvalue

[deep arrow]

Like row reduction?

[outer galleon]

i calculate determinant with elementary operations yes

[deep arrow]

Okay fine

So now

0+M=20

So m =20

[outer galleon]

nono

0+k+m = 20

[deep arrow]

3×3 matrices

I see

[outer galleon]

but you're told k is a nonzero eigenvalue

do we know that k is not 20?

if yes, then there must be three distinct eigenvalues

otherwise the eigenvalues could also be 0,0, 20

[deep arrow]

So we don't know anything about K,M

[deep arrow]

No

But options are weird why?

[outer galleon]

can you exclude k = 20?

[outer galleon]

the options don't matter, you have to take m = 20 - k once you show there are three distinct eigenvalues

any other choice would be wrong

[deep arrow]

And how will we prove it?

Ohh wait

Can we use cofactors?

[deep arrow]

+close