- #1
RK1992
- 89
- 0
if we say that an eigenvector u is a vector whose direction is unchanged by the matrix transformation A:
Au=λu=λIu
rearranging:
Au - λIu = 0
u(A-λI)=0
we know the u=0 is obviously a solution. and i also appreciate that if A-λI=0 then A=λI which is just a scalar multiplied by the identity so is also trivial.
but in class, my teacher said "so obviously, the other solution is when the matrix (A-λI) is singular" but didnt justify it and i don't see why its "obvious" that a matrix being singular means that it must satisfy that equation.
can anyone explain why det(A-λI)=0 is necessary? i am happy with using it but i don't understand why it must work and i find it frustrating not knowing why it does.
BTW we have never learned about vector spaces and things like that, is it possible to explain it without these? i looked on khan academy and they exxplained it using vector spaces and linear independence and that's not on my a level further maths syllabus :/
thanks for reading :)
Au=λu=λIu
rearranging:
Au - λIu = 0
u(A-λI)=0
we know the u=0 is obviously a solution. and i also appreciate that if A-λI=0 then A=λI which is just a scalar multiplied by the identity so is also trivial.
but in class, my teacher said "so obviously, the other solution is when the matrix (A-λI) is singular" but didnt justify it and i don't see why its "obvious" that a matrix being singular means that it must satisfy that equation.
can anyone explain why det(A-λI)=0 is necessary? i am happy with using it but i don't understand why it must work and i find it frustrating not knowing why it does.
BTW we have never learned about vector spaces and things like that, is it possible to explain it without these? i looked on khan academy and they exxplained it using vector spaces and linear independence and that's not on my a level further maths syllabus :/
thanks for reading :)
Last edited: