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> A complex number λ is called an eigenvalue of T if T −λI is not injective. Uhm, what the intuition behind _that_?

• zodiac 9y ago
  It is equivalent to "A complex number λ is called an eigenvalue of T if Tx = λx for some nonzero x"
  • pacala 9y ago
    Thanks! Looking up wikipedia, "In mathematics, an injective function or injection or one-to-one function is a function that preserves distinctness". From that definition, "T −λI is not injective" means that there exists 2 vectors, y and z, such that (T −λI)y = (T −λI)z. Giving T(y - z) = λI(y - z). Via x = (y - z), I get to your (usual) definition. The question I have is _why_ use the injective-based definition instead of the usual well-known one? Is there some further insight down the road?
    • Koshkin 9y ago
      The merit in using the purely functional language is in that it opens the door for application of powerful methods and concepts of Category Theory. (Note also that even the very notion of a function being injective can be defined in a purely functional manner, i.e. without any reference to sets or their elements.)
      • zodiac 9y ago
        Linear Algebra is a great place to get exposed to some category-theoretical concepts (maybe not when first learning linear algebra, though). My linear algebra prof used the category-theoretical definitions of subspace and quotient space (eg S is a subspace of L if there exists an injective linear map from S to L, a quotient space if there is an injective linear map) and tensor product (universal property)
    • zodiac 9y ago
      With the injectivity-based definition, you can ask "how injective is it?" in various ways (eg, what is the dimension of the nullspace of T - λI; the Ax = λx definition is rather binary and corresponds to simply "T - λI has nullity > 0"). This is what is meant by "eigenvalue multiplicity" (there are actually at least two different inequivalent ways to measure this, geometric multiplicity and algebraic multiplicity). Also for finite-dimensional vector space a lot of concepts are equivalent, eg "a linear operator L is invertible <=> L is injective", which are not in infinite-dimensional vector spaces (eg https://math.stackexchange.com/a/2447563/21437 https://math.stackexchange.com/a/2447563/21437). Studying T - λI turns out to be more useful in this case (eg https://en.wikipedia.org/wiki/Decomposition_of_spectrum_(functional_analysis) https://en.wikipedia.org/wiki/Decomposition_of_spectrum_(fun...)
• clickok 9y ago
  It's another way of saying that T - λI is not a one-to-one map, meaning that (T - λI) is non-invertible. So there are vectors x and y such that (T - λI)x = (T - λI)y for y ≠ x. Which means that (T - λI)(x - y) = 0, and in general because of linearity every scalar multiple of (x - y) also maps to zero. Letting (x - y) = v, we could get (T - λI)v = 0 ==> Tv = λv, which is perhaps a more familiar definition. So, it's a neat way of expressing the concept, but I'm not sure what it buys you in terms of improving one's intuition.
• tzahola 9y ago
  It’s equivalent to the usual definition, but spares you a few steps when proving the existence of eigenvalues. I.e. you only need to prove that T - hI maps some nonzero vector to zero.
• sgdpk 9y ago
  You can think of it in these two steps: 1. λI stretches every vector (the whole space, really) by a factor λ. 2. Saying that the function is not injective means you lose information: when you apply it on some object and get a result, you can't trace back what was the original object, as there may be several. (There is no inverse function, then). In linear algebra, this only happens because there is some direction of space where all the vectors get collapsed to zero. In short, T-λI collapses some line of vectors to zero. So, when you took the effect of λI from T, you make it a lossy transformation in some direction. This means that _in that direction_ T had the effect of stretching all vectors by a factor of λ. You gain some geometric understanding of T. It is sort of intuitive, but the language may obscure it a little if you are not used to it.
  • Myrmornis 9y ago
    Thanks! > So, when you took the effect of λI from T, If I understand right, you’re saying that there’s an interpretation in terms of the geometry of the T transformation, of subtracting this diagonal matrix from T. Multiplication of matrices is composition of transformations, I get that, but I’m not so sure what adddition/subtraction is.
    • sgdpk 9y ago
      Yes, that's right. Addition is just applying the transformations separately to the same vector and adding the result. So what this is saying is that if you apply λI to a vector in that particular direction, then there is nothing left to add to get the effect of T. Ideally you would like to do this for all n directions of space, and that way you completely describe what T does in simpler terms: it just stretches things differently in different directions. It's not always possible though. The matrices that allow this are called diagonalizable and the process of finding the stretch factors (eigenvalues) is called diagonalization. Just a caveat: if an eigenvalue is complex, the effect is not as simple as a stretch, but the interpretation is very similar.
      • Myrmornis 9y ago
        Thanks very much. I'm glad I asked. Clearly, I had failed to really internalize what it means to be a linear operator!