Can one eigenvector belong to multiple eigenvalues?
Thanks for the A2A.The first thing to remember is that a matrix is a name for a function. Let's say our matrix is [math]A \in \mathbb{R}^{n \times n}[/math] (square because only square matrices have eigenvalues and eigenvectors). Then the function [math]L_A(x) = A \cdot x[/math] is a function from [math]\mathbb{R}^n[/math] to itself. And a function is, by definition, a rule for taking some input and returning some specific output.So we have this function [math]L_A[/math]. And suppose that [math]x[/math] is an eigenvector. What is the definition of "eigenvector"? What makes [math]x[/math] an eigenvector is that, for some scalar [math]c \in \mathbb{R}[/math], [math]L_A(x) = c x[/math]. And this [math]c[/math] is called the eigenvalue associated to x.OK, so your question is, why can't there be multiple eigenvalues associated to the same eigenvector? And the answer is, because of what we said above about what it means to be a function: there's only one output for a given input. Suppose [math]c_1 x = L_A(x) = c_2 x[/math]. Then by the axioms of vector spaces, [math]\begin{matrix}\vec{0} &= c_1 x - c_2 x \\& = (c_1 - c_2) x \end{matrix}[/math]so [math]c_1 - c_2 = 0[/math]. In other words, only one scalar can possibly work.
What is the multiple zero and multiplicity of f(x)=(x-3)(x-3)(x+5)?
multiple zero: refers to the values of the variable in an equation that result in a zero solution when the variable is set to that value. f(x) = (x-3)(x-3)(x+5) 0 = (x-3)(x-3)(x+5) x-3 = 0 or x-3 = 0 or x + 5 = 0 x = 3 or x = 3 or x = -5 multiplicity: How many times does the solution appear in the original equation? From above: 3 has a multiplicity of 2 and -5 has a multiplicity of 1 (simple root)
If I have a 2x2 matrix and I want to find the geometric multiplicity of an eigenvalue (a.m.), why can't I say that that it is always at least 1, and possibly more?
Because the zero vector is, by definition of an eigenvector, not an eigenvector.The definition of an eigenvalue and eigenvector are as follows:If [math]Mx=\lambda x[/math] for some nonzero vector [math]x[/math], then [math]\lambda[/math] is said to be an eigenvalue of [math]M[/math] with associated eigenvector [math]x[/math].However, to confound matters slightly, the zero vector is in the vector space produced by the span of a basis of eigenvectors of a particular eigenvalue. (Indeed, it is present in any vector space, by definition of what a vector space is.)
Multiplicity at the leftmost zero?
what does this mean??!?! I'm a calc student for goodness sakes this multiplicity and factoring crap should be second nature, but I'm not sure what my teacher is asking for on this homework problem the equation is (x^3 - 10x^2 + 25x) i was asked to factor x[(x-5)^2] find zeros... 0,5 find multiplicity at the leftmost zero wich is where i am completely lost.... I understand multiplicity but at leftmost zero i don't understand the terminology
Why do irreducible polynomials in Q [x] have distinct roots? Generalize if possible.
Suppose that f ∈ Q[x], f≠0 has a multiple (complex or real) root. Then the derivative fʹ has the same root (with smaller multiplicity), and GCD(f,fʹ) is a non-constant polynomial, divisor of f, of degree at most the degree of fʹ (hence smaller than the degree of f). Hence f has a non trivial factor, hence f is reducible.The generalization is the theorem: if K is a perfect field, then an irreducible polynomial in K[x] does not have multiple roots.A field is perfect if the derivative of any irreducible polynomial is never 0; or equivalently, a field is not perfect if it has characteristics p>0 and the map π:K→K, π(a)=a^p is not surjective, i.e. an element exists that has no p-th root.Indeed, consider the polynomial f=x^p-a; it is irreducible in K[x] but has a root (of multiplicity p) in the algebraic closure. And indeed fʹ=0
If a matrix has a zero eigenvalue, how do you know its adjugate matrix also has a zero eigenvalue?
We know much more than that. If [math]M[/math] is an [math]n\times n[/math] singular matrix, thus having a zero eigenvalue, we not only know that the adjugate of [math]M[/math], [math]\operatorname{adj}(M)[/math], is singular, but that the multiplicity of the zero eigenvalue of [math]\operatorname{adj}(M)[/math] is huge: it is either [math]n-1[/math] or [math]n[/math]. In fact, if the multiplicity of the zero eigenvalue of [math]M[/math] is [math]1[/math], then [math]\operatorname{adj}(M)[/math] has rank [math]1[/math] (so the adjugate has nullity [math]n-1[/math]); otherwise, [math]\operatorname{adj}(M)[/math] is the zero matrix, having nullity [math]n[/math].I actually proved this result in my Ph.D. thesis, even though it is a well-known result in linear algebra. I decided to prove it nonetheless in order to show how the determinantal rank can be used to simplify proofs. The determinantal rank of a matrix is the size of the largest square submatrix having nonzero determinant; this can be shown to always be equal to the usual matrix rank.Here’s a photo of the thesis proof, taken from the submitted version (that is, not the finalized one — that is why there are some pencil scribbles).
Why does [math]x^5=0[/math] has five answers when all of them are the same?
The Fundamental Theorem of Algebra states:Every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots.The concept of multiplicity allows it to be stated this way. Multiplicity just means the number of times a particular root appears as a factor of the polynomial in question.Multiplicity is useful when talking about the behavior of a function at a particular root. Namely, the higher the multiplicity of a root, the “flatter” the function is at the point. That is, the higher the multiplicity, the further you can get from that root while the function value stays close to zero.As an example, [math]y=x[/math] is within 0.1 of zero at every value of x within 0.1 of zero.On the other hand, [math]y=x^5[/math] is within 0.1 of zero at every value of x within 0.63 of zero. You can go six times as far away from the root while staying close to zero!Multiplicity can also tell you other things about the behavior of the function around the root. Even multiplicity implies the root is at a relative extremum, while odd multiplicity implies that it is not (though it may yet be an inflection point if the multiplicity is at least 3).
When are roots equal?
This is a pretty generic question.If we have a polynomial equation [math]f(x)=0[/math] where [math]f[/math] factors into[math]f(x) = (x-c)^2 g(x) [/math]then we say [math]c[/math] is a multiple root or repeated root of [math]f(x)=0[/math]. If we just talk about [math]f(x),[/math] we can say [math]f[/math] has a zero of multiplicity two at [math]x=c[/math]. I probably should say at least two, since [math]g(x)[/math] may also have some additional factors of [math](x-c)[/math]. Let’s assume it doesn’t.Repeated zeros have interesting properties, especially when in the denominator of a fraction. Any zero in a denominator is called a pole. If it was our [math]f(x)[/math] was in the denominator, we’d say the fraction has a second order pole at [math]x=c[/math].One other thing about multiplicities. A multiplicity at [math]x=c[/math] implies the derivative is zero at [math]x=c[/math]. To take an easy example, let [math]f(x)=x^2[/math] which has a zero of multiplicity two at [math]x=0[/math]. And indeed, [math]f’(x)=2x[/math] is zero at [math]x=0[/math]. Let’s see if we can show it about[math]f(x) = (x-c)^2 g(x) [/math][math]f’(x) = (x-c)^2 g’(x) + 2(x-c) g(x) = (x-c)((x-c)g'(x) + 2g(x))[/math]Since [math]f’(x)[/math] has [math](x-c)[/math] as a factor, [math]f’(c)=0[/math] as stated.