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First, we'll start with the definition of an eigenvalue. Let v be a non-zero vector and A be a linear transformation acting on v. k is an eigenvalue of the linear transformation A if the following equation is satisfied:
Av = kv
Meaning the linear transformation has just scaled the vector, v, not changed its direction, by the value, k.
By definition, two matrices, A and B, are similar if B = TAT-1, where T is the change of basis matrix.
Let w be some vector that has had its base changed via Tv.
Therefore v = T-1w
We want to show that Bw = kv
Bw = TAT-1w = TAv = Tkv = kTv= kw
Q.E.D.
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Prove that two equilateral triangles are always similar?
Two triangles are similar if: 1) 3 angles of 1 triangle are the same as 3 angles of the other or 2) 3 pairs of corresponding sides are in the same ratio or 3) An angle of 1 triangle is the same as the angle of the other triangle and the sides containing these angles are in the same ratio. So if they are both equilateral, then they both have three 60 degree angles since equilateral triangles are equiangular as well. Then number 1 above tell us by AAA, they are similar.
Prove that the trace of a matrix A is equal to the sum of its eigenvalues?
Given a matrix A=([a,b],[c,d]), the trace of A is a+d, and the det of A is ad-bc.By using the characteristic equation, and representing the eigenvalues with x, we have the equationx2-(a+d)x+(ad-bc)=0Which, using the formula for quadratic equations, gives us the eigenvalues as,x1=[(a+d)+√((a+d)2-4(ad-bc))]/2x2=[(a+d)-√((a+d)2-4(ad-bc))]/2now by adding the two eigenvalues together we get:x1+x2=(a+d)/2+[√((a+d)2-4(ad-bc))]/2+ (a+d)/2-[√((a+d)2-4(ad-bc))]/2The square roots cancel each other out being the same value with opposite signs, leaving us with:x1+x2=(a+d)/2+(a+d)/2x1+x2= 2(a+d)/2x1+x2=(a+d)x1+x2=trace(A)Q.E.D.General proofThe above answer only works for 2x2 matrices. I'm going to answer it for nxn matrices. (Not mxn; the question only makes sense when the matrix is square.) The proof uses the following ingredients:(1) Every nxn matrix is conjugate to an upper-triangular matrix(2) If A is upper-triangular, then tr(A) is the sum of the eigenvalues of A(3) If A and B are conjugate, then tr(A) = tr(B)(4) If A and B are conjugate, then A and B have the same characteristic polynomial (and hence the same sum-of-eigenvalues)If these are all true, then we can do the following: Given a matrix A, find an upper-triangular matrix U conjugate to A; then (letting s(A) denote the sum of the eigenvalues of A) s(A) = s(U) = tr(U) = tr(A).Now to prove (1), (2), (3) and (4):(1) This is an inductive process. First you prove that your matrix is conjugate to one with a 0 in the bottom-left corner. Then you prove that this, in turn, is conjugate to one with 0s at the bottom-left and the one above it. And so on. Eventually you get a matrix with no nonzero entries below the leading diagonal, i.e. an upper-triangular matrix.(2) Suppose A is upper-triangular, with elements a1, a2, ... , an along the leading diagonal. Let f(t) be the characteristic polynomial of A. So f(t) = det(tI-A). Note that tI-A is also upper-triangular. Therefore its determinant is simply the product of the elements in its leading diagonal. So f(t) = det(tI-A) = (t-a1) * ... * (t-an). And its eigenvalues are a1, ... , an. So the sum of the eigenvalues is a1 + ... + an, which is the sum of the diagonal elements in A.(3) This is best proved using Summation Convention. Summation convention is a strange but rather useful trick. Basically, the calculations I've written below aren't true as they're written. For each expression, you need to sum over all possible values of the subscripts. For example, where it says b_ii, it really means b11 + b22 + ... + bnn. Where it says bil deltali, it means (b11 delta11 + ... + b1n deltan1) + ... + (bn1 delta1n + ... + bnn deltann). Oh, and deltakj=1 if k=j, and 0 otherwise.Suppose B = PAP-1. Let's say the element in row j and column k of A is ajk. Similarly, say the (i,j) element of P is pij, the (k,l) element of P-1 is p*kl, and the (i,l) element of B is bil. Then:bil = pij ajk p*klAnd the trace of B is given by:tr(B) = bii= bil deltali= p*kl deltali pij ajk= p*kl plj ajk= deltakj ajk (since p and p* are inverses)= ajj= tr(A)(4) Again, suppose B = PAP-1. Then, for any scalar t, we have tI-B = P(tI-A)P-1. Hence det(tI-B) = det(P).det(tI-A).det(P-1). Since det(P).det(P-1)=det(PP-1)=det(I)=1, we have det(tI-B) = det(tI-A).
What is the theme for The Outsiders when the teens fight?
the theme is that people had to fight to prove that they were the same no matter what class they were
How is Canada and Germany similar?
they have the same school system.
Do Ricola and Red Robin use the same melody?
No. Similar. But no.
