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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 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).
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.
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.
Do similar matrices have the same eigenvalues?
Yes, similar matrices have the same eigenvalues.
Do similar matrices have the same eigenvectors?
No, in general they do not. They have the same eigenvalues but not the same eigenvectors.
How can you prove that similar matrices have the same trace?
you tell me
What is the condition for the addition of matrices?
The matrices must have the same dimensions.
Can matrices of the same dimension be multiplied?
No. The number of columns of the first matrix needs to be the same as the number of rows of the second.So, matrices can only be multiplied is their dimensions are k*l and l*m. If the matrices are of the same dimension then the number of rows are the same so that k = l, and the number of columns are the same so that l = m. And therefore both matrices are l*l square matrices.
Do commutative matrices have the same eigenvectors?
It is true that diagonalizable matrices A and B commute if and only if they are simultaneously diagonalizable. This result can be found in standard texts (e.g. Horn and Johnson, Matrix Analysis, 1999, Theorem 1.3.12.) One direction of the if and only if proof is straightforward, but the other direction is more technical: If A and B are diagonalizable matrices of the same order, and have the same eigenvectors, then, without loss of generality, we can write their diagonalizations as A = VDV-1 and B = VLV-1, where V is the matrix composed of the basis eigenvectors of A and B, and D and L are diagonal matrices with the corresponding eigenvalues of A and B as their diagonal elements. Since diagonal matrices commute, DL = LD. So, AB = VDV-1VLV-1 = VDLV-1 = VLDV-1 = VLV-1VDV-1 = BA. The reverse is harder to prove, but one online proof is given below as a related link. The proof in Horn and Johnson is clear and concise. Consider the particular case that B is the identity, I. If A = VDV-1 is a diagonalization of A, then I = VIV-1 is a diagonalization of I; i.e., A and I have the same eigenvectors.
How can you use ratios of adjacent sides to prove if two rectangles are similar?
You can use ratios of adjacent sides to prove if two rectangles are similar by comparing to see if the ratios are the same
What is linear combination in matrices?
If X1, X2 , ... , Xn are matrices of the same dimensions and a1, a2, ... an are constants, then Y = a1*X1 + a2*X2 + ... + an,*Xn is a linear combination of the X matrices.
What are the two conditions that need to be true to prove that two polygons are similar?
That the sides are of the same ratio and that the interior angles are the same.
How do you find eigenvalues of a 3 by 3 matrix?
Call your matrix A, the eigenvalues are defined as the numbers e for which a nonzero vector v exists such that Av = ev. This is equivalent to requiring (A-eI)v=0 to have a non zero solution v, where I is the identity matrix of the same dimensions as A. A matrix A-eI with this property is called singular and has a zero determinant. The determinant of A-eI is a polynomial in e, which has the eigenvalues of A as roots. Often setting this polynomial to zero and solving for e is the easiest way to compute the eigenvalues of A.
What must be true in order to add matrices?
They must have the same dimensions.
How does transformations prove that two triangles are similar?
By enlargement on the Cartesian plane and that their 3 interior angles will remain the same
