you tell me
They must have the same dimensions.
No. You can only add matrices of the same size.
Yes. In general, two matrices of the same size can be added.
No,they are not similar
it is similar, easy as that. congruent is same shape same size ans similar is same shape different size similar=two figures that have the same shape, but not necessarily the same size
Yes, similar matrices have the same eigenvalues.
No, in general they do not. They have the same eigenvalues but not the same eigenvectors.
The matrices must have the same dimensions.
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.
You can use ratios of adjacent sides to prove if two rectangles are similar by comparing to see if the ratios are the same
That the sides are of the same ratio and that the interior angles are the same.
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.
They must have the same dimensions.
By enlargement on the Cartesian plane and that their 3 interior angles will remain the same
If both matrices have the same number of columns and rows ex: {1 2 3 4} can not be added with {5 4} b/c they dont have the same amount of numbers
You either show that the corresponding angles are equal or that the lengths of corresponding sides are in the same ratio.
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 = kvMeaning 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-1wWe want to show that Bw = kvBw = TAT-1w = TAv = Tkv = kTv= kwQ.E.D.