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They have the same diagonal values with larger one having zeros padded on the diagonal. $\endgroup$ – Mateus Sampaio Oct 22 '14 at 21:43 F. Similar matrices always have exactly the same eigenvalues. However, all eigenvectors are nonzero scalar multiples of (1,0) T, so its geometric multiplicity is only 1. A and A^T will not have the same eigenspaces, i.e. The diagonal values must be the same, since SS T and S T S have the same diagonal values, and these are just the eigenvalues of AA T and A T A. The matrices AAT and ATA have the same nonzero eigenvalues. This pattern keeps going, because the eigenvectors stay in their own directions (Figure 6.1) and never get mixed. They can however be related, as for example if one is a scalar multiple of another. So, the above two equations show the unitary diagonalizations of AA T and A T A. d) Conclude that if Ahas distinct real eigenvalues, then AB= BAif and only if there is a matrix Tso that both T 1ATand T 1BTare in canonical form, and this form is diagonal. The eigenvector .1;1/ is unchanged by R. The second eigenvector is .1; 1/—its signs are reversed by R. If two matrices are similar, they have the same eigenvalues and the same number of independent eigenvectors (but probably not the same eigenvectors). So we have shown that ##A - \lambda I## is invertible iff ##A^T - \lambda I## is also invertible. When A is squared, the eigenvectors stay the same. Show that A and A T have the same eigenvalues. When we diagonalize A, we’re ﬁnding a diagonal matrix Λ that is similar to A. Hence they are all mulptiples of (1;0;0). Presumably you mean a *square* matrix. These eigenvectors that correspond to the same eigenvalue may have no relation to one another. However, in my opinion, this is not a proof proving why A 2 and A have the same eigenvectors but rather why λ is squared on the basis that the matrices share the same eigenvectors. Example 3 The reﬂection matrix R D 01 10 has eigenvalues1 and 1. The eigenvalues of A 100are 1 = 1 and (1 2) 100 = very small number. Show that for any square matrix A, Atand A have the same characteristic polynomial and hence the same eigenvalues. Scalar multiples of the same matrix has the same eigenvectors. ST and TS always have the same eigenvalues but not the same eigenvectors! Show that: a. See the answer. EX) Imagine one of the elements in eigenVector V[i,j] is equal to a+bi calculated by approach A. If two matrices have the same n distinct eigenvalues, they’ll be similar to the same diagonal matrix. 24)If A is an n x n matrix, then A and A T have the same eigenvectors. Answer to: Do a and a^{T} have the same eigenvectors? Some of your past answers have not been well-received, and you're in danger of being blocked from answering. The next matrix R (a reﬂection and at the same time a permutation) is also special. eigenvectors of AAT and ATA. Do they necessarily have the same eigenvectors? If someone can prove that A 2 and A have the same eigenvectors by using equations A 2 y=αy and Ax=λx, and proceeding to prove y=x, I will be very much convinced that these two matrices have the same eigenvectors. @Colin T Bowers: I didn't,I asked a question and looking for the answer. So this shows that they have the same eigenvalues. Does this imply that A and its transpose also have the same eigenvectors? I really do n't see how it will help here case we are only going to get single. Same eigenvectors and TS always have exactly the same characteristic polynomial and the same.... Always have exactly the same eigenvalues 100are 1 = 1 and ( 2. 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