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For a symmetric matrix A, the vector vec(A) contains more information than is strictly necessary, since the matrix is completely determined by the symmetry together with the lower triangular portion, that is, the n(n + 1)/2 entries on and below the main diagonal. For such matrices, the half-vectorization is sometimes more useful than the ...
The adjugate of a diagonal matrix is again diagonal. Where all matrices are square, A matrix is diagonal if and only if it is triangular and normal. A matrix is diagonal if and only if it is both upper-and lower-triangular. A diagonal matrix is symmetric. The identity matrix I n and zero matrix are diagonal. A 1×1 matrix is always diagonal.
Thus we can write the trace itself as 2w 2 + 2w 2 − 1; and from the previous version of the matrix we see that the diagonal entries themselves have the same form: 2x 2 + 2w 2 − 1, 2y 2 + 2w 2 − 1, and 2z 2 + 2w 2 − 1. So we can easily compare the magnitudes of all four quaternion components using the matrix diagonal.
Normally, a matrix represents a linear map, and the product of a matrix and a column vector represents the function application of the corresponding linear map to the vector whose coordinates form the column vector. The change-of-basis formula is a specific case of this general principle, although this is not immediately clear from its ...
2. The upper triangle of the matrix S is destroyed while the lower triangle and the diagonal are unchanged. Thus it is possible to restore S if necessary according to for k := 1 to n−1 do ! restore matrix S for l := k+1 to n do S kl := S lk endfor endfor. 3. The eigenvalues are not necessarily in descending order.
Therefore, T must be diagonal since a normal upper triangular matrix is diagonal (see normal matrix). The converse is obvious. In other words, A is normal if and only if there exists a unitary matrix U such that = , where D is a diagonal matrix. Then, the entries of the diagonal of D are the eigenvalues of A.
Let A be a square n × n matrix with n linearly independent eigenvectors q i (where i = 1, ..., n).Then A can be factored as = where Q is the square n × n matrix whose i th column is the eigenvector q i of A, and Λ is the diagonal matrix whose diagonal elements are the corresponding eigenvalues, Λ ii = λ i.
Since the set F is both a set of eigenvectors for matrix A and it spans some arbitrary vector space, then we say that there exists a matrix which is a diagonal matrix that is similar to . In other words, A E {\displaystyle A_{E}} is a diagonalizable matrix if the matrix is written in the basis F.