The eigenvalues of symmetric matrices are real. Hence λ equals its conjugate, which means that λ is real. Theorem 2. The eigenvectors of a symmetric matrix A corresponding to different eigenvalues are orthogonal to each other.

1) Eigenvectors can always be scaled. So if v is an eigenvector then so is av for a∈k∗. So if each eigenvalue has multiplicity one a basis of eigenvectors is automatically orthogonal (and can be made orthonormal as above). In general we need to find an orthogonal basis of each eigenspace first, e.g. by Gram-Schmidt.

A set of vectors S is orthonormal if every vector in S has magnitude 1 and the set of vectors are mutually orthogonal. The set of vectors { u1, u2, u3} is orthonormal. Proposition An orthogonal set of non-zero vectors is linearly independent.

If v is an eigenvector for AT and if w is an eigenvector for A, and if the corresponding eigenvalues are different, then v and w must be orthogonal. Of course in the case of a symmetric matrix, AT = A, so this says that eigenvectors for A corresponding to different eigenvalues must be orthogonal.

Eigenvectors corresponding to distinct eigenvalues are linearly independent. As a consequence, if all the eigenvalues of a matrix are distinct, then their corresponding eigenvectors span the space of column vectors to which the columns of the matrix belong.

If A is an n x n symmetric matrix, then any two eigenvectors that come from distinct eigenvalues are orthogonal. If we take each of the eigenvalues to be unit vectors, then the we have the following corollary. Symmetric matrices with n distinct eigenvalues are orthogonally diagonalizable.

Symmetric matrices can never have complex eigenvalues.

crucial properties: ▶ All eigenvalues of a real symmetric matrix are real. orthogonal. complex matrices of type A ∈ Cn×n, where C is the set of complex numbers z = x + iy where x and y are the real and imaginary part of z and i = √ −1.

Diagonalizable means the matrix has n distinct eigenvectors (for n by n matrix). After reading eigenvalue and eigenvector part of textbook, I conclude that every symmetric matrix is diagonalizable.

Theorem: Every real n × n symmetric matrix A is orthogonally diagonalizable Theorem: Every complex n × n Hermitian matrix A is unitarily diagonalizable. Theorem: Every complex n × n normal matrix A is unitarily diagonalizable.

If A is a real symmetric matrix, then any two eigenvectors corresponding to different eigenvalues are orthogonal. Then P is an orthogonal matrix if and only if the columns of P are orthogonal and have unit length.

Thus AAT is symmetric. And symmetric matrices are diagonalizable.

The matrices AAT and ATA have the same nonzero eigenvalues. Section 6.5 showed that the eigenvectors of these symmetric matrices are orthogonal.

A square matrix is said to be orthogonally diagonalizable if there exist an orhtogonal matrix P such that P−1AP is a diagonal matrix. A square matrix A is orthogonally diagonalizable ⇔ A is symmetric. Notice that the condition in (2) is more strict than (1) in that (2) ⟹ (1).

Equivalently, a square matrix is symmetric if and only if there exists an orthogonal matrix S such that ST AS is diagonal. That is, a matrix is orthogonally diagonalizable if and only if it is symmetric. A non-symmetric matrix which admits an orthonormal eigenbasis.

Solution: Since the matrix in question is not invertible, one of its eigenvalues must be 0. Choose any λ = 0 to be the other eigenvalue. By definition, A is diagonalizable, but it’s not invertible since det(A) = 0.

Then A = PΛP−1 where Λ is a diagonal matrix with eigenvalues of A on the diagonal. But (b) shows that all eigenvalues of A are zeros. Hence Λ = 0. Therefore nilpotent matrix A is not diagonalizable unless A = 0.

The fundamental theorem of algebra applied to the characteristic polynomial shows that there are always n n n complex eigenvalues, counted with multiplicity. But this does not mean that every square matrix is diagonalizable over the complex numbers.

Orthogonal diagonalization. A real square matrix A is orthogonally diagonalizable if there exist an orthogonal matrix U and a diagonal matrix D such that A=UDUT. Orthogonalization is used quite extensively in certain statistical analyses. An orthogonally diagonalizable matrix is necessarily symmetric.

The matrix (01−10) is orthogonal, but it is not diagonalizable over R (it has no real eigenvalues). In fact, every real orthogonal matrix is diagonalizable over C; this follows from the spectral theorem.

No, there are plenty of matrices with repeated eigenvalues which are diagonalizable. The easiest example is A=[1001]. since A is a diagonal matrix. Therefore, the only n×n matrices with all eigenvalues the same and are diagonalizable are multiples of the identity.

2 Answers. If you want the number of real eigenvalues counted with multiplicity, then the answer is no: the characteristic polynomial of a real 3×3 matrix is a real polynomial of degree 3, and therefore has either 1 or 3 real roots if these roots are counted with multiplicity.

The zero matrix is a diagonal matrix, and thus it is diagonalizable. However, the zero matrix is not invertible as its determinant is zero.

Every matrix is not diagonalisable. Take for example non-zero nilpotent matrices. The Jordan decomposition tells us how close a given matrix can come to diagonalisability.